Problem 3.14. Timoshenko beam subjected to moment load

Determine the deflection function of a beam subjected to uniformly distributed bending moment shown in the Figure. Take the shear deformation into account. One end of the beam is hinged the other is built-in. Solve the differential equation system

a) analytically, see Example 25,

b) using the force method (see Example 27),

c) *analytically, using Eq.(3-61) (See Example 110),

Give the maximum rotation and the maximum deflection with the following data: length of the beam is L= 2 m, uniformly distributed moment load is m = 10.0 kNm/m. Shear stiffness of the cross section is S = 3.5×105 kN, the bending stiffness is EI = 1.71×104 kNm2.

Solve Problem

Solve

Maximum deflection, vmax[mm]=Check displacement vector.

Rotation at the hinged support, χ(L) [×10-5]

Show functions

vχ=mAEILx22x36LASxmAEIxx22L,    whereA=m11+3EIL2S

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Steps

Step by step

Solution a)

Show Steps of Solution a)

Follow steps of Example 25

Deflection function of the beam can be expressed from the equilibrium, geometrical and material equations.

Equations are given in Table 9.

Step 1. Determine the shear force and the bending moment functions from the equilibrium equations.

Check internal forces

dVdx=0      V=C1dMdxV=m      M=m+Vdx=m+C1x+C2

Step 2. Express the deformation functions from the material equations.

Check deformations

M=EIκ      κ=1EIm+C1x+1EIC2V=Sγ      γ=C1S

Step 3. Derive the displacement functions from the geometrical equations.

Check displacements

κ=dχdx      χ=κdx=1EIm+C1x22+C2x+C3γ=dvdxχ      v=χ+γdx=1EIm+C1x22+C2x+C3+C1Sdx                           v=1EIm+C1x36+C2x22+C3x+C1Sx+C4

Step 4. Determine constants C1, C2, C3, C4 from boundary conditions.

Check boundary conditions

See Table 10

v(0)=0      C4=0χ(0)=0     C3=0M(L)=0=1EIdχdx=1EIm+C1L+1EIC2      C2=m+C1L,v(L)=0=1EIm+C1L361EIC2L22+C1SL     C1=mL23EI11S+L23EI       C2=mL+mL33EI11S+L23EIC1=mL23EI11SL23EI =10.0×2.023×1.713×104113.5×105+2.023×1.713×104=9.65 kN,C2=m+C1L=(10.09.65)2=0.707 kNm

Step 5. Substitute the constants to obtain the displacement functions. Calculate maximum displacements.

Check maximum displacements

The function of the rotation of the cross section is

χ=1EIm+C1x22+1EIC2x=1EIm9.65x22+1EI0.707x=1.03×105x2+4.13×105x

It reaches its maximum value at the hinged support:

χ(L)=4.13×105

The deflection function is

v=1EIm+C1x36+C2x22+C1Sx=3.44×106x3+2.07×105x22.76×105x

The maximum locates where the first derivative is equal to 0:

dvdx=3.44×106×3x2+2.07×105×2x2.76×105      x=0.845 mvmax=3.44×106×1.333+2.07×105×1.3322.76×105=0.0106 mm

Solution b)

Show Steps of Solution b)

Force method is presented in Section 9.1

Step 1. Choose a primary structure.

Show primary structure

The primary structure is chosen to be a cantilever, thus the redundant is the support reaction of the hinge.

Step 2. Write the deflection function of the primary structure from the original load and from the redundant, respectively.

Check deflections

Displacement function from the moment load is

v0(x)=mLEIx22x36Lχ0(x)=mLEIxx22L

Displacement function from the unit load is

v1(x)=LEIx22x36L1Sxχ1(x)=dvdxγ=LEIxx22L1S+1S

Step 3. Determine redundant from the compatibility condition.

Check redundant

Compatibility condition is written to the end of the cantilever, where there is no deflection of the original structure:

a0=v0L,   a1=v1(L)a0+a1X=0      X=a0a1=mL33EI1L33EI+LS=m11+3EIL2S

Step 4. Give the displacement functions.

Check displacement functions

v(x)=LmXEIx22x36LXSx, χ(x)=LmXEIxx22L,      where    X=m1+3EISL2  

Step 5. Calculate maximum displacements.

Check maximum displacements

The maximum rotation occurs at the hinged support:

χ(L)=LmXEIxx22L=2.010.09641.71×1042.02.022×2.0=4.13×105

The maximum deflection and its location is given below:

dv(x)dx=LmXEIxx22LXS=0      x= 0.845m  vmax=LmXEIx22x36LXSx=0.0106 mm

Solution c)

Show Steps of Solution c)

Follow steps of Example 110 in the Appendix


Step 1. Write the differential equation system of the Timoshenko beam for the given moment load.

