\hppSrSSKJrJrJrJr SSKJr SSKJ r SSK J r J r SSK Jr SSKJr SSKJr SS KJr SS KJr SS KJr SS KJr SS KJrJrJr SSKJ r SSK!J"r" SSK#J$r$J%r% SSK&J'r' SSK(J)r) SSK*J+r+ SSK,J-r- SSK.J/r/ SSK0J1r1 SSK2r2SS/0r3\)"SSS/0S9r4"SS5r5"S S!\55r6g)"zd This module can be used to solve 2D beam bending problems with singularity functions in mechanics. )SSymboldiffsymbols)Add)Expr) DerivativeFunction)Mul)Eq)sympify)linsolve)dsolve)solve)sstr)SingularityFunction Piecewise factorial) integrate)limit)plotPlotGrid)GeometryEntity) import_module)Interval)lambdify)doctest_depends_on)iterableN)z Beam.drawzBeam.plot_bending_momentzBeam.plot_deflectionzBeam.plot_ild_momentzBeam.plot_ild_shearzBeam.plot_shear_forcezBeam.plot_shear_stresszBeam.plot_slope matplotlibnumpyfromlistarange) import_kwargsc(\rSrSrSr\"S5\"S5S\"S54SjrSr\S 5r \S 5r \S 5r \S 5r \S 5r \S5r\S5r\S5r\S5r\R$S5r\S5r\R$S5r\S5r\R$S5r\S5r\R$S5r\S5r\R$S5r\S5r\R$S5r\S5r\S5r\R$S5r\S 5r\R$S!5r\S"5r\R$S#5r\S$5r\R$S%5rSOS&jrSOS'jrS(rS)r S*r!SPS,jr"SPS-jr#S.r$\S/5r%\S05r&S1r'S2r(S3r)S4r*S5r+S6r,S7r-S8r.S9r/S:r0SPS;jr1SPS<jr2SPS=jr3SPS>jr4SPS?jr5SPS@jr6SAr7SBr8SPSCjr9SDr:SPSEjr;SFr"SHSI9SQSJj5r?SKr@SLrASMrBSNrCg+)RBeam-a0 A Beam is a structural element that is capable of withstanding load primarily by resisting against bending. Beams are characterized by their cross sectional profile(Second moment of area), their length and their material. .. note:: A consistent sign convention must be used while solving a beam bending problem; the results will automatically follow the chosen sign convention. However, the chosen sign convention must respect the rule that, on the positive side of beam's axis (in respect to current section), a loading force giving positive shear yields a negative moment, as below (the curved arrow shows the positive moment and rotation): .. image:: allowed-sign-conventions.png Examples ======== There is a beam of length 4 meters. A constant distributed load of 6 N/m is applied from half of the beam till the end. There are two simple supports below the beam, one at the starting point and another at the ending point of the beam. The deflection of the beam at the end is restricted. Using the sign convention of downwards forces being positive. >>> from sympy.physics.continuum_mechanics.beam import Beam >>> from sympy import symbols, Piecewise >>> E, I = symbols('E, I') >>> R1, R2 = symbols('R1, R2') >>> b = Beam(4, E, I) >>> b.apply_load(R1, 0, -1) >>> b.apply_load(6, 2, 0) >>> b.apply_load(R2, 4, -1) >>> b.bc_deflection = [(0, 0), (4, 0)] >>> b.boundary_conditions {'bending_moment': [], 'deflection': [(0, 0), (4, 0)], 'shear_force': [], 'slope': []} >>> b.load R1*SingularityFunction(x, 0, -1) + R2*SingularityFunction(x, 4, -1) + 6*SingularityFunction(x, 2, 0) >>> b.solve_for_reaction_loads(R1, R2) >>> b.load -3*SingularityFunction(x, 0, -1) + 6*SingularityFunction(x, 2, 0) - 9*SingularityFunction(x, 4, -1) >>> b.shear_force() 3*SingularityFunction(x, 0, 0) - 6*SingularityFunction(x, 2, 1) + 9*SingularityFunction(x, 4, 0) >>> b.bending_moment() 3*SingularityFunction(x, 0, 1) - 3*SingularityFunction(x, 2, 2) + 9*SingularityFunction(x, 4, 1) >>> b.slope() (-3*SingularityFunction(x, 0, 2)/2 + SingularityFunction(x, 2, 3) - 9*SingularityFunction(x, 4, 2)/2 + 7)/(E*I) >>> b.deflection() (7*x - SingularityFunction(x, 0, 3)/2 + SingularityFunction(x, 2, 4)/4 - 3*SingularityFunction(x, 4, 3)/2)/(E*I) >>> b.deflection().rewrite(Piecewise) (7*x - Piecewise((x**3, x >= 0), (0, True))/2 - 3*Piecewise(((x - 4)**3, x >= 4), (0, True))/2 + Piecewise(((x - 2)**4, x >= 2), (0, True))/4)/(E*I) Calculate the support reactions for a fully symbolic beam of length L. There are two simple supports below the beam, one at the starting point and another at the ending point of the beam. The deflection of the beam at the end is restricted. The beam is loaded with: * a downward point load P1 applied at L/4 * an upward point load P2 applied at L/8 * a counterclockwise moment M1 applied at L/2 * a clockwise moment M2 applied at 3*L/4 * a distributed constant load q1, applied downward, starting from L/2 up to 3*L/4 * a distributed constant load q2, applied upward, starting from 3*L/4 up to L No assumptions are needed for symbolic loads. However, defining a positive length will help the algorithm to compute the solution. >>> E, I = symbols('E, I') >>> L = symbols("L", positive=True) >>> P1, P2, M1, M2, q1, q2 = symbols("P1, P2, M1, M2, q1, q2") >>> R1, R2 = symbols('R1, R2') >>> b = Beam(L, E, I) >>> b.apply_load(R1, 0, -1) >>> b.apply_load(R2, L, -1) >>> b.apply_load(P1, L/4, -1) >>> b.apply_load(-P2, L/8, -1) >>> b.apply_load(M1, L/2, -2) >>> b.apply_load(-M2, 3*L/4, -2) >>> b.apply_load(q1, L/2, 0, 3*L/4) >>> b.apply_load(-q2, 3*L/4, 0, L) >>> b.bc_deflection = [(0, 0), (L, 0)] >>> b.solve_for_reaction_loads(R1, R2) >>> print(b.reaction_loads[R1]) (-3*L**2*q1 + L**2*q2 - 24*L*P1 + 28*L*P2 - 32*M1 + 32*M2)/(32*L) >>> print(b.reaction_loads[R2]) (-5*L**2*q1 + 7*L**2*q2 - 8*L*P1 + 4*L*P2 + 32*M1 - 32*M2)/(32*L) AxCaczXlX l[U[5(aX0lO SUlX0lXPlXplX`l////S.Ul SUl X@l /Ul /Ul /Ul/Ul/Ul/Ul/Ul0Ul0UlSUlSUlSUlSUlg)aInitializes the class. Parameters ========== length : Sympifyable A Symbol or value representing the Beam's length. elastic_modulus : Sympifyable A SymPy expression representing the Beam's Modulus of Elasticity. It is a measure of the stiffness of the Beam material. It can also be a continuous function of position along the beam. second_moment : Sympifyable or Geometry object Describes the cross-section of the beam via a SymPy expression representing the Beam's second moment of area. It is a geometrical property of an area which reflects how its points are distributed with respect to its neutral axis. It can also be a continuous function of position along the beam. Alternatively ``second_moment`` can be a shape object such as a ``Polygon`` from the geometry module representing the shape of the cross-section of the beam. In such cases, it is assumed that the x-axis of the shape object is aligned with the bending axis of the beam. The second moment of area will be computed from the shape object internally. area : Symbol/float Represents the cross-section area of beam variable : Symbol, optional A Symbol object that will be used as the variable along the beam while representing the load, shear, moment, slope and deflection curve. By default, it is set to ``Symbol('x')``. base_char : String, optional A String that will be used as base character to generate sequential symbols for integration constants in cases where boundary conditions are not sufficient to solve them. ild_variable : Symbol, optional A Symbol object that will be used as the variable specifying the location of the moving load in ILD calculations. By default, it is set to ``Symbol('a')``. N) deflectionslopebending_moment shear_forcerF)lengthelastic_modulus isinstancer cross_section second_momentvariable ild_variable _base_char_boundary_conditions_loadarea_applied_supports_applied_rotation_hinges_applied_sliding_hinges_rotation_hinge_symbols_sliding_hinge_symbols_support_as_loads_applied_loads_reaction_loads_ild_reactions _ild_shear _ild_moment_original_load _joined_beam)selfr0r1r4r:r5 base_charr6s ^/var/www/auris/envauris/lib/python3.13/site-packages/sympy/physics/continuum_mechanics/beam.py__init__ Beam.__init__sX . m^ 4 4!. !%D !.  (#35VXik$l!  !#(*%')$')$&(#!# !  !cUR(a URO URnSR[UR5[UR 5[U55nU$)NzBeam({}, {}, {}))_cross_section_second_momentformatr_length_elastic_modulus)rHshape_descriptionstr_sols rJ__str__ Beam.__str__sR373F3FD//DL_L_$++D,>TEZEZ@[]abs]turMcUR$)z,Returns the reaction forces in a dictionary.)rBrHs rJreaction_loadsBeam.reaction_loadss###rMcUR$)z Returns the value for the rotation jumps in rotation hinges in a dictionary. The rotation jump is the rotation (in radian) in a rotation hinge. This can be seen as a jump in the slope plot. )_rotation_jumpsrYs rJrotation_jumpsBeam.rotation_jumpss###rMcUR$)z Returns the deflection jumps in sliding hinges in a dictionary. The deflection jump is the deflection (in meters) in a sliding hinge. This can be seen as a jump in the deflection plot. )_deflection_jumpsrYs rJdeflection_jumpsBeam.deflection_jumpss%%%rMcUR$)z"Returns the I.L.D. shear equation.)rDrYs rJ ild_shearBeam.ild_shearsrMcUR$)z3Returns the I.L.D. reaction forces in a dictionary.)rCrYs rJ ild_reactionsBeam.ild_reactions"""rMcUR$)z Returns the I.L.D. rotation jumps in rotation hinges in a dictionary. The rotation jump is the rotation (in radian) in a rotation hinge. This can be seen as a jump in the slope plot. )_ild_rotations_jumpsrYs rJild_rotation_jumpsBeam.ild_rotation_jumpss(((rMcUR$)z Returns the I.L.D. deflection jumps in sliding hinges in a dictionary. The deflection jump is the deflection (in meters) in a sliding hinge. This can be seen as a jump in the deflection plot. )_ild_deflection_jumpsrYs rJild_deflection_jumpsBeam.ild_deflection_jumpss)))rMcUR$)z#Returns the I.L.D. moment equation.)rErYs rJ ild_momentBeam.ild_moment srMcUR$)zLength of the Beam.)rRrYs rJr0 Beam.lengths||rMc$[U5UlgN)r rR)rHls rJr0rws qz rMcUR$z"Cross-sectional area of the Beam. _arearYs rJr: Beam.areazzrMc$[U5Ulgryr r~rHr*s rJr:r QZ rMcUR$)a% A symbol that can be used as a variable along the length of the beam while representing load distribution, shear force curve, bending moment, slope curve and the deflection curve. By default, it is set to ``Symbol('x')``, but this property is mutable. Examples ======== >>> from sympy.physics.continuum_mechanics.beam import Beam >>> from sympy import symbols >>> E, I, A = symbols('E, I, A') >>> x, y, z = symbols('x, y, z') >>> b = Beam(4, E, I) >>> b.variable x >>> b.variable = y >>> b.variable y >>> b = Beam(4, E, I, A, z) >>> b.variable z ) _variablerYs rJr5 Beam.variable#s2~~rMcP[U[5(aXlg[S5e)Nz'The variable should be a Symbol object.)r2rr TypeError)rHvs rJr5r>s a NIJ JrMcUR$zYoung's Modulus of the Beam. )rSrYs rJr1Beam.elastic_modulusEs$$$rMc$[U5Ulgry)r rSrHes rJr1rJs ' rMcUR$z#Second moment of area of the Beam. rPrYs rJr4Beam.second_momentNrjrMcrSUl[U[5(a [S5e[ U5Ulg)Nz>To update cross-section geometry use `cross_section` attribute)rOr2r ValueErrorr rP)rHis rJr4rSs/" a ( (]^ ^")!*D rMcUR$)zCross-section of the beam)rOrYs rJr3Beam.cross_section[rjrMcNU(aUR5SUlXlg)Nr)second_moment_of_arearPrO)rHss rJr3r`s! "#"9"9";A">D rMcUR$)a Returns a dictionary of boundary conditions applied on the beam. The dictionary has three keywords namely moment, slope and deflection. The value of each keyword is a list of tuple, where each tuple contains location and value of a boundary condition in the format (location, value). Examples ======== There is a beam of length 4 meters. The bending moment at 0 should be 4 and at 4 it should be 0. The slope of the beam should be 1 at 0. The deflection should be 2 at 0. >>> from sympy.physics.continuum_mechanics.beam import Beam >>> from sympy import symbols >>> E, I = symbols('E, I') >>> b = Beam(4, E, I) >>> b.bc_deflection = [(0, 2)] >>> b.bc_slope = [(0, 1)] >>> b.boundary_conditions {'bending_moment': [], 'deflection': [(0, 2)], 'shear_force': [], 'slope': [(0, 1)]} Here the deflection of the beam should be ``2`` at ``0``. Similarly, the slope of the beam should be ``1`` at ``0``. r8rYs rJboundary_conditionsBeam.boundary_conditionsfs6(((rMc URS$Nr/rrYs rJbc_shear_forceBeam.bc_shear_forces((77rMc XRS'grr)rHsf_bcss rJrrs39!!-0rMc URS$Nr.rrYs rJbc_bending_momentBeam.bc_bending_moments(()9::rMc XRS'grr)rHbm_bcss rJrrs6<!!"23rMc URS$Nr-rrYs rJbc_slope Beam.bc_slopes((11rMc XRS'grr)rHs_bcss rJrrs-2!!'