   # Solve Problem 16.9 for the loading shown in Fig. P16.9 and a settlement of 25 mm at support C . FIG. P16.9, P16.15

#### Solutions

Chapter
Section
Chapter 16, Problem 15P
Textbook Problem
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## Solve Problem 16.9 for the loading shown in Fig. P16.9 and a settlement of 25 mm at support C.FIG. P16.9, P16.15 To determine

Find the reaction and plot the shear and bending moment diagram.

### Explanation of Solution

Fixed end moment:

Formula to calculate the relative stiffness for fixed support IL and for roller support (34)(IL).

Formula to calculate the fixed moment for point load with equal length are PL8.

Formula to calculate the fixed moment for point load with unequal length are Pab2L2 and Pa2bL2.

Formula to calculate the fixed moment for UDL is WL212.

Formula to calculate the fixed moment for deflection is 6EIΔL2

Calculation:

Consider the flexural rigidity EI of the beam is constant.

Show the free body diagram of the entire beam as in Figure 1.

Refer Figure 1,

Calculate the relative stiffness KBA for member AB of the beam as below:

KBA=I8

Calculate the relative stiffness KBC for member BC of the beam as below:

KBC=I8

Calculate the relative stiffness KCB for member BC of the beam as below:

KCB=I8

Calculate the relative stiffness KCE for member CE of the beam as below:

KCE=I8

In the above beam joint B and C are free to rotate. Hence, calculate the distribution factor at joint B and C.

Calculate the distribution factor DFBA for member AB of the beam.

DFBA=KBAKBA+KBC

Substitute I8 for KBA and I8 for KBC.

DFBA=I8I8+I8=0.5

Calculate the distribution factor DFBC for member BC of the beam.

DFBC=KBCKBA+KBC

Substitute I8 for KBA and I8 for KBC.

DFBC=I8I8+I8=0.5

Check for sum of distribution factor as below:

DFBA+DFBC=1

Substitute 0.5 for DFBA and 0.5 for DFBC.

0.5+0.5=1

Hence, OK.

Calculate the distribution factor DFCB for member CB of the beam.

DFCB=KCBKCE+KCB

Substitute I8 for KCB and I8 for KCE.

DFCB=I8I8+I8=0.5

Calculate the distribution factor DFCE for member EC of the beam.

DFCE=KCEKCB+KCE

Substitute I8 for KCB and I8 for KCE.

DFCE=I8I8+I8=0.5

Check for sum of distribution factor as below:

DFCB+DFCE=1

Substitute 0.5 for DFCB and 0.5 for DFCE.

0.5+0.5=1

Hence, OK.

Calculate the fixed end moment for AB.

FEMAB=20×8212=106.7kNm

Calculate the fixed end moment for BA.

FEMBA=106.7kNm

Calculate the fixed end moment for BC.

FEMBC=20×8212+6(70)(800)(0.025)82=106.7+131.3=238kNm

Calculate the fixed end moment for CB.

FEMCB=20×8212+6(70)(800)(0.025)82=106.7+131.3=24.6kNm

Calculate the fixed end moment for CE.

FEMCE=60×886(70)(800)(0.025)82=71.3kNm

Calculate the fixed end moment for EC.

FEMEC=60×886(70)(800)(0.025)82=191.3kNm

Show the calculation of final moments using moment distribution method as in Table 1.

Consider the member AB of the beam:

Show the section free body diagram of the member AB, BC and CE as in Figure 2.

Calculate the vertical reaction at the left end of the joint B by taking moment about point A.

+MA=0By,L(8)20×(8)×(82)+68.7183=0By,L(8)=754.3By,L=754.38By,L=94.3kN

Calculate the horizontal reaction at point A by resolving the horizontal equilibrium.

+Fx=0Ax=0

Calculate the vertical reaction at point A by resolving the vertical equilibrium

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