# Solve Problem 16.4 for the loading shown in Fig. Pl6.4 and the support settlements of 1 in. at B and 1 4 in. at C . FIG. P16.4, P16.7

### Structural Analysis

6th Edition
KASSIMALI + 1 other
Publisher: Cengage,
ISBN: 9781337630931

Chapter
Section

### Structural Analysis

6th Edition
KASSIMALI + 1 other
Publisher: Cengage,
ISBN: 9781337630931
Chapter 16, Problem 7P
Textbook Problem
1 views

## Solve Problem 16.4 for the loading shown in Fig. Pl6.4 and the support settlements of 1 in. at B and 1 4 in. at C.FIG. P16.4, P16.7

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.

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=I36

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

KBC=(34)I24=I32

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

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

DFBA=KBAKBA+KBC

Substitute I36 for KBA and I32 for KBC.

DFBA=I36I36+I32=0.471

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

DFBC=KBCKBA+KBC

Substitute I36 for KBA and I32 for KBC.

DFBC=I32I36+I32=0.529

Check for sum of distribution factor as below:

DFBA+DFBC=1

Substitute 0.471 for DFBA and 0.529 for DFBC.

0.471+0.529=1

Hence, OK.

Calculate the fixed end moment for AB.

FEMAB=2×36212=216kft

Calculate the fixed end moment for BA.

FEMBA=2×36212=216kft

Calculate the fixed end moment for BC.

FEMBC=2×24212=96kft

Calculate the fixed end moment for CB.

FEMCB=2×24212=96kft

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

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

Consider the member AB of the beam:

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

+MA=0By,L(36)2×(36)×(362)347.5+72=0By,L(36)=1020.5By,L=1020.536By,L=28.3k

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.

+Fy=0Ay(2×36)+By,L=0Ay72+28

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