# Solve Problem 16.17 for the loading shown in Fig. P16.17 and a settlement of 50 mm at support D. FIG. P16.17, P16.21

### 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 21P
Textbook Problem
19 views

## Solve Problem 16.17 for the loading shown in Fig. P16.17 and a settlement of 50 mm at support D.FIG. P16.17, P16.21

To determine

Find the member end moments and reactions for the frames.

### 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 UVL are WL230 and WL220.

Formula to calculate the fixed moment for deflection is 6EIΔL2

Calculation:

Consider the flexural rigidity EI of the frame is constant.

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

Refer Figure 1,

Calculate the relative stiffness KCA for member AC of the frame as below:

KCA=I6

Calculate the relative stiffness KCD for member CD of the frame as below:

KCD=(34)I9=I12

Calculate the distribution factor DFCA for member CA of the frame.

DFCA=KCAKCA+KCD

Substitute I6 for KCA and I12 for KCD.

DFCA=I6I6+I12=23

Calculate the distribution factor DFCD for member CD of the frame.

DFCD=KCDKCA+KCD

Substitute 3I40 for KCA and I5 for KCD.

DFCD=I12I6+I12=13

Check for sum of distribution factor as below:

DFCA+DFCD=1

Substitute 23 for DFCA and 13 for DFCD.

(23)+(13)=1

Hence, OK.

Calculate the fixed end moment for AC.

FEMAC=75×68=56

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