*rMc URS$Nr,rrYs rJ bc_deflectionBeam.bc_deflections((66rMc XRS'grr)rHd_bcss rJrrs27!!,/rMcURnURnURUR-nURUR:wa [S5eURUR:wa3[ URX0R:*4URX5:*45nO URnUS:Xa[ XTXc5nSUlU$US:Xa0[ XTXc5nSUlURUR5 U$g)a This method joins two beams to make a new composite beam system. Passed Beam class instance is attached to the right end of calling object. This method can be used to form beams having Discontinuous values of Elastic modulus or Second moment. Parameters ========== beam : Beam class object The Beam object which would be connected to the right of calling object. via : String States the way two Beam object would get connected - For axially fixed Beams, via="fixed" - For Beams connected via rotation hinge, via="hinge" Examples ======== There is a cantilever beam of length 4 meters. For first 2 meters its moment of inertia is `1.5*I` and `I` for the other end. A pointload of magnitude 4 N is applied from the top at its free end. >>> from sympy.physics.continuum_mechanics.beam import Beam >>> from sympy import symbols >>> E, I = symbols('E, I') >>> R1, R2 = symbols('R1, R2') >>> b1 = Beam(2, E, 1.5*I) >>> b2 = Beam(2, E, I) >>> b = b1.join(b2, "fixed") >>> b.apply_load(20, 4, -1) >>> b.apply_load(R1, 0, -1) >>> b.apply_load(R2, 0, -2) >>> b.bc_slope = [(0, 0)] >>> b.bc_deflection = [(0, 0)] >>> b.solve_for_reaction_loads(R1, R2) >>> b.load 80*SingularityFunction(x, 0, -2) - 20*SingularityFunction(x, 0, -1) + 20*SingularityFunction(x, 4, -1) >>> b.slope() (-((-80*SingularityFunction(x, 0, 1) + 10*SingularityFunction(x, 0, 2) - 10*SingularityFunction(x, 4, 2))/I + 120/I)/E + 80.0/(E*I))*SingularityFunction(x, 2, 0) - 0.666666666666667*(-80*SingularityFunction(x, 0, 1) + 10*SingularityFunction(x, 0, 2) - 10*SingularityFunction(x, 4, 2))*SingularityFunction(x, 0, 0)/(E*I) + 0.666666666666667*(-80*SingularityFunction(x, 0, 1) + 10*SingularityFunction(x, 0, 2) - 10*SingularityFunction(x, 4, 2))*SingularityFunction(x, 2, 0)/(E*I) z@Joining beams with different Elastic modulus is not implemented.fixedThingeN) r5r1r0NotImplementedErrorr4rr%rGapply_rotation_hinge)rHbeamviar(E new_lengthnew_second_momentnew_beams rJjoin Beam.joinsV MM  [[4;;.   4#7#7 7%&hi i   !3!3 3 )4+=+=q++~*N%)%7%7$G!I !% 2 2  '>J+<@H$(H !O '>J+<@H$(H !  ) )$++ 6O rMc[U5nURRX45 US;aG[S[ U5-5nUR X1S5 UR RUS45 O[S[ U5-5n[S[ U5-5nUR X1S5 UR XAS5 UR RUS45 URRUS45 URRXASS45 URRX1SS45 US;aU$UW4$)aa This method applies support to a particular beam object and returns the symbol of the unknown reaction load(s). Parameters ========== loc : Sympifyable Location of point at which support is applied. type : String Determines type of Beam support applied. To apply support structure with - zero degree of freedom, type = "fixed" - one degree of freedom, type = "pin" - two degrees of freedom, type = "roller" Returns ======= Symbol or tuple of Symbol The unknown reaction load as a symbol. - Symbol(reaction_force) if type = "pin" or "roller" - Symbol(reaction_force), Symbol(reaction_moment) if type = "fixed" Examples ======== There is a beam of length 20 meters. A moment of magnitude 100 Nm is applied in the clockwise direction at the end of the beam. A pointload of magnitude 8 N is applied from the top of the beam at a distance of 10 meters. There is one fixed support at the start of the beam and a roller at the end. Using the sign convention of upward forces and clockwise moment being positive. >>> from sympy.physics.continuum_mechanics.beam import Beam >>> from sympy import symbols >>> E, I = symbols('E, I') >>> b = Beam(20, E, I) >>> p0, m0 = b.apply_support(0, 'fixed') >>> p1 = b.apply_support(20, 'roller') >>> b.apply_load(-8, 10, -1) >>> b.apply_load(100, 20, -2) >>> b.solve_for_reaction_loads(p0, m0, p1) >>> b.reaction_loads {M_0: 20, R_0: -2, R_20: 10} >>> b.reaction_loads[p0] -2 >>> b.load 20*SingularityFunction(x, 0, -2) - 2*SingularityFunction(x, 0, -1) - 8*SingularityFunction(x, 10, -1) + 100*SingularityFunction(x, 20, -2) + 10*SingularityFunction(x, 20, -1) pinrollerR_rM_N) r r;appendrstr apply_loadrrr@rHloctype reaction_loadreaction_moments rJ apply_supportBeam.apply_supports#fcl %%sk2 $ $"4C=1M OOM 3    % %sAh /"4C=1M$T#c(]3O OOM 3 OOO" 5    % %sAh / MM #q *  " " ) )?T*J K %%}2t&DE $ $  /1 1rMcURn[U[5(dU$[[ UR 55H9nXR USR S::dM'UR USs $ g)zO Helper function that returns the Second moment (I) at a location in the beam. rN)r4r2rrangelenargs)rHrIrs rJ_get_I Beam._get_I/sg   !Y''H3qvv;'&&)A,++A..66!9Q<'(rMcZ[U5nURnURU5n[S[ U5-5nUR R U5 URR U5 URX#-U-US5 URR US45 U$)a This method applies a rotation hinge at a single location on the beam. Parameters ---------- loc : Sympifyable Location of point at which hinge is applied. Returns ======= Symbol The unknown rotation jump multiplied by the elastic modulus and second moment as a symbol. Examples ======== There is a beam of length 15 meters. Pin supports are placed at distances of 0 and 10 meters. There is a fixed support at the end. There are two rotation hinges in the structure, one at 5 meters and one at 10 meters. A pointload of magnitude 10 kN is applied on the hinge at 5 meters. A distributed load of 5 kN works on the structure from 10 meters to the end. Using the sign convention of upward forces and clockwise moment being positive. >>> from sympy.physics.continuum_mechanics.beam import Beam >>> from sympy import Symbol >>> E = Symbol('E') >>> I = Symbol('I') >>> b = Beam(15, E, I) >>> r0 = b.apply_support(0, type='pin') >>> r10 = b.apply_support(10, type='pin') >>> r15, m15 = b.apply_support(15, type='fixed') >>> p5 = b.apply_rotation_hinge(5) >>> p12 = b.apply_rotation_hinge(12) >>> b.apply_load(-10, 5, -1) >>> b.apply_load(-5, 10, 0, 15) >>> b.solve_for_reaction_loads(r0, r10, r15, m15) >>> b.reaction_loads {M_15: -75/2, R_0: 0, R_10: 40, R_15: -5} >>> b.rotation_jumps {P_12: -1875/(16*E*I), P_5: 9625/(24*E*I)} >>> b.rotation_jumps[p12] -1875/(16*E*I) >>> b.bending_moment() -9625*SingularityFunction(x, 5, -1)/24 + 10*SingularityFunction(x, 5, 1) - 40*SingularityFunction(x, 10, 1) + 5*SingularityFunction(x, 10, 2)/2 + 1875*SingularityFunction(x, 12, -1)/16 + 75*SingularityFunction(x, 15, 0)/2 + 5*SingularityFunction(x, 15, 1) - 5*SingularityFunction(x, 15, 2)/2 P_r) r r1rrrr<rr>rr)rHrrr rotation_jumps rJrBeam.apply_rotation_hinge;sdcl   KK tCH}-  %%,,S1 $$++M:  -sB7 %%sAh/rMcZ[U5nURnURU5n[S[ U5-5nUR R U5 URR U5 URX#-U-US5 URR US45 U$)a; This method applies a sliding hinge at a single location on the beam. Parameters ---------- loc : Sympifyable Location of point at which hinge is applied. Returns ======= Symbol The unknown deflection jump multiplied by the elastic modulus and second moment as a symbol. Examples ======== There is a beam of length 13 meters. A fixed support is placed at the beginning. There is a pin support at the end. There is a sliding hinge at a location of 8 meters. A pointload of magnitude 10 kN is applied on the hinge at 5 meters. Using the sign convention of upward forces and clockwise moment being positive. >>> from sympy.physics.continuum_mechanics.beam import Beam >>> b = Beam(13, 20, 20) >>> r0, m0 = b.apply_support(0, type="fixed") >>> s8 = b.apply_sliding_hinge(8) >>> r13 = b.apply_support(13, type="pin") >>> b.apply_load(-10, 5, -1) >>> b.solve_for_reaction_loads(r0, m0, r13) >>> b.reaction_loads {M_0: -50, R_0: 10, R_13: 0} >>> b.deflection_jumps {W_8: 85/24} >>> b.deflection_jumps[s8] 85/24 >>> b.bending_moment() 50*SingularityFunction(x, 0, 0) - 10*SingularityFunction(x, 0, 1) + 10*SingularityFunction(x, 5, 1) - 4250*SingularityFunction(x, 8, -2)/3 >>> b.deflection() -SingularityFunction(x, 0, 2)/16 + SingularityFunction(x, 0, 3)/240 - SingularityFunction(x, 5, 3)/240 + 85*SingularityFunction(x, 8, 0)/24 W_r) r r1rrrr=rr?rr)rHrrrdeflection_jumps rJapply_sliding_hingeBeam.apply_sliding_hingexsVcl   KK  C1 $$++C0 ##**?; /b9 ""C8,rMNc TURn[U5n[U5n[U5nURRXX445 U=RU[ XRU5-- slU=R U[ XRU5-- slU(aURXQX#USS9 gg)aS This method adds up the loads given to a particular beam object. Parameters ========== value : Sympifyable The value inserted should have the units [Force/(Distance**(n+1)] where n is the order of applied load. Units for applied loads: - For moments, unit = kN*m - For point loads, unit = kN - For constant distributed load, unit = kN/m - For ramp loads, unit = kN/m/m - For parabolic ramp loads, unit = kN/m/m/m - ... so on. start : Sympifyable The starting point of the applied load. For point moments and point forces this is the location of application. order : Integer The order of the applied load. - For moments, order = -2 - For point loads, order =-1 - For constant distributed load, order = 0 - For ramp loads, order = 1 - For parabolic ramp loads, order = 2 - ... so on. end : Sympifyable, optional An optional argument that can be used if the load has an end point within the length of the beam. Examples ======== There is a beam of length 4 meters. A moment of magnitude 3 Nm is applied in the clockwise direction at the starting point of the beam. A point load of magnitude 4 N is applied from the top of the beam at 2 meters from the starting point and a parabolic ramp load of magnitude 2 N/m is applied below the beam starting from 2 meters to 3 meters away from the starting point of the beam. >>> from sympy.physics.continuum_mechanics.beam import Beam >>> from sympy import symbols >>> E, I = symbols('E, I') >>> b = Beam(4, E, I) >>> b.apply_load(-3, 0, -2) >>> b.apply_load(4, 2, -1) >>> b.apply_load(-2, 2, 2, end=3) >>> b.load -3*SingularityFunction(x, 0, -2) + 4*SingularityFunction(x, 2, -1) - 2*SingularityFunction(x, 2, 2) + 2*SingularityFunction(x, 3, 0) + 4*SingularityFunction(x, 3, 1) + 2*SingularityFunction(x, 3, 2) applyrN)r5r rArr9rrF _handle_end)rHvaluestartorderendr(s rJrBeam.apply_loadsn MM ""E%#=> e/%@@@  u%85%III    QuSw  G rMc URn[U5n[U5n[U5nXX44UR;abU=RU[ XRU5--slU=R U[ XRU5--slURR XX445 O Sn[U5eU(aURXQX#USS9 gg)a This method removes a particular load present on the beam object. Returns a ValueError if the load passed as an argument is not present on the beam. Parameters ========== value : Sympifyable The magnitude of an applied load. start : Sympifyable The starting point of the applied load. For point moments and point forces this is the location of application. order : Integer The order of the applied load. - For moments, order= -2 - For point loads, order=-1 - For constant distributed load, order=0 - For ramp loads, order=1 - For parabolic ramp loads, order=2 - ... so on. end : Sympifyable, optional An optional argument that can be used if the load has an end point within the length of the beam. Examples ======== There is a beam of length 4 meters. A moment of magnitude 3 Nm is applied in the clockwise direction at the starting point of the beam. A pointload of magnitude 4 N is applied from the top of the beam at 2 meters from the starting point and a parabolic ramp load of magnitude 2 N/m is applied below the beam starting from 2 meters to 3 meters away from the starting point of the beam. >>> from sympy.physics.continuum_mechanics.beam import Beam >>> from sympy import symbols >>> E, I = symbols('E, I') >>> b = Beam(4, E, I) >>> b.apply_load(-3, 0, -2) >>> b.apply_load(4, 2, -1) >>> b.apply_load(-2, 2, 2, end=3) >>> b.load -3*SingularityFunction(x, 0, -2) + 4*SingularityFunction(x, 2, -1) - 2*SingularityFunction(x, 2, 2) + 2*SingularityFunction(x, 3, 0) + 4*SingularityFunction(x, 3, 1) + 2*SingularityFunction(x, 3, 2) >>> b.remove_load(-2, 2, 2, end = 3) >>> b.load -3*SingularityFunction(x, 0, -2) + 4*SingularityFunction(x, 2, -1) z4No such load distribution exists on the beam object.removerN) r5r rAr9rrFrrr)rHrrrrr(msgs rJ remove_loadBeam.remove_loads^ MM % %)<)< < JJ% 3Ae DD DJ   5) ??H| LM ##q (9(9!5[(I 3AA >)??H|)LM#) X 1eai( qvva|00%K@ 3AA > ??H| LM ##q (9(9!5[(I 3AA >)??H|)LM#)rMcUR$)aB Returns a Singularity Function expression which represents the load distribution curve of the Beam object. Examples ======== There is a beam of length 4 meters. A moment of magnitude 3 Nm is applied in the clockwise direction at the starting point of the beam. A point load of magnitude 4 N is applied from the top of the beam at 2 meters from the starting point and a parabolic ramp load of magnitude 2 N/m is applied below the beam starting from 3 meters away from the starting point