Check differential equation system

Eq.(3-61)

Sd2dx2SddxSddxSEId2dx2vχy=0m

Step 2. Find a particular solution.

Check particular solution

Sd2vdx2+Sdχydx=0      dvdx=χySdvdx+SχyEIdχydx=m  EId2χydx2=m  χy=mEIx22+Cx+D, v=m6EIx3+Cx22+E

In a particular solution the constants can be chosen arbitrarily, thus let us set all constants equal to zero:

  χy=mEIx22, v=m2EIx32

Step 3.  Assume the solution of the homogeneous equation system in exponential form.

Check assumed displacement functions

v0(x)χy0(x)=AiBieλix

Step 3. Introduce the i-th displacement function into the homogeneous differential equation system. Solve the eigenvalue problem.

Check eigenvalue problem

Sλi2SλiSλiSEIλi2viχyieλix=00

The above problem results in four zero eigenvalues:

λ1,2,3,4=0

Because of the multiplicity of the roots, the solution function results in

vχ=A0B0+A1B1x+A2B2x2+A3B3x3

Now the vector of the solution functions is substituted into the homogeneous differential equation:

Sd2dx2SddxSddxSEId2dx2A0B0+A1B1x+A2B2x2+A3B3x3=00

Which is equivalent to the following equations:

SB12SA2+(2SB26SA3)x+3B2x2=0B0SA1S2EIB2+(SB1+2A26EIB3)x+(SB23SA3)x2+SB3x3=0

The above equations are fulfilled by any value of x only when all the coefficients of the polynomials are zero. These conditions result in the following relationship between the constants:

B0=A1+6EISA3B1=2A2B2=3A3B3=0

Step 4. Write the general solution.

Check general solution

vχ=A0A1+6EISA3+A12A2x+A23A3x2+A30x3mEIx36x22==10A0+x1A1+x22xA2+x30A3x3mEIx366EIS+3x2

Step 5. Determine constants of the displacement functions from the boundary conditions.

Check boundary conditions

At the built-in end of the beam:

v(0)=0, χy(0)=0

At the hinged end of the beam:

v(L)=0, M(L)=0      M=EIκz=EIdχydx      dχy(L)dx=0

After straightforward algebraic manipulation the four conditions result in:

A0=0A1=mS11+3EISL2A2=mL2EI3EISL21+3EISL2A3=m6EI11+3EISL2

Step 6. Give the displacement functions. Calculate the maximum deflections.

Check results

Substituting the constants to the general solution the displacement vectors  can be rearranged in the same form, as they are derived in Solutions b):

v(x)=LmXEIx22x36LXSx, χ(x)=LmXEIxx22L,      where    X=m1+3EISL2  

The maximum rotation occurs at the hinged support:

χ(L)=LmXEIxx22L=2.010.09641.71×1042.02.022×2.0=4.13×105

The maximum deflection and its location is given below:

dv(x)dx=LmXEIxx22LXS=0      x= 0.845m  vmax=LmXEIx22x36LXSx=0.0106 mm

Results

Show worked out solution

Solution a)

Show Solution a)

Follow steps of Example 25

Deflection function of the beam can be expressed from the equilibrium, geometrical and material equations.

Equations are given in Table 9.

First the shear force and the bending moment functions are determined from the equilibrium equations.

dVdx=0      V=C1dMdxV=m      M=m+Vdx=m+C1x+C2

The deformation functions are expressed from the material equations:

M=EIκ      κ=1EIm+C1x+1EIC2V=Sγ      γ=C1S

The displacement functions are derived from the geometrical equations.

κ=dχdx      χ=κdx=1EIm+C1x22+C2x+C3γ=dvdxχ      v=χ+γdx=1EIm+C1x22+C2x+C3+C1Sdx                           v=1EIm+C1x36+C2x22+C3x+C1Sx+C4

where constants C1, C2, C3, C4 can be determined from boundary conditions.

