Problem 3.5. Bent beam with different supports

Write the differential equation of a beam subjected to uniformly distributed vertical load when one end is hinged the other is built-in.
Bending stiffness of the beam is EI.

Derive the deflection in the function of p and L.

(Neglect shear deformations.)

Solve Problem

Solve

Derive the deflection function.

Check expression

v=348pL2x2EI548pL2x3EI+px424EI

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Steps

Step by step

A bent beam with different end supports is solved in Examples 3.3 and 3.4.

Step 1. Write the differential equation of the bent beam and give the general solution.

Check differential equation

Eq.(3-27), when the distributed moment load is zero

EId4vdx4=p

Check general solution

Eq.(D-79) in the Appendix

v=C1+C2x+C3x2+C4x3+pEIx424

Step 2. Write the boundary conditions. Express the constants.

Check constants

See Table 3.2.

The boundary conditions are:

At the left support 

v(0)=0      C1=0v(0)=0      C2=0

At the right support

v(L)=0    v(L)=C3L2+C4L3+pL424EI=0M(L)=0  M(L)=EId2vdx2=EI2C3+6C4L+pL22EI=0   C3, C4

From the above conditions C3 and C4 can be expressed as:C3=348pL2EI, C4=548pLEI

Step 3. Substitute the constants and give the deflection function.

Check result

v=348pL2x2EI548pLx3EI+px424EI

Results

Worked out solution

A bent beam with different end supports is solved in Examples 3.3 and 3.4.

The differential equation of the bent beam is

EId4vdx4=p

Eq.(3-27), when the distributed moment load is zero

The general solution has the following form

v=C1+C2x+C3x2+C4x3+pEIx424

Eq.(D-79) in the Appendix

The boundary conditions are:

See Table 3.2.

At the left support 

v(0)=0      C1=0v(0)=0      C2=0

At the right support

v(L)=0    v(L)=C3L2+C4L3+pL424EI=0M(L)=0  M(L)=EId2vdx2=EI2C3+6C4L+pL22EI=0   C3, C4

From the above conditions C3 and C4 can be expressed as:C3=348pL2EI, C4=548pLEI

Substituting the constants the deflection function results in

v=348pL2x2EI548pLx3EI+px424EI