of the beam. >>> from sympy.physics.continuum_mechanics.beam import Beam >>> from sympy import symbols >>> E, I = symbols('E, I') >>> b = Beam(4, E, I) >>> b.apply_load(-3, 0, -2) >>> b.apply_load(4, 2, -1) >>> b.apply_load(-2, 3, 2) >>> b.load -3*SingularityFunction(x, 0, -2) + 4*SingularityFunction(x, 2, -1) - 2*SingularityFunction(x, 3, 2) )r9rYs rJload Beam.loadRs2zzrMcUR$)a7 Returns a list of all loads applied on the beam object. Each load in the list is a tuple of form (value, start, order, end). Examples ======== There is a beam of length 4 meters. A moment of magnitude 3 Nm is applied in the clockwise direction at the starting point of the beam. A pointload of magnitude 4 N is applied from the top of the beam at 2 meters from the starting point. Another pointload of magnitude 5 N is applied at same position. >>> from sympy.physics.continuum_mechanics.beam import Beam >>> from sympy import symbols >>> E, I = symbols('E, I') >>> b = Beam(4, E, I) >>> b.apply_load(-3, 0, -2) >>> b.apply_load(4, 2, -1) >>> b.apply_load(5, 2, -1) >>> b.load -3*SingularityFunction(x, 0, -2) + 9*SingularityFunction(x, 2, -1) >>> b.applied_loads [(-3, 0, -2, None), (4, 2, -1, None), (5, 2, -1, None)] )rArYs rJ applied_loadsBeam.applied_loadsms4"""rMctURnURn[S5n[S5n[UR5n[UR 5n[ UR5X#5n[ UR5X#5n /n /n /n /n URSHTupUR5RX.5U- n[SUR55nU RU5 MV URSHTupUR5RX.5U- n[SUR55nU RU5 MV [UR5U5U-nURSH*upURX.5U- nU RU5 M, [UU5U-nURSH*upURX.5U- nU RU5 M, [[!X/U -U -U -U -XE4U-U-U-5RS 5nS [#U5-nU[#U5-nUS UnUUUnUUS n[%['UU55Ul[%['UU55Ul[%['UU55UlUR.RUR(5UlUR.RUR*5UlUR.RUR,5Ulg ) a  Solves for the reaction forces. Examples ======== There is a beam of length 30 meters. A moment of magnitude 120 Nm is applied in the clockwise direction at the end of the beam. A pointload of magnitude 8 N is applied from the top of the beam at the starting point. There are two simple supports below the beam. One at the end and another one at a distance of 10 meters from the start. The deflection is restricted at both the supports. Using the sign convention of upward forces and clockwise moment being positive. >>> from sympy.physics.continuum_mechanics.beam import Beam >>> from sympy import symbols >>> E, I = symbols('E, I') >>> R1, R2 = symbols('R1, R2') >>> b = Beam(30, E, I) >>> b.apply_load(-8, 0, -1) >>> b.apply_load(R1, 10, -1) # Reaction force at x = 10 >>> b.apply_load(R2, 30, -1) # Reaction force at x = 30 >>> b.apply_load(120, 30, -2) >>> b.bc_deflection = [(10, 0), (30, 0)] >>> b.load R1*SingularityFunction(x, 10, -1) + R2*SingularityFunction(x, 30, -1) - 8*SingularityFunction(x, 0, -1) + 120*SingularityFunction(x, 30, -2) >>> b.solve_for_reaction_loads(R1, R2) >>> b.reaction_loads {R1: 6, R2: 2} >>> b.load -8*SingularityFunction(x, 0, -1) + 6*SingularityFunction(x, 10, -1) + 120*SingularityFunction(x, 30, -2) + 2*SingularityFunction(x, 30, -1) C3C4r/c3l# UH*n[SUR55(aM&Uv M, g7f)c38# UHoRv M g7fry is_infinite.0nums rJ :Beam.solve_for_reaction_loads....&b#r?rr/r.r8rsumrrrlistrrdictziprBr]rar9)rH reactionsr(rzrrr^rb shear_curve moment_curveshear_force_eqsbending_moment_eqs slope_eqsdeflection_eqspositionreqsnew_eqs slope_curvedeflection_curvesolutionreaction_indexrotation_indexreaction_solutionrotation_solutiondeflection_solutions rJsolve_for_reaction_loadsBeam.solve_for_reaction_loadss J MM KK D\ D\t;;< !CbbbG  " "7 + H $889IJOH%%',,Q9EACbbbG  % %g . K   3 3 5q9B> #88AOH""1/%7C   S ! B%[!4r9#88FOH"''4u)'#n*==$Q~6$^NC&~7#C 3D$EF#C8I$JK!%c*:>> from sympy.physics.continuum_mechanics.beam import Beam >>> from sympy import symbols >>> E, I = symbols('E, I') >>> R1, R2 = symbols('R1, R2') >>> b = Beam(30, E, I) >>> b.apply_load(-8, 0, -1) >>> b.apply_load(R1, 10, -1) >>> b.apply_load(R2, 30, -1) >>> b.apply_load(120, 30, -2) >>> b.bc_deflection = [(10, 0), (30, 0)] >>> b.solve_for_reaction_loads(R1, R2) >>> b.shear_force() 8*SingularityFunction(x, 0, 0) - 6*SingularityFunction(x, 10, 0) - 120*SingularityFunction(x, 30, -1) - 2*SingularityFunction(x, 30, 0) )r5rrrHr(s rJr/Beam.shear_forces!> MM$))Q'''rMc UR5nURnURn/nUHEn[U[5(aURSnUR URS5 MG [ [U55nUR5 /n/n[U5GH$upU S:XaM[[S5X$US- :*4URR[5X):4[S5S45n [X5n /n U H,n U R [UR!X-555 M. U R#XHS- U /5 U [[%XXHS- S55[[%XU S55/- n ['U 5nUR U5 UR XR)U55 GM' [ [/[U55n['U5nXgR)U5n U U4$![*a [%XXHS- S5n[%XU S5nUR!X$US- U -S- 5UU-S- :Xa5UU:wa/UR#UU/5 UR#XHS- U /5 GMUR U5 UR [-XHS- U 55 GMf=f zBReturns maximum Shear force and its coordinate in the Beam object.rrrnanT+-r)r/r5rr2r rrsetsort enumeraterfloatr9rewriterabsrextendrmaxindexrrmap)rHr"r(terms singularityterm intervals shear_valuesrr shear_slopepointsvalpoint max_shear initial_shear final_shear maximum_shears rJmax_shear_forceBeam.max_shear_forces&&( MM   D$$$yy}   tyy| ,3{+,   k*DAAv D'uqac:J7J(KTZZM_M_`iMjlmloLpsxy~sBFsGH {.#EJJs;#3#3A#=>?$ {Q3/34E++c2BCHI3uU`efhkOlKmnnH ##I.   )(+8C\23 L) ,,];<}%%' D %kkA#6F L #KAs; ##AAaC(81(%BCC Ds8DH  BK1KKcNURn[UR5U5$)a^ Returns a Singularity Function expression which represents the bending moment curve of the Beam object. Examples ======== There is a beam of length 30 meters. A moment of magnitude 120 Nm is applied in the clockwise direction at the end of the beam. A pointload of magnitude 8 N is applied from the top of the beam at the starting point. There are two simple supports below the beam. One at the end and another one at a distance of 10 meters from the start. The deflection is restricted at both the supports. Using the sign convention of upward forces and clockwise moment being positive. >>> from sympy.physics.continuum_mechanics.beam import Beam >>> from sympy import symbols >>> E, I = symbols('E, I') >>> R1, R2 = symbols('R1, R2') >>> b = Beam(30, E, I) >>> b.apply_load(-8, 0, -1) >>> b.apply_load(R1, 10, -1) >>> b.apply_load(R2, 30, -1) >>> b.apply_load(120, 30, -2) >>> b.bc_deflection = [(10, 0), (30, 0)] >>> b.solve_for_reaction_loads(R1, R2) >>> b.bending_moment() 8*SingularityFunction(x, 0, 1) - 6*SingularityFunction(x, 10, 1) - 120*SingularityFunction(x, 30, 0) - 2*SingularityFunction(x, 30, 1) )r5rr/r6s rJr.Beam.bending_moment4s#> MM))+Q//rMc UR5nURnURn/nUHEn[U[5(aURSnUR URS5 MG [ [U55nUR5 /n/n[U5GH(upU S:XaM[[S5X$US- :*4UR5R[5X):4[S5S45n [X5n /n U H,n U R [UR!X-555 M. U R#XHS- U /5 U [[%XXHS- S55[[%XU S55/- n ['U 5nUR U5 UR XR)U55 GM+ [ [/[U55n['U5nXgR)U5n U U4$![*a [%XXHS- S5n[%XU S5nUR!X$US- U -S- 5UU-S- :Xa5UU:wa/UR#UU/5 UR#XHS- U /5 GMUR U5 UR [-XHS- U 55 GM"f=fr9)r.r5rr2r rrr=r>r?rr@r/rArrBrrCrrDrErrrF)rH bending_curver(rGrHrIrJ moment_valuesrr moment_sloperMrNrO max_momentinitial_moment final_momentmaximum_moments rJ max_bmomentBeam.max_bmomentVs++- MM"" D$$$yy}   tyy| ,3{+,   k*DAAv D(5\1AE(:#:;%%'// :AEB5\4( * |/#EJJs=#5#5a#?@A$ {Q3/34E-K!4DcJKSQVWdijloQpMqrr X $$Z0   *(=!>?!+@Sm45 ]+--n=>~&&' D!&}qS9I3!O$]q#>  %%aac*:Q*>)AB~XdGdfgFgglzKmK!((.,)GH$$kA#&6%:;;!((6$$XkA#.>%BCC Ds8DHBK 1K  K c[SUR5R55n[[ S5UR S:*4XR UR :4[ S5S45n[UR[5UR [RS9nU$![a nS[U5;a [S5eeSnAff=f) a Returns a Set of point(s) with zero bending moment and where bending moment curve of the beam object changes its sign from negative to positive or vice versa. Examples ======== There is is 10 meter long overhanging beam. There are two simple supports below the beam. One at the start and another one at a distance of 6 meters from the start. Point loads of magnitude 10KN and 20KN are applied at 2 meters and 4 meters from start respectively. A Uniformly distribute load of magnitude of magnitude 3KN/m is also applied on top starting from 6 meters away from starting point till end. Using the sign convention of upward forces and clockwise moment being positive. >>> from sympy.physics.continuum_mechanics.beam import Beam >>> from sympy import symbols >>> E, I = symbols('E, I') >>> b = Beam(10, E, I) >>> b.apply_load(-4, 0, -1) >>> b.apply_load(-46, 6, -1) >>> b.apply_load(10, 2, -1) >>> b.apply_load(20, 4, -1) >>> b.apply_load(3, 6, 0) >>> b.point_cflexure() [10/3] c3h# UH(oRSRSS:aM$Uv M* g7f)rrrN)rrs rJr&Beam.point_cflexure..s2)s9S#[c[cde[f[k[klm[nqr[r##9Ss#2 2r:rTdomainz"An expression is already zero whenzXThis method cannot be used when a whole region of the bending moment line is equal to 0.N) rr.rrr@r5r0rrArRealsrr)rHnon_singular_bending_momentr#rMrs rJpoint_cflexureBeam.point_cflexures@'*)s9L9L9N9S9S)s&s#!%, q0@!A,mmDKK.GHut$&  <// :DMM"#''+F # 3s1v=)+STT  s46B,, C6CCc bURnURnURnURS(d[ UR 5U5$[ U[5(GaUR(Ga URnSnSnSn[[U55HnUS:waXHS- SRSn[R*U- [UR5XHS- XU45-n U[U5S- :wa:XVU -[!XS5-Xi-[!XUSRSS5-- - nOXVU -[!XS5-- nU R#XUSRS5nM U$[%S5n [[RX#-- UR5-U5*U -n /n URSH*upU R#X5U- nU R'U5 M, [)[+X55nU R#U USS05n U $)al Returns a Singularity Function expression which represents the slope the elastic curve of the Beam object. Examples ======== There is a beam of length 30 meters. A moment of magnitude 120 Nm is applied in the clockwise direction at the end of the beam. A pointload of magnitude 8 N is applied from the top of the beam at the starting point. There are two simple supports below the beam. One at the end and another one at a distance of 10 meters from the start. The deflection is restricted at both the supports. Using the sign convention of upward forces and clockwise moment being positive. >>> from sympy.physics.continuum_mechanics.beam import Beam >>> from sympy import symbols >>> E, I = symbols('E, I') >>> R1, R2 = symbols('R1, R2') >>> b = Beam(30, E, I) >>> b.apply_load(-8, 0, -1) >>> b.apply_load(R1, 10, -1) >>> b.apply_load(R2, 30, -1) >>> b.apply_load(120, 30, -2) >>> b.bc_deflection = [(10, 0), (30, 0)] >>> b.solve_for_reaction_loads(R1, R2) >>> b.slope() (-4*SingularityFunction(x, 0, 2) + 3*SingularityFunction(x, 10, 2) + 120*SingularityFunction(x, 30, 1) + SingularityFunction(x, 30, 2) + 4000/3)/(E*I) r-rrr)r5r1r4r8rr,r2rrGrrrrOnerr.rrrrrr)rHr(rrrr- prev_slopeprev_endr slope_valuerr+bc_eqsr(rr) constantss rJr- Beam.slopes'@ MM     ((1)1- - a # #(9(9(966DEJH3t9%6#aCy|003H uufQhy1D1D1Ftwqz1QTUabSc'dd D A %;68KAYZ8[[#13Fqq'RS*//Z[J\^_3``aaE;68KAYZ8[[[E(--aa1CD &L D\ T-@-@-B!BAFFK #88AOH""1/%7C MM#  B&-. !&&IaLO'<= rMc  URnURnURnURS(GdURS(Gd[ U[ 5(GaTUR (GaBURnSnSnSnSn[[U55GHn U S:waXIS- SRSn[R*U- [UR5XIS- XU45-n XZ-n [XXq45n U [U5S- :wa:XU -[XS5-Xl-[XU SRSS5-- - nOXU -[XS5-- nU RXU SRS5nU RXU SRS5nGM U$UR n [#U S-5n[RX#-- [[UR5U5*U5-USU--US-$URS(d7UR n [#U S-5n[UR%5U5U-$URS(GdZURS(GaE[ U[ 5(GaTUR (GaBURnSnSnSnSn[[U55GHn U S:waXIS- SRSn[R*U- [UR5XIS- XU45-n XZ-n [XXq45n U [U5S- :wa:XU -[XS5-Xl-[XU SRSS5-- - nOXU -[XS5-- nU RXU SRS5nU RXU SRS5nGM U$UR n [#U S-5unn[UR5U5*U-n[UU5U-n/nURSH,unnURUU5U- nUR'U5 M. [)[+UUU455nURUUSSUUSS05n[RX#-- U-$[ U[ 5(GaSUR (GaAURnSnSnSnSn[[U55GHn U S:waXIS- SRSn[RU- [UR5XIS- XU45-n XZ-n [XXq45n U [U5S- :wa:XU -[XS5-Xl-[XU SRSS5-- - nOXU -[XS5-- nU RXU SRS5nU RXU SRS5nGM U$[-S5n[UR%5U5U-n/nURSH,unnURUU5U- nUR'U5 M. [)[+UU55nURUUSS05nU$)a Returns a Singularity Function expression which represents the elastic curve or deflection of the Beam object. Examples ======== There is a beam of length 30 meters. A moment of magnitude 120 Nm is applied in the clockwise direction at the end of the beam. A pointload of magnitude 8 N is applied from the top of the beam at the starting point. There are two simple supports below the beam. One at the end and another one at a distance of 10 meters from the start. The deflection is restricted at both the supports. Using the sign convention of upward forces and clockwise moment being positive. >>> from sympy.physics.continuum_mechanics.beam import Beam >>> from sympy import symbols >>> E, I = symbols('E, I') >>> R1, R2 = symbols('R1, R2') >>> b = Beam(30, E, I) >>> b.apply_load(-8, 0, -1) >>> b.apply_load(R1, 10, -1) >>> b.apply_load(R2, 30, -1) >>> b.apply_load(120, 30, -2) >>> b.bc_deflection = [(10, 0), (30, 0)] >>> b.solve_for_reaction_loads(R1, R2) >>> b.deflection() (4000*x/3 - 4*SingularityFunction(x, 0, 3)/3 + SingularityFunction(x, 10, 3) + 60*SingularityFunction(x, 30, 2) + SingularityFunction(x, 30, 3)/3 - 12000)/(E*I) r,r-rrz3:54r)r5r1r4r8r2rrGrrrrrlrr.rrr7rr-rrrr)rHr(rrrrmprev_defrnr,rrorecent_segment_slopedeflection_valuerIrqconstantrrr+r,rpr(rr)s rJr,Beam.deflections@ MM     ((66t?X?XY`?a?a!Y''D,=,=,=vv  s4y)AAv#'!9Q<#4#4Q#7#$55&(9T5H5H5J47ST:5UXYefWg+h"hK+5+C('01E8GW'X$CIM)"2B'BDWXYefDg&g':>CK(a82= F#'#<#<\#J%&++Ax85@ c"$KXfr2h78I/44b)A,q/2yYZ|\]5_` 55!