See Table 10

v(0)=0      C4=0χ(0)=0     C3=0M(L)=0=1EIdχdx=1EIm+C1L+1EIC2      C2=m+C1L,v(L)=0=1EIm+C1L361EIC2L22+C1SL     C1=mL23EI11S+L23EI       C2=mL+mL33EI11S+L23EIC1=mL23EI11SL23EI =10.0×2.023×1.713×104113.5×105+2.023×1.713×104=9.65 kN,C2=m+C1L=(10.09.65)2=0.707 kNm By substituting the constants we obtain the displacement functions. The function of the rotation of the cross section is

χ=1EIm+C1x22+1EIC2x=1EIm9.65x22+1EI0.707x=1.03×105x2+4.13×105x

It reaches its maximum value at the hinged support:

χ(L)=4.13×105

The deflection function is

v=1EIm+C1x36+C2x22+C1Sx=3.44×106x3+2.07×105x22.76×105x

The maximum locates where the first derivative is equal to 0:

dvdx=3.44×106×3x2+2.07×105×2x2.76×105      x=0.845 mvmax=3.44×106×1.333+2.07×105×1.3322.76×105=0.0106 mm

Solution b)

Show Solution b)

Force method is presented in Section 9.1

The primary structure is chosen to be a cantilever, thus the redundant is the support reaction of the hinge.

The deflection functions of the primary structure from the original load and from the redundant are given below:

Displacement function from the moment load is

v0(x)=mLEIx22x36Lχ0(x)=mLEIxx22L

Displacement function from the unit load is

v1(x)=LEIx22x36L1Sxχ1(x)=dvdxγ=LEIxx22L1S+1S

The redundant is determined from the compatibility condition.The compatibility condition is written to the end of the cantilever, where there is no deflection of the original structure:

a0=v0L,   a1=v1(L)a0+a1X=0      X=a0a1=mL33EI1L33EI+LS=m11+3EIL2S

Thee displacement functions are

v(x)=LmXEIx22x36LXSx, χ(x)=LmXEIxx22L,      where    X=m1+3EISL2  

Maximum displacements and their locations are given below. The maximum rotation occurs at the hinged support:

χ(L)=LmXEIxx22L=2.010.09641.71×1042.02.022×2.0=4.13×105

The maximum deflection and its location is given below:

dv(x)dx=LmXEIxx22LXS=0      x= 0.845m  vmax=LmXEIx22x36LXSx=0.0106 mm

Solution c)

Show Solution c)

Follow steps of Example 110 in the Appendix


The differential equation system of the Timoshenko beam for the given moment load is written in the following form:

Equation (3-61)

Sd2dx2SddxSddxSEId2dx2vχy=0m

First a particular solution is given:

Sd2vdx2+Sdχydx=0      dvdx=χySdvdx+SχyEIdχydx=m  EId2χydx2=m  χy=mEIx22+Cx+D, v=m6EIx3+Cx22+E

In a particular solution the constants can be chosen arbitrarily, thus let us set all constants equal to zero:

  χy=mEIx22, v=m2EIx32

The solution of the homogeneous equation system is assumed in exponential form.

v0(x)χy0(x)=AiBieλix

Introducing the i-th displacement function into the homogeneous differential equation system we get an eigenvalue problem.

Sλi2SλiSλiSEIλi2viχyieλix=00

The above problem results in four zero eigenvalues:

λ1,2,3,4=0

Because of the multiplicity of the roots, the solution function results in

vχ=A0B0+A1B1x+A2B2x2+A3B3x3

Now the vector of the solution functions is substituted into the homogeneous differential equation:

Sd2dx2SddxSddxSEId2dx2A0B0+A1B1x+A2B2x2+A3B3x3=00

Which is equivalent to the following equations:

SB12SA2+(2SB26SA3)x+3B2x2=0B0SA1S2EIB2+(SB1+2A26EIB3)x+(SB23SA3)x2+SB3x3=0

The above equations are fulfilled by any value of x only when all the coefficients of the polynomials are zero. These conditions result in the following relationship between the constants:

B0=A1+6EISA3B1=2A2B2=3A3B3=0

The general solution is

vχ=A0A1+6EISA3+A12A2x+A23A3x2+A30x3mEIx36x22==10A0+x1A1+x22xA2+x30A3x3mEIx366EIS+3x2

Constants of the displacement functions are determined from the boundary conditions.

At the built-in end of the beam:

v(0)=0, χy(0)=0

At the hinged end of the beam:

v(L)=0, M(L)=0      M=EIκz=EIdχydx      dχy(L)dx=0

After straightforward algebraic manipulation the four conditions result in:

A0=0A1=mS11+3EISL2A2=mL2EI3EISL21+3EISL2A3=m6EI11+3EISL2

Substituting the constants to the general solution the displacement vectors  can be rearranged in the same form, as they are derived in Solutions b):

v(x)=LmXEIx22x36LXSx, χ(x)=LmXEIxx22L,      where    X=m1+3EISL2  

The maximum rotation occurs at the hinged support:

χ(L)=LmXEIxx22L=2.010.09641.71×1042.02.022×2.0=4.13×105

The maximum deflection and its location is given below:

dv(x)dx=LmXEIxx22LXS=0      x= 0.845m  vmax=LmXEIx22x36LXSx=0.0106 mm