#;// / a # #(9(9(966DJHHJ3t9%6#aCy|003HeeAgi0C0C0Edgaj0PST`aRb&cc '1'?$#,-AxCS#T D A %.>#>@STUab@c"c#68KATUwWXz_`Oacd8ee#ffJ.>#>@STUab@c"ccJ(--aa1CD +00GAJOOA4FG&  D\$TZZ\15:#88FOHe"''84u O3sK01 { q +&G,,W56@ @ Ps"%Dc<UR5UR- $)zO Returns an expression representing the Shear Stress curve of the Beam object. r/r~rYs rJ shear_stressBeam.shear_stresss !$**,,rMc UR5nURnURnUc0nUR[5HnXS:wdM XQ;dM[ SU-5e XA;aXn[ URU5USU4SSSSS9$)a Returns a plot of shear stress present in the beam object. Parameters ========== subs : dictionary Python dictionary containing Symbols as key and their corresponding values. Examples ======== There is a beam of length 8 meters and area of cross section 2 square meters. A constant distributed load of 10 KN/m is applied from half of the beam till the end. There are two simple supports below the beam, one at the starting point and another at the ending point of the beam. A pointload of magnitude 5 KN is also applied from top of the beam, at a distance of 4 meters from the starting point. Take E = 200 GPa and I = 400*(10**-6) meter**4. Using the sign convention of downwards forces being positive. .. plot:: :context: close-figs :format: doctest :include-source: True >>> from sympy.physics.continuum_mechanics.beam import Beam >>> from sympy import symbols >>> R1, R2 = symbols('R1, R2') >>> b = Beam(8, 200*(10**9), 400*(10**-6), 2) >>> b.apply_load(5000, 2, -1) >>> b.apply_load(R1, 0, -1) >>> b.apply_load(R2, 8, -1) >>> b.apply_load(10000, 4, 0, end=8) >>> b.bc_deflection = [(0, 0), (8, 0)] >>> b.solve_for_reaction_loads(R1, R2) >>> b.plot_shear_stress() Plot object containing: [0]: cartesian line: 6875*SingularityFunction(x, 0, 0) - 2500*SingularityFunction(x, 2, 0) - 5000*SingularityFunction(x, 4, 1) + 15625*SingularityFunction(x, 8, 0) + 5000*SingularityFunction(x, 8, 1) for x over (0.0, 8.0) zvalue of %s was not passed.rz Shear Stress $\mathrm{x}$z$\tau$rtitlexlabelylabel line_color)rr5r0atomsrrrr)rHrrr(r0syms rJplot_shear_stressBeam.plot_shear_stresssZ((* MM <D%%f-CxCO !>!CDD. >\F\&&t,q!Vn_Y rMc VUR5nUc0nUR[5H'nX0R:XaMX1;dM[ SU-5e UR U;aXR nO UR n[ URU5URSU4SSSSS9$)a Returns a plot for Shear force present in the Beam object. Parameters ========== subs : dictionary Python dictionary containing Symbols as key and their corresponding values. Examples ======== There is a beam of length 8 meters. A constant distributed load of 10 KN/m is applied from half of the beam till the end. There are two simple supports below the beam, one at the starting point and another at the ending point of the beam. A pointload of magnitude 5 KN is also applied from top of the beam, at a distance of 4 meters from the starting point. Take E = 200 GPa and I = 400*(10**-6) meter**4. Using the sign convention of downwards forces being positive. .. plot:: :context: close-figs :format: doctest :include-source: True >>> from sympy.physics.continuum_mechanics.beam import Beam >>> from sympy import symbols >>> R1, R2 = symbols('R1, R2') >>> b = Beam(8, 200*(10**9), 400*(10**-6)) >>> b.apply_load(5000, 2, -1) >>> b.apply_load(R1, 0, -1) >>> b.apply_load(R2, 8, -1) >>> b.apply_load(10000, 4, 0, end=8) >>> b.bc_deflection = [(0, 0), (8, 0)] >>> b.solve_for_reaction_loads(R1, R2) >>> b.plot_shear_force() Plot object containing: [0]: cartesian line: 13750*SingularityFunction(x, 0, 0) - 5000*SingularityFunction(x, 2, 0) - 10000*SingularityFunction(x, 4, 1) + 31250*SingularityFunction(x, 8, 0) + 10000*SingularityFunction(x, 8, 1) for x over (0.0, 8.0) Value of %s was not passed.r Shear Forcer $\mathrm{V}$grr/rrr5rr0rr)rHrr/rr0s rJplot_shear_forceBeam.plot_shear_forcesV&&( <D$$V,Cmm# !>!CDD - ;;$ ++&F[[FK$$T*T]]Av,Fm&3P PrMc VUR5nUc0nUR[5H'nX0R:XaMX1;dM[ SU-5e UR U;aXR nO UR n[ URU5URSU4SSSSS9$)a Returns a plot for Bending moment present in the Beam object. Parameters ========== subs : dictionary Python dictionary containing Symbols as key and their corresponding values. Examples ======== There is a beam of length 8 meters. A constant distributed load of 10 KN/m is applied from half of the beam till the end. There are two simple supports below the beam, one at the starting point and another at the ending point of the beam. A pointload of magnitude 5 KN is also applied from top of the beam, at a distance of 4 meters from the starting point. Take E = 200 GPa and I = 400*(10**-6) meter**4. Using the sign convention of downwards forces being positive. .. plot:: :context: close-figs :format: doctest :include-source: True >>> from sympy.physics.continuum_mechanics.beam import Beam >>> from sympy import symbols >>> R1, R2 = symbols('R1, R2') >>> b = Beam(8, 200*(10**9), 400*(10**-6)) >>> b.apply_load(5000, 2, -1) >>> b.apply_load(R1, 0, -1) >>> b.apply_load(R2, 8, -1) >>> b.apply_load(10000, 4, 0, end=8) >>> b.bc_deflection = [(0, 0), (8, 0)] >>> b.solve_for_reaction_loads(R1, R2) >>> b.plot_bending_moment() Plot object containing: [0]: cartesian line: 13750*SingularityFunction(x, 0, 1) - 5000*SingularityFunction(x, 2, 1) - 5000*SingularityFunction(x, 4, 2) + 31250*SingularityFunction(x, 8, 1) + 5000*SingularityFunction(x, 8, 2) for x over (0.0, 8.0) rrBending Momentr $\mathrm{M}$brr.rrr5rr0rr)rHrr.rr0s rJplot_bending_momentBeam.plot_bending_momentsV,,. <D!''/Cmm# !>!CDD 0 ;;$ ++&F[[FN''- q&/IQa&3P PrMc VUR5nUc0nUR[5H'nX0R:XaMX1;dM[ SU-5e UR U;aXR nO UR n[ URU5URSU4SSSSS9$)a Returns a plot for slope of deflection curve of the Beam object. Parameters ========== subs : dictionary Python dictionary containing Symbols as key and their corresponding values. Examples ======== There is a beam of length 8 meters. A constant distributed load of 10 KN/m is applied from half of the beam till the end. There are two simple supports below the beam, one at the starting point and another at the ending point of the beam. A pointload of magnitude 5 KN is also applied from top of the beam, at a distance of 4 meters from the starting point. Take E = 200 GPa and I = 400*(10**-6) meter**4. Using the sign convention of downwards forces being positive. .. plot:: :context: close-figs :format: doctest :include-source: True >>> from sympy.physics.continuum_mechanics.beam import Beam >>> from sympy import symbols >>> R1, R2 = symbols('R1, R2') >>> b = Beam(8, 200*(10**9), 400*(10**-6)) >>> b.apply_load(5000, 2, -1) >>> b.apply_load(R1, 0, -1) >>> b.apply_load(R2, 8, -1) >>> b.apply_load(10000, 4, 0, end=8) >>> b.bc_deflection = [(0, 0), (8, 0)] >>> b.solve_for_reaction_loads(R1, R2) >>> b.plot_slope() Plot object containing: [0]: cartesian line: -8.59375e-5*SingularityFunction(x, 0, 2) + 3.125e-5*SingularityFunction(x, 2, 2) + 2.08333333333333e-5*SingularityFunction(x, 4, 3) - 0.0001953125*SingularityFunction(x, 8, 2) - 2.08333333333333e-5*SingularityFunction(x, 8, 3) + 0.00138541666666667 for x over (0.0, 8.0) rrSloper$\theta$mrr-rrr5rr0rr)rHrr-rr0s rJ plot_slopeBeam.plot_slopeRsV  <D;;v&Cmm# !>!CDD ' ;;$ ++&F[[FEJJt$t}}a&@&{sL LrMc VUR5nUc0nUR[5H'nX0R:XaMX1;dM[ SU-5e UR U;aXR nO UR n[ URU5URSU4SSSSS9$)a Returns a plot for deflection curve of the Beam object. Parameters ========== subs : dictionary Python dictionary containing Symbols as key and their corresponding values. Examples ======== There is a beam of length 8 meters. A constant distributed load of 10 KN/m is applied from half of the beam till the end. There are two simple supports below the beam, one at the starting point and another at the ending point of the beam. A pointload of magnitude 5 KN is also applied from top of the beam, at a distance of 4 meters from the starting point. Take E = 200 GPa and I = 400*(10**-6) meter**4. Using the sign convention of downwards forces being positive. .. plot:: :context: close-figs :format: doctest :include-source: True >>> from sympy.physics.continuum_mechanics.beam import Beam >>> from sympy import symbols >>> R1, R2 = symbols('R1, R2') >>> b = Beam(8, 200*(10**9), 400*(10**-6)) >>> b.apply_load(5000, 2, -1) >>> b.apply_load(R1, 0, -1) >>> b.apply_load(R2, 8, -1) >>> b.apply_load(10000, 4, 0, end=8) >>> b.bc_deflection = [(0, 0), (8, 0)] >>> b.solve_for_reaction_loads(R1, R2) >>> b.plot_deflection() Plot object containing: [0]: cartesian line: 0.00138541666666667*x - 2.86458333333333e-5*SingularityFunction(x, 0, 3) + 1.04166666666667e-5*SingularityFunction(x, 2, 3) + 5.20833333333333e-6*SingularityFunction(x, 4, 4) - 6.51041666666667e-5*SingularityFunction(x, 8, 3) - 5.20833333333333e-6*SingularityFunction(x, 8, 4) for x over (0.0, 8.0) rr Deflectionr$\delta$rrr,rrr5rr0rr)rHrr,rr0s rJplot_deflectionBeam.plot_deflectionsX__& <D##F+Cmm# !>!CDD , ;;$ ++&F[[FJOOD)DMM1f+E&{"$ $rMc ~URnURnUc0nUR5R[5H'nX@R:XaMXA;dM[ SU-5e X!;aXn[ UR5RU5USU4SSSSSS9n[ UR5RU5USU4S SS S SS9n[ UR5RU5USU4S SS SSS9n[ UR5RU5USU4SSSSSS9n[SSXVXx5$)a Returns a subplot of Shear Force, Bending Moment, Slope and Deflection of the Beam object. Parameters ========== subs : dictionary Python dictionary containing Symbols as key and their corresponding values. Examples ======== There is a beam of length 8 meters. A constant distributed load of 10 KN/m is applied from half of the beam till the end. There are two simple supports below the beam, one at the starting point and another at the ending point of the beam. A pointload of magnitude 5 KN is also applied from top of the beam, at a distance of 4 meters from the starting point. Take E = 200 GPa and I = 400*(10**-6) meter**4. Using the sign convention of downwards forces being positive. .. plot:: :context: close-figs :format: doctest :include-source: True >>> from sympy.physics.continuum_mechanics.beam import Beam >>> from sympy import symbols >>> R1, R2 = symbols('R1, R2') >>> b = Beam(8, 200*(10**9), 400*(10**-6)) >>> b.apply_load(5000, 2, -1) >>> b.apply_load(R1, 0, -1) >>> b.apply_load(R2, 8, -1) >>> b.apply_load(10000, 4, 0, end=8) >>> b.bc_deflection = [(0, 0), (8, 0)] >>> b.solve_for_reaction_loads(R1, R2) >>> axes = b.plot_loading_results() rrrrrrFrrrrshowrrrrrrrrrr) r0r5r,rrrrr/rr.r-r) rHrr0r5rax1ax2ax3ax4s rJplot_loading_resultsBeam.plot_loading_resultssVR== <D??$**62Cmm# !>!CDD 3 >\F4##%**408Q2G&!/4&&(--d3h65J)//!/4::<$$T*Xq&,A !/4??$))$/(Av1F%ok!/1c11rMcURnURnURU[X#S5--n[ XB5*n[ XR5nXV4$)z Helper function for I.L.D. It takes the unsubstituted copy of the load equation and uses it to calculate shear force and bending moment equations. r)r5r6r9rr)rHrr(r*rr/r.s rJ_solve_for_ild_equationsBeam._solve_for_ild_equations sS MM   zzE$7b$AAA )) ";2**rMc dURU5up4URnURnURn[ UR 5n[ UR 5n [S5n [S5n [X5U5U[USS5[XvS5- -- n [XEU5X[USS5-[USS5- [XvS5--- n /n/n/n/nURSH|unnUR5RUU5U- n[SUR55nU[U*U*S5[U*SS5- -nUU- nURU5 M~ URSHunnUR!5RUU5U- n[SUR55nUU[USS5-[USS5- [UUS5--nUU- nURU5 M [#XE5U -nURS HVunnURUU5U- U[U*SS5U[U*SS5--S --S - -nURU5 MX [#UU5U -nURS HVunnURUU5U- U[U*SS5U[U*SS5--S --S - -nURU5 MX [%['X/U-U-U-U-X4U-U-U -5RS5nS [)U5-nU[)U5-nUS UnUUUn UUSn![+[-UU55Ul[+[-UU 55Ul[+[-U U!55Ulg)aZ Determines the Influence Line Diagram equations for reaction forces under the effect of a moving load. Parameters ========== value : Integer Magnitude of moving load reactions : The reaction forces applied on the beam. Warning ======= This method creates equations that can give incorrect results when substituting a = 0 or a = l, with l the length of the beam. Examples ======== There is a beam of length 10 meters. There are two simple supports below the beam, one at the starting point and another at the ending point of the beam. Calculate the I.L.D. equations for reaction forces under the effect of a moving load of magnitude 1kN. Using the sign convention of downwards forces being positive. .. plot:: :context: close-figs :format: doctest :include-source: True >>> from sympy import symbols >>> from sympy.physics.continuum_mechanics.beam import Beam >>> E, I = symbols('E, I') >>> R_0, R_10 = symbols('R_0, R_10') >>> b = Beam(10, E, I) >>> p0 = b.apply_support(0, 'pin') >>> p10 = b.apply_support(10, 'roller') >>> b.solve_for_ild_reactions(1,R_0,R_10) >>> b.ild_reactions {R_0: -SingularityFunction(a, 0, 0) + SingularityFunction(a, 0, 1)/10 - SingularityFunction(a, 10, 1)/10, R_10: -SingularityFunction(a, 0, 1)/10 + SingularityFunction(a, 10, 0) + SingularityFunction(a, 10, 1)/10} rrrrr/c3l# UH*n[SUR55(aM&Uv M, g7f)c38# UHoRv M g7fryr r s rJr9Beam.solve_for_ild_reactions...`Di`hY\__`hrNrrs rJr/Beam.solve_for_ild_reactions..`&!j#Di`c`h`hDiAi##rr.c3l# UH*n[SUR55(aM&Uv M, g7f)c38# UHoRv M g7fryr r s rJrrgrrNrrs rJrrgrrr-rr,N)rr5r0r6rr>r?rrrr8r/rrrrr.rrrrrr rCrlrp)"rHrr!r/r.r(rzr*r^rbrrr"r#r$r%r&r'r(rNr)eqs_without_inf shear_sinceqs_with_shear_sinc moment_sinceqs_with_moment_sincr+r,r-r.r/r0r1r2s" rJsolve_for_ild_reactionsBeam.solve_for_ild_reactionss\'+&C&CE&J# MM KK   t;;< !   " "#6 7 F"667GHMHc%%',,Q9C?C!!j!jjO8.A!Q.J#J%8Aq%A$BDWXY[cefDg$hiK#2[#@  % %&: ; I 2R7 !66w?MHc""1h/#5ATVWUWYZ\]A^ailBCACEFHImJbJBJMNAN9NQR9RRC   S !@%[!4r9!66|DMHc"''84s:UFY[\Z\^_abFcfnrEGHFHJKMNrOgOGOTUFU>UXY>YYC  ! !# &E;"="ORd"dgp"p$2#357Hy4H>4Y\l4lnnrnrtuwxS^+'#n*==$Q~6$^NC&~7"3y2C#DE$(^=N)O$P!%)#.>@S*T%U"rMc UR(d [S5eURn/nUc0nURHFnURUR[5HnXR:wdM XQ;dM[SU-5e MH UR R[5HnXR:wdM XQ;dM[SU-5e URHXnUR [URURU5USUR RU54SX$SSS95 MZ [[U5S/UQ76$) aC Plots the Influence Line Diagram of Reaction Forces under the effect of a moving load. This function should be called after calling solve_for_ild_reactions(). Parameters ========== subs : dictionary Python dictionary containing Symbols as key and their corresponding values. Warning ======= The values for a = 0 and a = l, with l the length of the beam, in the plot can be incorrect. Examples ======== There is a beam of length 10 meters. A point load of magnitude 5KN is also applied from top of the beam, at a distance of 4 meters from the starting point. There are two simple supports below the beam, located at the starting point and at a distance of 7 meters from the starting point. Plot the I.L.D. equations for reactions at both support points under the effect of a moving load of magnitude 1kN. Using the sign convention of downwards forces being positive. .. plot:: :context: close-figs :format: doctest :include-source: True >>> from sympy import symbols >>> from sympy.physics.continuum_mechanics.beam import Beam >>> E, I = symbols('E, I') >>> R_0, R_7 = symbols('R_0, R_7') >>> b = Beam(10, E, I) >>> p0 = b.apply_support(0, 'roller') >>> p7 = b.apply_support(7, 'roller') >>> b.apply_load(5,4,-1) >>> b.solve_for_ild_reactions(1,R_0,R_7) >>> b.ild_reactions {R_0: -SingularityFunction(a, 0, 0) + SingularityFunction(a, 0, 1)/7 - 3*SingularityFunction(a, 10, 0)/7 - SingularityFunction(a, 10, 1)/7 - 15/7, R_7: -SingularityFunction(a, 0, 1)/7 + 10*SingularityFunction(a, 10, 0)/7 + SingularityFunction(a, 10, 1)/7 - 20/7} >>> b.plot_ild_reactions() PlotGrid object containing: Plot[0]:Plot object containing: [0]: cartesian line: -SingularityFunction(a, 0, 0) + SingularityFunction(a, 0, 1)/7 - 3*SingularityFunction(a, 10, 0)/7 - SingularityFunction(a, 10, 1)/7 - 15/7 for a over (0.0, 10.0) Plot[1]:Plot object containing: [0]: cartesian line: -SingularityFunction(a, 0, 1)/7 + 10*SingularityFunction(a, 10, 0)/7 + SingularityFunction(a, 10, 1)/7 - 20/7 for a over (0.0, 10.0) ztI.L.D. reaction equations not found. Please use solve_for_ild_reactions() to generate the I.L.D. reaction equations.rrzI.L.D. for ReactionsblueFrr) rCrr6rrrRrrrrr)rHrr*ildplotsreactionrs rJplot_ild_reactionsBeam.plot_ild_reactionss7x""TU U    <D++H**84::6B8$%BC%GHHC, <<%%f-CxCO !>!CDD.++H OOD!4!4X!>!C!CD!I 4<<$$T* +3I&uF G, H q4844rMcURnURnURnURU5upxU[ XtU5- n [ XtU5[ XtU5- U- n UH?n U R XR U 5n U R XR U 5n MA XU - [XaS5-- U[U*SS5-- U[XeS5--n Xlg)aG Determines the Influence Line Diagram equations for shear at a specified point under the effect of a moving load. Parameters ========== distance : Integer Distance of the point from the start of the beam for which equations are to be determined value : Integer Magnitude of moving load reactions : The reaction forces applied on the beam. Warning ======= This method creates equations that can give incorrect results when substituting a = 0 or a = l, with l the length of the beam. Examples ======== There is a beam of length 12 meters. There are two simple supports below the beam, one at the starting point and another at a distance of 8 meters. Calculate the I.L.D. equations for Shear at a distance of 4 meters under the effect of a moving load of magnitude 1kN. Using the sign convention of downwards forces being positive. .. plot:: :context: close-figs :format: doctest :include-source: True >>> from sympy import symbols >>> from sympy.physics.continuum_mechanics.beam import Beam >>> E, I = symbols('E, I') >>> R_0, R_8 = symbols('R_0, R_8') >>> b = Beam(12, E, I) >>> p0 = b.apply_support(0, 'roller') >>> p8 = b.apply_support(8, 'roller') >>> b.solve_for_ild_reactions(1, R_0, R_8) >>> b.solve_for_ild_shear(4, 1, R_0, R_8) >>> b.ild_shear -(-SingularityFunction(a, 0, 0) + SingularityFunction(a, 12, 0) + 2)*SingularityFunction(a, 4, 0) - SingularityFunction(-a, 0, 0) - SingularityFunction(a, 0, 0) + SingularityFunction(a, 0, 1)/8 + SingularityFunction(a, 12, 0)/2 - SingularityFunction(a, 12, 1)/8 + 1 rN) r5r0r6rrrrCrrD) rHdistancerr!r(rzr*r/_ shear_curve1 shear_curve2rshear_eqs rJsolve_for_ild_shearBeam.solve_for_ild_shearsh MM KK   66u= u[X>> ka053RRV[[ !H',,X6I6I(6STL',,X6I6I(6STL"!<$?CVWXdeCf#ff11"a;;<>CFYZ[`aFb>bc#rMc UR(d [S5eURnURnUc0nURR [ 5HnXC:wdM XA;dM[SU-5e URR [ 5HnXC:wdM XA;dM[SU-5e [ URRU5USU4SSSSSS 9$) aA Plots the Influence Line Diagram for Shear under the effect of a moving load. This function should be called after calling solve_for_ild_shear(). Parameters ========== subs : dictionary Python dictionary containing Symbols as key and their corresponding values. Warning ======= The values for a = 0 and a = l, with l the length of the beam, in the plot can be incorrect. Examples ======== There is a beam of length 12 meters. There are two simple supports below the beam, one at the starting point and another at a distance of 8 meters. Plot the I.L.D. for Shear at a distance of 4 meters under the effect of a moving load of magnitude 1kN. Using the sign convention of downwards forces being positive. .. plot:: :context: close-figs :format: doctest :include-source: True >>> from sympy import symbols >>> from sympy.physics.continuum_mechanics.beam import Beam >>> E, I = symbols('E, I') >>> R_0, R_8 = symbols('R_0, R_8') >>> b = Beam(12, E, I) >>> p0 = b.apply_support(0, 'roller') >>> p8 = b.apply_support(8, 'roller') >>> b.solve_for_ild_reactions(1, R_0, R_8) >>> b.solve_for_ild_shear(4, 1, R_0, R_8) >>> b.ild_shear -(-SingularityFunction(a, 0, 0) + SingularityFunction(a, 12, 0) + 2)*SingularityFunction(a, 4, 0) - SingularityFunction(-a, 0, 0) - SingularityFunction(a, 0, 0) + SingularityFunction(a, 0, 1)/8 + SingularityFunction(a, 12, 0)/2 - SingularityFunction(a, 12, 1)/8 + 1 >>> b.plot_ild_shear() Plot object containing: [0]: cartesian line: -(-SingularityFunction(a, 0, 0) + SingularityFunction(a, 12, 0) + 2)*SingularityFunction(a, 4, 0) - SingularityFunction(-a, 0, 0) - SingularityFunction(a, 0, 0) + SingularityFunction(a, 0, 1)/8 + SingularityFunction(a, 12, 0)/2 - SingularityFunction(a, 12, 1)/8 + 1 for a over (0.0, 12.0) ziI.L.D. shear equation not found. Please use solve_for_ild_shear() to generate the I.L.D. shear equations.rrzI.L.D. for Shear $\mathrm{a}$rrTr)rDrrRr6rrrr)rHrrzr*rs rJplot_ild_shearBeam.plot_ild_shearsnIJ J LL    <D??((0CxCO !>!CDD1<<%%f-CxCO !>!CDD.DOO((.Aq BT%o&VZ\ \rMc2URnURnURnURU5upxX![ USS5-[ USS5- [ XaS5--[ XU5- n [ XU5[ XU5- X%[ USS5-[ USS5- [ XeS5--- n UH?n U R XRU 5n U R XRU 5n MA XU - [ XaS5-- n Xlg)aB Determines the Influence Line Diagram equations for moment at a specified point under the effect of a moving load. Parameters ========== distance : Integer Distance of the point from the start of the beam for which equations are to be determined value : Integer Magnitude of moving load reactions : The reaction forces applied on the beam. Warning ======= This method creates equations that can give incorrect results when substituting a = 0 or a = l, with l the length of the beam. Examples ======== There is a beam of length 12 meters. There are two simple supports below the beam, one at the starting point and another at a distance of 8 meters. Calculate the I.L.D. equations for Moment at a distance of 4 meters under the effect of a moving load of magnitude 1kN. Using the sign convention of downwards forces being positive. .. plot:: :context: close-figs :format: doctest :include-source: True >>> from sympy import symbols >>> from sympy.physics.continuum_mechanics.beam import Beam >>> E, I = symbols('E, I') >>> R_0, R_8 = symbols('R_0, R_8') >>> b = Beam(12, E, I) >>> p0 = b.apply_support(0, 'roller') >>> p8 = b.apply_support(8, 'roller') >>> b.solve_for_ild_reactions(1, R_0, R_8) >>> b.solve_for_ild_moment(4, 1, R_0, R_8) >>> b.ild_moment -(4*SingularityFunction(a, 0, 0) - SingularityFunction(a, 0, 1) + SingularityFunction(a, 4, 1))*SingularityFunction(a, 4, 0) - SingularityFunction(a, 0, 1)/2 + SingularityFunction(a, 4, 1) - 2*SingularityFunction(a, 12, 0) - SingularityFunction(a, 12, 1)/2 rrN) r5r0r6rrrrrCrE) rHrrr!r(rzr*rmoment moment_curve1 moment_curve2r moment_eqs rJsolve_for_ild_momentBeam.solve_for_ild_momentjs?h MM KK   11%8 *=aA*FFI\]^`acdIee!4Q!!D EFHMfYaHbc v!,U6h-GG (;Aq!(D$DGZ[\^_abGc$c&9!&B%CDD "H)..x9L9LX9VWM)..x9L9LX9VWM""]%BFYZ[ghFi$ii $rMc UR(d [S5eURnUc0nURR[5HnX2:wdM X1;dM[SU-5e UR R[5HnX2:wdM X1;dM[SU-5e [ URRU5USUR 4SSSSSS 9$) a5 Plots the Influence Line Diagram for Moment under the effect of a moving load. This function should be called after calling solve_for_ild_moment(). Parameters ========== subs : dictionary Python dictionary containing Symbols as key and their corresponding values. Warning ======= The values for a = 0 and a = l, with l the length of the beam, in the plot can be incorrect. Examples ======== There is a beam of length 12 meters. There are two simple supports below the beam, one at the starting point and another at a distance of 8 meters. Plot the I.L.D. for Moment at a distance of 4 meters under the effect of a moving load of magnitude 1kN. Using the sign convention of downwards forces being positive. .. plot:: :context: close-figs :format: doctest :include-source: True >>> from sympy import symbols >>> from sympy.physics.continuum_mechanics.beam import Beam >>> E, I = symbols('E, I') >>> R_0, R_8 = symbols('R_0, R_8') >>> b = Beam(12, E, I) >>> p0 = b.apply_support(0, 'roller') >>> p8 = b.apply_support(8, 'roller') >>> b.solve_for_ild_reactions(1, R_0, R_8) >>> b.solve_for_ild_moment(4, 1, R_0, R_8) >>> b.ild_moment -(4*SingularityFunction(a, 0, 0) - SingularityFunction(a, 0, 1) + SingularityFunction(a, 4, 1))*SingularityFunction(a, 4, 0) - SingularityFunction(a, 0, 1)/2 + SingularityFunction(a, 4, 1) - 2*SingularityFunction(a, 12, 0) - SingularityFunction(a, 12, 1)/2 >>> b.plot_ild_moment() Plot object containing: [0]: cartesian line: -(4*SingularityFunction(a, 0, 0) - SingularityFunction(a, 0, 1) + SingularityFunction(a, 4, 1))*SingularityFunction(a, 4, 0) - SingularityFunction(a, 0, 1)/2 + SingularityFunction(a, 4, 1) - 2*SingularityFunction(a, 12, 0) - SingularityFunction(a, 12, 1)/2 for a over (0.0, 12.0) zlI.L.D. moment equation not found. Please use solve_for_ild_moment() to generate the I.L.D. moment equations.rrzI.L.D. for MomentrrrTr)rErr6rrrRrr)rHrr*rs rJplot_ild_momentBeam.plot_ild_momentsnLM M    <D##))&1CxCO !>!CDD2<<%%f-CxCO !>!CDD.D$$))$/!Q 1EM`%o&W[] ]rM)r )modulesc[(d [S5e[[UR5[UR 5- 5nU(d"[ SU55(a [S5eURn[UR[5(aZ[URR[55n[RUS5nURR!U5nO0nURnUS- n/nUR#SXVSS.5 UR%XU5uppn UR'XT5upX~- nX- n UR(H1nXR- n[+U5U-nU U/US- //S S S S ./- n M3 UR,H1nXR- n[+U5U-nU U/US- //S SS S ./- n M3 U*SU-4nU (a}[/[/U S5[/SU555n[1[1U S5[1SU555nUUS:d UUS:aU- 5S-nUU- UU-4n[5Xj-Xk-USU4U*XV-4UXUSU SSS9 nU$)a4 Returns a plot object representing the beam diagram of the beam. In particular, the diagram might include: * the beam. * vertical black arrows represent point loads and support reaction forces (the latter if they have been added with the ``apply_load`` method). * circular arrows represent moments. * shaded areas represent distributed loads. * the support, if ``apply_support`` has been executed. * if a composite beam has been created with the ``join`` method and a hinge has been specified, it will be shown with a white disc. The diagram shows positive loads on the upper side of the beam, and negative loads on the lower side. If two or more distributed loads acts along the same direction over the same region, the function will add them up together. .. note:: The user must be careful while entering load values. The draw function assumes a sign convention which is used for plotting loads. Given a right handed coordinate system with XYZ coordinates, the beam's length is assumed to be along the positive X axis. The draw function recognizes positive loads(with n>-2) as loads acting along negative Y direction and positive moments acting along positive Z direction. Parameters ========== pictorial: Boolean (default=True) Setting ``pictorial=True`` would simply create a pictorial (scaled) view of the beam diagram. On the other hand, ``pictorial=False`` would create a beam diagram with the exact dimensions on the plot. Examples ======== .. plot:: :context: close-figs :format: doctest :include-source: True >>> from sympy.physics.continuum_mechanics.beam import Beam >>> from sympy import symbols >>> P1, P2, M = symbols('P1, P2, M') >>> E, I = symbols('E, I') >>> b = Beam(50, 20, 30) >>> b.apply_load(-10, 2, -1) >>> b.apply_load(15, 26, -1) >>> b.apply_load(P1, 10, -1) >>> b.apply_load(-P2, 40, -1) >>> b.apply_load(90, 5, 0, 23) >>> b.apply_load(10, 30, 1, 50) >>> b.apply_load(M, 15, -2) >>> b.apply_load(-M, 30, -2) >>> p50 = b.apply_support(50, "pin") >>> p0, m0 = b.apply_support(0, "fixed") >>> p20 = b.apply_support(20, "roller") >>> p = b.draw() # doctest: +SKIP >>> p # doctest: +ELLIPSIS,+SKIP Plot object containing: [0]: cartesian line: 25*SingularityFunction(x, 5, 0) - 25*SingularityFunction(x, 23, 0) + SingularityFunction(x, 30, 1) - 20*SingularityFunction(x, 50, 0) - SingularityFunction(x, 50, 1) + 5 for x over (0.0, 50.0) [1]: cartesian line: 5 for x over (0.0, 50.0) ... >>> p.show() # doctest: +SKIP z-To use this function numpy module is requiredc3v# UH/n[USR5S:=(a USS:v M1 g7f)rrN)r free_symbols)r rzs rJrBeam.draw..I s7"_Y^TUC!(9(9$:Q$>#OQqTQY#OY^s79zl`pictorial=False` requires numerical distributed loads. Instead, symbolic loads were found. Cannot continue. rrbrown)xywidthheight facecolorrorwhitermarker markersizecolor| g?y2c30# UH oSSv M g7frrNr rs rJrrr +K 1dGAJ y1c30# UH oSSv M g7frrrs rJrrs rrrrg?F) xlimylim annotationsmarkers rectanglesrfillaxisr)r ImportErrorrr=rr@rrr5r2r0rrrrfromkeysrr _draw_load_draw_supportsr<r@r=minrDrBr)rH pictorialloadsr(rzr0rrrrload_eqload_eq1rsupport_markerssupport_rectanglesrratiox_posr_min_maxoffset sing_plots rJdraw Beam.drawsTuMN NS++,s43I3I/JJKs"_Y^"___#$ $ MM dkk4 ( (T[[&&v./A a$A[[%%a(FA[[F V]^_7;yZ[7\4 g.2.A.A&.L+( "00C++%E%L6)E %6A:, 7#TU_fgh hG1 //C++%E%L6)E 5'FQJIyy{!!#///rMc  [[UR5[UR5- 5nUS- nURn/n/n/n Sn /n Sn [ R n [ R nSnSnSnUGHtnU(aUSRU5nOUSnUSS:Xa}URUS5nUcUSUS<S US<S 3- nU(a'URSUS4UUS U-- 4S S S SS.S.5 MURSUU4UUS -4S S S SS.S.5 MUSS:XaURUS5nUcUSUS<S US<S 3- nURUS5(aURU/US- //SSS.5 GMURU/US- //SSS.5 GM6USS:dGMBUunnnnURU5nUcUSU<SU<SU<S 3- nU(dU(aUS:aSSU- -OUS- nU U[UUU5-- n U(aqUS:aUUU--O US- UU--n[SUS-5HEnU URUU5RUUU- 5[UUU5-[U5- -n MG [U [5(a U R n OU 4n [U Vs/sHnURU5PM sn6n GMkU(aUS:aSSU- -OUS- nU [#U5[UUU5-- n U(azUS:a[#U5UU--O US- UU--n[SUS-5HEnU URUU5RUUU- 5[UUU5-[U5- -n MG [U [5(a U R n OU 4n U Vs/sHnURU5PM nn[U6*U- nGMw [%U5S:a[&R("UU-5 [*R-S[/U5S5n/X]R3[45-5"U5n[1U/X^R3[45-5"U5n[U[*R65(dU[*R9U5-n[U[*R65(dU[*R9U5-nUUUSSS.nXxXU4$s snfs snf)NrrzPlease, note that this schematic view might not be in agreement with the sign convention used by the Beam class for load-related computations, because it was not possible to determine the sign (hence, the direction) of the following loads: rrrz * Point load z located at  rg?black)r headlength headwidthr)textrxytext arrowpropsrz * Moment z$\circlearrowright$)rrrz$\circlearrowleft$z* Distributed load z from z to gMbP? darkkhaki)r(rrrzorder)rr=rr@r5rZerorr"rrrrrr2rrrBrwarningswarnr r"r@rrArndarray ones_like)rHrr0rzrrr(rr load_args scaled_load load_args1 scaled_load1rrr warning_head warning_bodyrposilnrrrrf2rxxyy1yy2s rJr Beam._draw_load s[S++,s43I3I/JJK MM     &&66!  DAw||A1gAw"},,T!W5; aRVWXRY$ZZL&&raCQWZ[\bZbQbKc|PQ`apwsx(yz&&rf RUW]^_W_Q`y|MN]^mtpu(vwaB,,T!W5; T!WdSTg$VVL))$q'22NNSEF1H:+>J`oq#rsNNSEF1H:+>J_np#qraA+/(ueS,,U3; uV[]`$aaL!27QuW q5),>q#+,N,?3,J-KKTUV<-XYK"5"+s33$/$4$4 &1N !y#Ay!AFF1Iy#ABG!27QuW q CJ/B1eU/S$SSL49QJSZ50F1HQPUXDU!&q%!)!4A(RWWQ]-?-?3;-O,?3,J.KKTUV<.XYL"5",44%1%6%6 '3_ 3=>:aq :H> #X7HWZ | q MM,5 6 \\!U6]E 2sF__Y%??@DsF%5%5i%@@A"E#u}}-- 5??2& &C#u}}-- 5??2& &Cs#2/W<r]rPr?r@rr:r3r1r6r0r4r5rry)T)D__name__ __module__ __qualname____firstlineno____doc__rrKrVpropertyrZr^rbrerhrmrqrtr0setterr:r5r1r4r3rrrrrrrrrrrrrrrr3r/rTr.r`rir-r,r}rrrrrrrrrrrrrrrrr"r r __static_attributes__rrMrJr%r%-s[zEK3KZ`adZeqtDJKNDOG"R $$$$&&##))**   ]]"" [[  4__KK %%++##--##  ))888::;;==22__337788@DH2T (;z4lBHH>I@M@4##6U=n (D0&d 0D4'l0d@D B,-=@8Pt8Pt8Lt:$zA2F +gVRS5jD#LI\VF%PG]R +A,AH0 t=n3rMr%c ^\rSrSrSr\"S54U4Sjjr\S5r\RS5r\S5r \ RS5r \S 5r \ RS 5r \S 5r \S 5r \S 5rSrS0SjrS0SjrS1SjrSrSrSrSrSrSrSrSrSrSrSrSrS2SjrS3Sjr S2S jr!S3S!jr"S2S"jr#S3S#jr$S2S$jr%S3S%jr&S4S&jr'S2S'jr(S3S(jr)S)r*S*r+S+r,S,r-\-r.S-r/S.r0S/r1U=r2$)5Beam3Di! am This class handles loads applied in any direction of a 3D space along with unequal values of Second moment along different axes. .. note:: A consistent sign convention must be used while solving a beam bending problem; the results will automatically follow the chosen sign convention. This class assumes that any kind of distributed load/moment is applied through out the span of a beam. Examples ======== There is a beam of l meters long. A constant distributed load of magnitude q is applied along y-axis from start till the end of beam. A constant distributed moment of magnitude m is also applied along z-axis from start till the end of beam. Beam is fixed at both of its end. So, deflection of the beam at the both ends is restricted. >>> from sympy.physics.continuum_mechanics.beam import Beam3D >>> from sympy import symbols, simplify, collect, factor >>> l, E, G, I, A = symbols('l, E, G, I, A') >>> b = Beam3D(l, E, G, I, A) >>> x, q, m = symbols('x, q, m') >>> b.apply_load(q, 0, 0, dir="y") >>> b.apply_moment_load(m, 0, -1, dir="z") >>> b.shear_force() [0, -q*x, 0] >>> b.bending_moment() [0, 0, -m*x + q*x**2/2] >>> b.bc_slope = [(0, [0, 0, 0]), (l, [0, 0, 0])] >>> b.bc_deflection = [(0, [0, 0, 0]), (l, [0, 0, 0])] >>> b.solve_slope_deflection() >>> factor(b.slope()) [0, 0, x*(-l + x)*(-A*G*l**3*q + 2*A*G*l**2*q*x - 12*E*I*l*q - 72*E*I*m + 24*E*I*q*x)/(12*E*I*(A*G*l**2 + 12*E*I))] >>> dx, dy, dz = b.deflection() >>> dy = collect(simplify(dy), x) >>> dx == dz == 0 True >>> dy == (x*(12*E*I*l*(A*G*l**2*q - 2*A*G*l*m + 12*E*I*q) ... + x*(A*G*l*(3*l*(A*G*l**2*q - 2*A*G*l*m + 12*E*I*q) + x*(-2*A*G*l**2*q + 4*A*G*l*m - 24*E*I*q)) ... + A*G*(A*G*l**2 + 12*E*I)*(-2*l**2*q + 6*l*m - 4*m*x + q*x**2) ... - 12*E*I*q*(A*G*l**2 + 12*E*I)))/(24*A*E*G*I*(A*G*l**2 + 12*E*I))) True References ========== .. [1] https://homes.civil.aau.dk/jc/FemteSemester/Beams3D.pdf r(c>[TU]XXF5 X0lXPl/SQUl/SQUl0Ul/SQUl/SQUl/SQUl SUl g)a0Initializes the class. Parameters ========== length : Sympifyable A Symbol or value representing the Beam's length. elastic_modulus : Sympifyable A SymPy expression representing the Beam's Modulus of Elasticity. It is a measure of the stiffness of the Beam material. shear_modulus : Sympifyable A SymPy expression representing the Beam's Modulus of rigidity. It is a measure of rigidity of the Beam material. second_moment : Sympifyable or list A list of two elements having SymPy expression representing the Beam's Second moment of area. First value represent Second moment across y-axis and second across z-axis. Single SymPy expression can be passed if both values are same area : Sympifyable A SymPy expression representing the Beam's cross-sectional area in a plane perpendicular to length of the Beam. variable : Symbol, optional A Symbol object that will be used as the variable along the beam while representing the load, shear, moment, slope and deflection curve. By default, it is set to ``Symbol('x')``. rrrrN) superrK shear_modulusr: _load_vector_moment_load_vector_torsion_moment_load_Singularity_slope _deflection_angular_deflection)rHr0r1rXr4r:r5 __class__s rJrKBeam3D.__init__W sV6 -J* %#, !!* $#$ rMcUR$r)_shear_modulusrYs rJrXBeam3D.shear_modulus} rjrMc$[U5Ulgry)r rcrs rJrXrd s%ajrMcUR$rrrYs rJr4Beam3D.second_moment rjrMc[U[5(a"UVs/sHn[U5PM nnXlg[U5Ulgs snfry)r2rr rP)rHrr(s rJr4rg s> a  %&'QQA'"# ")!*D (sA cUR$r|r}rYs rJr: Beam3D.area rrMc$[U5Ulgryrrs rJr:rj rrMcUR$)z< Returns a three element list representing the load vector. )rYrYs rJ load_vectorBeam3D.load_vector s    rMcUR$)zA Returns a three element list representing moment loads on Beam. )rZrYs rJmoment_load_vectorBeam3D.moment_load_vector  '''rMcUR$)az Returns a dictionary of boundary conditions applied on the beam. The dictionary has two keywords namely slope and deflection. The value of each keyword is a list of tuple, where each tuple contains location and value of a boundary condition in the format (location, value). Further each value is a list corresponding to slope or deflection(s) values along three axes at that location. Examples ======== There is a beam of length 4 meters. The slope at 0 should be 4 along the x-axis and 0 along others. At the other end of beam, deflection along all the three axes should be zero. >>> from sympy.physics.continuum_mechanics.beam import Beam3D >>> from sympy import symbols >>> l, E, G, I, A, x = symbols('l, E, G, I, A, x') >>> b = Beam3D(30, E, G, I, A, x) >>> b.bc_slope = [(0, (4, 0, 0))] >>> b.bc_deflection = [(4, [0, 0, 0])] >>> b.boundary_conditions {'bending_moment': [], 'deflection': [(4, [0, 0, 0])], 'shear_force': [], 'slope': [(0, (4, 0, 0))]} Here the deflection of the beam should be ``0`` along all the three axes at ``4``. Similarly, the slope of the beam should be ``4`` along x-axis and ``0`` along y and z axis at ``0``. rrYs rJrBeam3D.boundary_conditions s:(((rMc~[UR5(dSUR-$[UR5$)ay Returns the polar moment of area of the beam about the X axis with respect to the centroid. Examples ======== >>> from sympy.physics.continuum_mechanics.beam import Beam3D >>> from sympy import symbols >>> l, E, G, I, A = symbols('l, E, G, I, A') >>> b = Beam3D(l, E, G, I, A) >>> b.polar_moment() 2*I >>> I1 = [9, 15] >>> b = Beam3D(l, E, G, I1, A) >>> b.polar_moment() 24 r)rr4rrYs rJ polar_momentBeam3D.polar_moment s7&**++T''' '4%%&&rMc URn[U5n[U5n[U5nUS:XaBUS:XdURS==U- ss'URS==U[ XRU5-- ss'gUS:XaBUS:XdURS==U- ss'URS==U[ XRU5-- ss'gUS:XdURS==U- ss'URS==U[ XRU5-- ss'g)a This method adds up the force load to a particular beam object. Parameters ========== value : Sympifyable The magnitude of an applied load. dir : String Axis along which load is applied. order : Integer The order of the applied load. - For point loads, order=-1 - For constant distributed load, order=0 - For ramp loads, order=1 - For parabolic ramp loads, order=2 - ... so on. r(rryrrN)r5r rYr\rrHrrrdirr(s rJrBeam3D.apply_load s$ MM #:B;!!!$-$  " "1 %/B1U/S)S S % CZB;!!!$-$  " "1 %/B1U/S)S S %B;!!!$-$  " "1 %/B1U/S)S S %rMc URn[U5n[U5n[U5nUS:XaUS:XdURS==U- ss'O?U[UR5;aURU==U- ss'OXRU'UR S==U[ XRU5-- ss'gUS:XaBUS:XdURS==U- ss'UR S==U[ XRU5-- ss'gUS:XdURS==U- ss'UR S==U[ XRU5-- ss'g)a This method adds up the moment loads to a particular beam object. Parameters ========== value : Sympifyable The magnitude of an applied moment. dir : String Axis along which moment is applied. order : Integer The order of the applied load. - For point moments, order=-2 - For constant distributed moment, order=-1 - For ramp moments, order=0 - For parabolic ramp moments, order=1 - ... so on. r(rrryrrN)r5r rZrr[r\rrzs rJapply_moment_loadBeam3D.apply_moment_load s-$ MM #:B;((+u4+D!5!566((/58/27((/  " "1 %/B1U/S)S S % CZB;((+u4+  " "1 %/B1U/S)S S %B;((+u4+  " "1 %/B1U/S)S S %rMcUS;aE[S[U5-5nX0RU'URR U/SQ45 g[S[U5-5n[S[U5-5nX4/URU'URR U/SQ45 UR R U/SQ45 g)NrrrVr)rrrBrrrrs rJrBeam3D.apply_support/ s $ $"4C=1M2?  /    % %sI&6 7"4C=1M$T#c(]3O3@2RD  /    % %sI&6 7 MM #y!1 2rMcURnURnURnUVs/sHn[XR5PM nnUVs/sHn[Xr5PM nn[ S5Hn UV s/sH7oU R U 5(dXR U 5(dM5U PM9 n n [ U 5S:XaMX[XiX#5n [XX#5n [[X/U 5RS5n[[X55nURnUH%nUU;dM UUUU:wdM[SU-5e URRU5 M gs snfs snfs sn f)a Solves for the reaction forces. Examples ======== There is a beam of length 30 meters. It it supported by rollers at of its end. A constant distributed load of magnitude 8 N is applied from start till its end along y-axis. Another linear load having slope equal to 9 is applied along z-axis. >>> from sympy.physics.continuum_mechanics.beam import Beam3D >>> from sympy import symbols >>> l, E, G, I, A, x = symbols('l, E, G, I, A, x') >>> b = Beam3D(30, E, G, I, A, x) >>> b.apply_load(8, start=0, order=0, dir="y") >>> b.apply_load(9*x, start=0, order=0, dir="z") >>> b.bc_deflection = [(0, [0, 0, 0]), (30, [0, 0, 0])] >>> R1, R2, R3, R4 = symbols('R1, R2, R3, R4') >>> b.apply_load(R1, start=0, order=-1, dir="y") >>> b.apply_load(R2, start=30, order=-1, dir="y") >>> b.apply_load(R3, start=0, order=-1, dir="z") >>> b.apply_load(R4, start=30, order=-1, dir="z") >>> b.solve_for_reaction_loads(R1, R2, R3, R4) >>> b.reaction_loads {R1: -120, R2: -120, R3: -1350, R4: -2700} rrz2Ambiguous solution for %s in different directions.N)r5r0r\rrhasrrrrrrr rBrupdate)rHrr(rzqr shear_curvesshear moment_curvesrrreactr"r#solsol_dictrZkeys rJr3Beam3D.solve_for_reaction_loads; sR6 MM KK  " "789qt $*q 9:FG,5,, GqA (`1!_-@-@-C-C}GWG[G[\]G^QE`5zQ 6K !118L+!Q-Q$%Y\_%_``   ' ' 1:G`sE(E--4E2%E2cURnURn[US*U5[US*U5[US*U5/$)zu Returns a list of three expressions which represents the shear force curve of the Beam object along all three axes. rrr)r5rYr)rHr(rs rJr/Beam3D.shear_forcek sL MM   1Q4%#Y!ua%8)QqTE1:MNNrMc(UR5S$)zI Returns expression of Axial shear force present inside the Beam object. r)r/rYs rJ axial_forceBeam3D.axial_forcet s!!$$rMcUR5SUR- UR5SUR- UR5SUR- /$)zv Returns a list of three expressions which represents the shear stress curve of the Beam object along all three axes. rrrrrYs rJrBeam3D.shear_stressz s[   "1%djj0$2B2B2DQ2G 2RTXTdTdTfghTijnjtjtTtuurMc<UR5UR- $)zD Returns expression of Axial stress present inside the Beam object. )rr~rYs rJ axial_stressBeam3D.axial_stress s!$**,,rMcURnURnUR5n[US*U5[US*US-U5[US*US- U5/$)zx Returns a list of three expressions which represents the bending moment curve of the Beam object along all three axes. rrr)r5rZr/r)rHr(rrs rJr.Beam3D.bending_moment sq MM  $ $  "1Q4%#Y!uuQx/?%C1Q4%%(*A.1 1rMc(UR5S$)zH Returns expression of Torsional moment present inside the Beam object. r)r.rYs rJtorsional_momentBeam3D.torsional_moment s""$Q''rMc 2URnSn[UR5HnX RU- nM [UR5R5 [UR5n[ X!US:*4SXS:45n[ [ U55SSH@nX RXFS- -nU[ SXUS- :*4X!XF:*4SXU:45- nMB [U5nXpRUR5-- Ul g)a Solves for the angular deflection due to the torsional effects of moments being applied in the x-direction i.e. out of or into the beam. Here, a positive torque means the direction of the torque is positive i.e. out of the beam along the beam-axis. Likewise, a negative torque signifies a torque into the beam cross-section. Examples ======== >>> from sympy.physics.continuum_mechanics.beam import Beam3D >>> from sympy import symbols >>> l, E, G, I, A, x = symbols('l, E, G, I, A, x') >>> b = Beam3D(20, E, G, I, A, x) >>> b.apply_moment_load(4, 4, -2, dir='x') >>> b.apply_moment_load(4, 8, -2, dir='x') >>> b.apply_moment_load(4, 8, -2, dir='x') >>> b.solve_for_torsion() >>> b.angular_deflection().subs(x, 3) 18/(G*I) rrN) r5rr[r>rrrrrXrvr_)rHr( sum_momentsrO pointsListtorque_diagramrintegrated_torque_diagrams rJsolve_for_torsionBeam3D.solve_for_torsion s%. MM $../E //6 6K0 T ! !"'')$../ "KJqM1A#BQVW=HXDYZs:'+A // Q3@ @K iA!A#,>(?+R\R_O_A`cdfgtuivfvbwx xN,%.n$=!$=?Q?QRVRcRcRe?e$f rMc  URnURnURnURnURn[ U[ 5(a USUSpvOU=pgURnURn URn [S5n [S5n [X8-[U "U5U5-U5U S-n [[U S5U "U55RSn[S5n[S5n[ [!UR#US5UR#X5/UU5RS5nUR#UUSUUS05nUR%U5nXR&S'UUR(S'[S5n[X7-[U "U5U5-U5[+U S*U5U--U S-n[[US55RSn[ [!UR#US5UR#X5/UU5RS5nUR#UUSUUS05nXH-[U "U5U5-U SU--U- XH-U-- n[[US5U "U55RSn[ [!UR#US5UR#X5/UU5RS5nUR#UUSUUS05UR&S'UR#UUS5UR(S'[X6-[U "U5U5-U5[+U SU5U- -U S-n[[US55RSn[ [!UR#US5UR#X5/UU5RS5nUR#UUSUUS05nXH-[U "U5U5-U SU--U- XH-U--n[[US55RSn[ [!UR#US5UR#X5/UU5RS5nUR#UUSUUS05UR&S'UR#UUS5UR(S'g) NrrdeflthetaC1C2C_ir)r5r0r1rXr4r2rr~rYrZr r rr rrrrrr^r]r)rHr(rzrGrI_yI_zr'rrrreqdef_xrrrqslope_xreq1slope_zeq2def_yslope_ydef_zs rJsolve_slope_deflectionBeam3D.solve_slope_deflection s MM KK         a  tQqTMC JJ  ))!JtAw22A 6a @r"ay$q'*//2 D\ D\(EJJq!$4ejj6F#GRPUUWXYZ  By|R ! =>**Q-#  A Umz%(A66:iaRS>TWZ>Z[^def^ggC$))!,(GLLA$6 Q8J#KRQSTYY[\]^ ,,9Q<IaLABc:d1gq)*T!WQY6SVY>YZ]cde]ffC$))!,(GLLA$6 Q8J#KRQSTYY[\]^ ,,9Q<IaLABc:d1gq)*T!WQY6!CDD6 ;;$ ++&F[[FK(--d3dmmQ5OX]fFGJfJ&/B3/FSXZ ZrMcUR5nUS:Xa"URSU5nUR5$US:Xa"URSU5nUR5$US:Xa"URSU5nUR5$URSU5nURSU5nURSU5n[SSX4U5$)a Returns a plot for Shear force along all three directions present in the Beam object. Parameters ========== dir : string (default : "all") Direction along which shear force plot is required. If no direction is specified, all plots are displayed. subs : dictionary Python dictionary containing Symbols as key and their corresponding values. Examples ======== There is a beam of length 20 meters. It is supported by rollers at both of its ends. A linear load having slope equal to 12 is applied along y-axis. A constant distributed load of magnitude 15 N is applied from start till its end along z-axis. .. plot:: :context: close-figs :format: doctest :include-source: True >>> from sympy.physics.continuum_mechanics.beam import Beam3D >>> from sympy import symbols >>> l, E, G, I, A, x = symbols('l, E, G, I, A, x') >>> b = Beam3D(20, E, G, I, A, x) >>> b.apply_load(15, start=0, order=0, dir="z") >>> b.apply_load(12*x, start=0, order=0, dir="y") >>> b.bc_deflection = [(0, [0, 0, 0]), (20, [0, 0, 0])] >>> R1, R2, R3, R4 = symbols('R1, R2, R3, R4') >>> b.apply_load(R1, start=0, order=-1, dir="z") >>> b.apply_load(R2, start=20, order=-1, dir="z") >>> b.apply_load(R3, start=0, order=-1, dir="y") >>> b.apply_load(R4, start=20, order=-1, dir="y") >>> b.solve_for_reaction_loads(R1, R2, R3, R4) >>> b.plot_shear_force() PlotGrid object containing: Plot[0]:Plot object containing: [0]: cartesian line: 0 for x over (0.0, 20.0) Plot[1]:Plot object containing: [0]: cartesian line: -6*x**2 for x over (0.0, 20.0) Plot[2]:Plot object containing: [0]: cartesian line: -15*x for x over (0.0, 20.0) r(ryrrr)lowerrrrrHr{rPxPyPzs rJrBeam3D.plot_shear_force4 shiik #:''T2B779  CZ''T2B779  CZ''T2B779 ''T2B''T2B''T2BAq""- -rMc UR5nUS:XaSnSnOUS:XaSnSnO US:XaSnS nUc0nUWR[5H'nX`R:wdMXb;dM[ S U-5e UR U;aX R nO UR n[ X4RU5URSU4S S U-S SU-WS9$)Nr(rrryrcrrrrFz!Bending Moment along %c directionrz$\mathrm{M(%c)}$rr)rHr{rr.rrrr0s rJ_plot_bending_momentBeam3D._plot_bending_moment| s,,. #:GE CZGE CZGE <D!'*008Cmm# !>!CDD9 ;;$ ++&F[[FN+00668R[`iLMPiP&/B3/FSXZ ZrMcUR5nUS:Xa"URSU5nUR5$US:Xa"URSU5nUR5$US:Xa"URSU5nUR5$URSU5nURSU5nURSU5n[SSX4U5$)a Returns a plot for bending moment along all three directions present in the Beam object. Parameters ========== dir : string (default : "all") Direction along which bending moment plot is required. If no direction is specified, all plots are displayed. subs : dictionary Python dictionary containing Symbols as key and their corresponding values. Examples ======== There is a beam of length 20 meters. It is supported by rollers at both of its ends. A linear load having slope equal to 12 is applied along y-axis. A constant distributed load of magnitude 15 N is applied from start till its end along z-axis. .. plot:: :context: close-figs :format: doctest :include-source: True >>> from sympy.physics.continuum_mechanics.beam import Beam3D >>> from sympy import symbols >>> l, E, G, I, A, x = symbols('l, E, G, I, A, x') >>> b = Beam3D(20, E, G, I, A, x) >>> b.apply_load(15, start=0, order=0, dir="z") >>> b.apply_load(12*x, start=0, order=0, dir="y") >>> b.bc_deflection = [(0, [0, 0, 0]), (20, [0, 0, 0])] >>> R1, R2, R3, R4 = symbols('R1, R2, R3, R4') >>> b.apply_load(R1, start=0, order=-1, dir="z") >>> b.apply_load(R2, start=20, order=-1, dir="z") >>> b.apply_load(R3, start=0, order=-1, dir="y") >>> b.apply_load(R4, start=20, order=-1, dir="y") >>> b.solve_for_reaction_loads(R1, R2, R3, R4) >>> b.plot_bending_moment() PlotGrid object containing: Plot[0]:Plot object containing: [0]: cartesian line: 0 for x over (0.0, 20.0) Plot[1]:Plot object containing: [0]: cartesian line: -15*x**2/2 for x over (0.0, 20.0) Plot[2]:Plot object containing: [0]: cartesian line: 2*x**3 for x over (0.0, 20.0) r(ryrrr)rrrrrs rJrBeam3D.plot_bending_moment shiik #:**35B779  CZ**35B779  CZ**35B779 **35B**35B**35BAq""- -rMc UR5nUS:XaSnSnOUS:XaSnSnO US:XaSnS nUc0nUWR[5H'nX`R:wdMXb;dM[ S U-5e UR U;aX R nO UR n[ X4RU5URSU4S S U-S SU-WS9$)Nr(rrryrrrrrrFzSlope along %c directionrz$\mathrm{\theta(%c)}$rr)rHr{rr-rrrr0s rJ _plot_slopeBeam3D._plot_slope s  #:GE CZGE CZGE <D>''/Cmm# !>!CDD0 ;;$ ++&F[[FEN''- q&/IRW_yz}_}&/G/KX]_ _rMcUR5nUS:Xa"URSU5nUR5$US:Xa"URSU5nUR5$US:Xa"URSU5nUR5$URSU5nURSU5nURSU5n[SSX4U5$)a Returns a plot for Slope along all three directions present in the Beam object. Parameters ========== dir : string (default : "all") Direction along which Slope plot is required. If no direction is specified, all plots are displayed. subs : dictionary Python dictionary containing Symbols as keys and their corresponding values. Examples ======== There is a beam of length 20 meters. It is supported by rollers at both of its ends. A linear load having slope equal to 12 is applied along y-axis. A constant distributed load of magnitude 15 N is applied from start till its end along z-axis. .. plot:: :context: close-figs :format: doctest :include-source: True >>> from sympy.physics.continuum_mechanics.beam import Beam3D >>> from sympy import symbols >>> l, E, G, I, A, x = symbols('l, E, G, I, A, x') >>> b = Beam3D(20, 40, 21, 100, 25, x) >>> b.apply_load(15, start=0, order=0, dir="z") >>> b.apply_load(12*x, start=0, order=0, dir="y") >>> b.bc_deflection = [(0, [0, 0, 0]), (20, [0, 0, 0])] >>> R1, R2, R3, R4 = symbols('R1, R2, R3, R4') >>> b.apply_load(R1, start=0, order=-1, dir="z") >>> b.apply_load(R2, start=20, order=-1, dir="z") >>> b.apply_load(R3, start=0, order=-1, dir="y") >>> b.apply_load(R4, start=20, order=-1, dir="y") >>> b.solve_for_reaction_loads(R1, R2, R3, R4) >>> b.solve_slope_deflection() >>> b.plot_slope() PlotGrid object containing: Plot[0]:Plot object containing: [0]: cartesian line: 0 for x over (0.0, 20.0) Plot[1]:Plot object containing: [0]: cartesian line: -x**3/1600 + 3*x**2/160 - x/8 for x over (0.0, 20.0) Plot[2]:Plot object containing: [0]: cartesian line: x**4/8000 - 19*x**2/172 + 52*x/43 for x over (0.0, 20.0) r(ryrrr)rrrrrs rJrBeam3D.plot_slope sjiik #:!!#t,B779  CZ!!#t,B779  CZ!!#t,B779 !!#t,B!!#t,B!!#t,BAq""- -rMc UR5nUS:XaSnSnOUS:XaSnSnO US:XaSnS nUc0nUWR[5H'nX`R:wdMXb;dM[ S U-5e UR U;aX R nO UR n[ X4RU5URSU4S S U-S SU-WS9$)Nr(rrryrrrrrrFzDeflection along %c directionrz$\mathrm{\delta(%c)}$rr)rHr{rr,rrrr0s rJ_plot_deflectionBeam3D._plot_deflectionJ s__& #:GE CZGE CZGE <Dg&,,V4Cmm# !>!CDD5 ;;$ ++&F[[FJ',,T2T]]Av4NW\eDEHeH&/G/KX]_ _rMcUR5nUS:Xa"URSU5nUR5$US:Xa"URSU5nUR5$US:Xa"URSU5nUR5$URSU5nURSU5nURSU5n[SSX4U5$)a* Returns a plot for Deflection along all three directions present in the Beam object. Parameters ========== dir : string (default : "all") Direction along which deflection plot is required. If no direction is specified, all plots are displayed. subs : dictionary Python dictionary containing Symbols as keys and their corresponding values. Examples ======== There is a beam of length 20 meters. It is supported by rollers at both of its ends. A linear load having slope equal to 12 is applied along y-axis. A constant distributed load of magnitude 15 N is applied from start till its end along z-axis. .. plot:: :context: close-figs :format: doctest :include-source: True >>> from sympy.physics.continuum_mechanics.beam import Beam3D >>> from sympy import symbols >>> l, E, G, I, A, x = symbols('l, E, G, I, A, x') >>> b = Beam3D(20, 40, 21, 100, 25, x) >>> b.apply_load(15, start=0, order=0, dir="z") >>> b.apply_load(12*x, start=0, order=0, dir="y") >>> b.bc_deflection = [(0, [0, 0, 0]), (20, [0, 0, 0])] >>> R1, R2, R3, R4 = symbols('R1, R2, R3, R4') >>> b.apply_load(R1, start=0, order=-1, dir="z") >>> b.apply_load(R2, start=20, order=-1, dir="z") >>> b.apply_load(R3, start=0, order=-1, dir="y") >>> b.apply_load(R4, start=20, order=-1, dir="y") >>> b.solve_for_reaction_loads(R1, R2, R3, R4) >>> b.solve_slope_deflection() >>> b.plot_deflection() PlotGrid object containing: Plot[0]:Plot object containing: [0]: cartesian line: 0 for x over (0.0, 20.0) Plot[1]:Plot object containing: [0]: cartesian line: x**5/40000 - 4013*x**3/90300 + 26*x**2/43 + 1520*x/903 for x over (0.0, 20.0) Plot[2]:Plot object containing: [0]: cartesian line: x**4/6400 - x**3/160 + 27*x**2/560 + 2*x/7 for x over (0.0, 20.0) r(ryrrr)rrrrrs rJrBeam3D.plot_deflectionh sliik #:&&sD1B779  CZ&&sD1B779  CZ&&sD1B779 &&sD1B&&sD1B&&sD1BAq""- -rMcUR5nUc0nURX5nURX5nURX5nUR X5n[ SSX4XV5$)a Returns a subplot of Shear Force, Bending Moment, Slope and Deflection of the Beam object along the direction specified. Parameters ========== dir : string (default : "x") Direction along which plots are required. If no direction is specified, plots along x-axis are displayed. subs : dictionary Python dictionary containing Symbols as key and their corresponding values. Examples ======== There is a beam of length 20 meters. It is supported by rollers at both of its ends. A linear load having slope equal to 12 is applied along y-axis. A constant distributed load of magnitude 15 N is applied from start till its end along z-axis. .. plot:: :context: close-figs :format: doctest :include-source: True >>> from sympy.physics.continuum_mechanics.beam import Beam3D >>> from sympy import symbols >>> l, E, G, I, A, x = symbols('l, E, G, I, A, x') >>> b = Beam3D(20, E, G, I, A, x) >>> subs = {E:40, G:21, I:100, A:25} >>> b.apply_load(15, start=0, order=0, dir="z") >>> b.apply_load(12*x, start=0, order=0, dir="y") >>> b.bc_deflection = [(0, [0, 0, 0]), (20, [0, 0, 0])] >>> R1, R2, R3, R4 = symbols('R1, R2, R3, R4') >>> b.apply_load(R1, start=0, order=-1, dir="z") >>> b.apply_load(R2, start=20, order=-1, dir="z") >>> b.apply_load(R3, start=0, order=-1, dir="y") >>> b.apply_load(R4, start=20, order=-1, dir="y") >>> b.solve_for_reaction_loads(R1, R2, R3, R4) >>> b.solve_slope_deflection() >>> b.plot_loading_results('y',subs) PlotGrid object containing: Plot[0]:Plot object containing: [0]: cartesian line: -6*x**2 for x over (0.0, 20.0) Plot[1]:Plot object containing: [0]: cartesian line: -15*x**2/2 for x over (0.0, 20.0) Plot[2]:Plot object containing: [0]: cartesian line: -x**3/1600 + 3*x**2/160 - x/8 for x over (0.0, 20.0) Plot[3]:Plot object containing: [0]: cartesian line: x**5/40000 - 4013*x**3/90300 + 26*x**2/43 + 1520*x/903 for x over (0.0, 20.0) rr)rrrrrr)rHr{rrrrrs rJrBeam3D.plot_loading_results skriik <D$$S/''2s)##C.1c11rMc UR5nUS:XaSnSnOUS:XaSnSnO US:XaSnS nUc0nUWR[5H'nX`R:wdMXb;dM[ S U-5e UR U;aX R nO UR n[ X4RU5URSU4S S U-S SU-WS9$)Nr(rrryrrrrrrFzShear stress along %c directionrz $\tau(%c)$r)rrrr5rr0rr)rHr{rrrrrr0s rJ_plot_shear_stressBeam3D._plot_shear_stress s((* #:GE CZGE CZGE <D(..v6Cmm# !>!CDD7 ;;$ ++&F[[FL)..t4t}}a6PY^gHILgL&}S/@UT TrMcUR5nUS:Xa"URSU5nUR5$US:Xa"URSU5nUR5$US:Xa"URSU5nUR5$URSU5nURSU5nURSU5n[SSX4U5$)a Returns a plot for Shear Stress along all three directions present in the Beam object. Parameters ========== dir : string (default : "all") Direction along which shear stress plot is required. If no direction is specified, all plots are displayed. subs : dictionary Python dictionary containing Symbols as key and their corresponding values. Examples ======== There is a beam of length 20 meters and area of cross section 2 square meters. It is supported by rollers at both of its ends. A linear load having slope equal to 12 is applied along y-axis. A constant distributed load of magnitude 15 N is applied from start till its end along z-axis. .. plot:: :context: close-figs :format: doctest :include-source: True >>> from sympy.physics.continuum_mechanics.beam import Beam3D >>> from sympy import symbols >>> l, E, G, I, A, x = symbols('l, E, G, I, A, x') >>> b = Beam3D(20, E, G, I, 2, x) >>> b.apply_load(15, start=0, order=0, dir="z") >>> b.apply_load(12*x, start=0, order=0, dir="y") >>> b.bc_deflection = [(0, [0, 0, 0]), (20, [0, 0, 0])] >>> R1, R2, R3, R4 = symbols('R1, R2, R3, R4') >>> b.apply_load(R1, start=0, order=-1, dir="z") >>> b.apply_load(R2, start=20, order=-1, dir="z") >>> b.apply_load(R3, start=0, order=-1, dir="y") >>> b.apply_load(R4, start=20, order=-1, dir="y") >>> b.solve_for_reaction_loads(R1, R2, R3, R4) >>> b.plot_shear_stress() PlotGrid object containing: Plot[0]:Plot object containing: [0]: cartesian line: 0 for x over (0.0, 20.0) Plot[1]:Plot object containing: [0]: cartesian line: -3*x**2 for x over (0.0, 20.0) Plot[2]:Plot object containing: [0]: cartesian line: -15*x/2 for x over (0.0, 20.0) r(ryrrr)rrrrrs rJrBeam3D.plot_shear_stressshiik #:((d3B779  CZ((d3B779  CZ((d3B779 ((d3B((d3B((d3BAq""- -rMcUR5nUS:XaSnOUS:XaSnOUS:XaSnUR5W(dg[[S5URS:*4UR UURUR :4[S5S 45n[UR[5UR[RS 9nURS5 URUR 5 UR5UnUVs/sHoeRURU5PM nn[[[U55n[!U5nXGR#U5U4$s snf) z( Helper function for max_shear_force(). r(rryrrrrr:Tre)rr/rr@r5rYr0rrArrgrrrrFrBrDrE) rHr{r load_curverMr"r(rKrPs rJ_max_shear_forceBeam3D._max_shear_force\sC iik #:G CZG CZG!'*e dmmQ.>?""7+T]]4;;-FGut$& z)))4dmm ww( a dkk"&&(1 DJKFq((:F KC\23  % )))45yAA Ls%E3c/nURURS55 URURS55 URURS55 U$)a Returns point of max shear force and its corresponding shear value along all directions in a Beam object as a list. solve_for_reaction_loads() must be called before using this function. Examples ======== There is a beam of length 20 meters. It is supported by rollers at both of its ends. A linear load having slope equal to 12 is applied along y-axis. A constant distributed load of magnitude 15 N is applied from start till its end along z-axis. .. plot:: :context: close-figs :format: doctest :include-source: True >>> from sympy.physics.continuum_mechanics.beam import Beam3D >>> from sympy import symbols >>> l, E, G, I, A, x = symbols('l, E, G, I, A, x') >>> b = Beam3D(20, 40, 21, 100, 25, x) >>> b.apply_load(15, start=0, order=0, dir="z") >>> b.apply_load(12*x, start=0, order=0, dir="y") >>> b.bc_deflection = [(0, [0, 0, 0]), (20, [0, 0, 0])] >>> R1, R2, R3, R4 = symbols('R1, R2, R3, R4') >>> b.apply_load(R1, start=0, order=-1, dir="z") >>> b.apply_load(R2, start=20, order=-1, dir="z") >>> b.apply_load(R3, start=0, order=-1, dir="y") >>> b.apply_load(R4, start=20, order=-1, dir="y") >>> b.solve_for_reaction_loads(R1, R2, R3, R4) >>> b.max_shear_force() [(0, 0), (20, 2400), (20, 300)] r(ryr)rr)rHrPs rJrTBeam3D.max_shear_force~s[F ..s34..s34..s34rMcUR5nUS:XaSnOUS:XaSnOUS:XaSnUR5W(dg[[S5URS:*4UR 5UURUR :4[S5S 45n[UR[5UR[RS 9nURS5 URUR 5 UR5UnUVs/sHoeRURU5PM nn[[[U55n[!U5nXGR#U5U4$s snf) z+ Helper function for max_bending_moment(). r(rryrrrrr:Tre)rr.rr@r5r/r0rrArrgrrrrFrBrDrE) rHr{rr"rMbending_moment_curver(bending_momentsmax_bending_moments rJ_max_bending_momentBeam3D._max_bending_momentsL iik #:G CZG CZG""$W-ut}}a/? @!!#G,dmmDKK.GHut$& {**95t}} ww( a dkk"#224W=PVWPV144T]]AFPVWs389 1,,-?@ACUVV X%E7c/nURURS55 URURS55 URURS55 U$)a Returns point of max bending moment and its corresponding bending moment value along all directions in a Beam object as a list. solve_for_reaction_loads() must be called before using this function. Examples ======== There is a beam of length 20 meters. It is supported by rollers at both of its ends. A linear load having slope equal to 12 is applied along y-axis. A constant distributed load of magnitude 15 N is applied from start till its end along z-axis. .. plot:: :context: close-figs :format: doctest :include-source: True >>> from sympy.physics.continuum_mechanics.beam import Beam3D >>> from sympy import symbols >>> l, E, G, I, A, x = symbols('l, E, G, I, A, x') >>> b = Beam3D(20, 40, 21, 100, 25, x) >>> b.apply_load(15, start=0, order=0, dir="z") >>> b.apply_load(12*x, start=0, order=0, dir="y") >>> b.bc_deflection = [(0, [0, 0, 0]), (20, [0, 0, 0])] >>> R1, R2, R3, R4 = symbols('R1, R2, R3, R4') >>> b.apply_load(R1, start=0, order=-1, dir="z") >>> b.apply_load(R2, start=20, order=-1, dir="z") >>> b.apply_load(R3, start=0, order=-1, dir="y") >>> b.apply_load(R4, start=20, order=-1, dir="y") >>> b.solve_for_reaction_loads(R1, R2, R3, R4) >>> b.max_bending_moment() [(0, 0), (20, 3000), (20, 16000)] r(ryr)rr)rHr`s rJrBeam3D.max_bending_moments[F 433C89433C89433C89rMcUR5nUS:XaSnOUS:XaSnOUS:XaSnUR5W(dg[[S5URS:*4UR 5UURUR :4[S5S 45n[UR[5UR[RS 9nURS5 URUR5 UR5UnUVs/sHoeRURU5PM nn[[[ U55n[#U5nXGR%U5U4$s snf) z& Helper function for max_Deflection() r(rryrrrrr:Tre)rr,rr@r5r-r0rrArrgrrRrrrFrBrDrE) rHr{rr+rMr,r(r{r|s rJ_max_deflectionBeam3D._max_deflections@ iik #:G CZG CZG )ut}}a/? @g& dkk(ABut$& {**95t}} ww( a dll#??,W5HNO1,,T]]A> O3sK01 k"((12G<< Prc/nURURS55 URURS55 URURS55 U$)a Returns point of max deflection and its corresponding deflection value along all directions in a Beam object as a list. solve_for_reaction_loads() and solve_slope_deflection() must be called before using this function. Examples ======== There is a beam of length 20 meters. It is supported by rollers at both of its ends. A linear load having slope equal to 12 is applied along y-axis. A constant distributed load of magnitude 15 N is applied from start till its end along z-axis. .. plot:: :context: close-figs :format: doctest :include-source: True >>> from sympy.physics.continuum_mechanics.beam import Beam3D >>> from sympy import symbols >>> l, E, G, I, A, x = symbols('l, E, G, I, A, x') >>> b = Beam3D(20, 40, 21, 100, 25, x) >>> b.apply_load(15, start=0, order=0, dir="z") >>> b.apply_load(12*x, start=0, order=0, dir="y") >>> b.bc_deflection = [(0, [0, 0, 0]), (20, [0, 0, 0])] >>> R1, R2, R3, R4 = symbols('R1, R2, R3, R4') >>> b.apply_load(R1, start=0, order=-1, dir="z") >>> b.apply_load(R2, start=20, order=-1, dir="z") >>> b.apply_load(R3, start=0, order=-1, dir="y") >>> b.apply_load(R4, start=20, order=-1, dir="y") >>> b.solve_for_reaction_loads(R1, R2, R3, R4) >>> b.solve_slope_deflection() >>> b.max_deflection() [(0, 0), (10, 495/14), (-10 + 10*sqrt(10793)/43, (10 - 10*sqrt(10793)/43)**3/160 - 20/7 + (10 - 10*sqrt(10793)/43)**4/6400 + 20*sqrt(10793)/301 + 27*(10 - 10*sqrt(10793)/43)**2/560)] r(ryr)rr)rHr|s rJr}Beam3D.max_deflectionsTJt++C01t++C01t++C01rM) r_r~r^r\rYrZrPrcr]r[r:rX)ryrJry)allN)r(N)3rKrLrMrNrOrrKrPrXrQr4r:rmrprrvrr~rr3r/rrrr.rrrr-r,rrrrrrrrrrrrrrTrrr`rr}rR __classcell__)r`s@rJrTrT! s3l!'s $%L##))##-- [[  !! 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