Draw the influence line for the force in member AL. LKJH3 mGA,ВCDEF- 4 m–--– 4 m-m-→-- 4 m-4 m4 m--+-4 mProbs. 6–45/46/47

Answered: L K J H 3 m G A, В C D E F - 4 m–--– 4…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Question

Draw the influence line for the force in member AL.

L
K
J
H
3 m
G
A,
В
C
D
E
F
- 4 m–--– 4 m-
m-→-- 4 m-
4 m
4 m-
-+-
4 m
Probs. 6–45/46/47

Expert Solution

Step 1: Diagram with sub-sections;

Diagram with sub-sections;

Step 2: For member EH & JE;

(a) For member EH

=> Consider the section 1-1

When the moving load of 1 KN is on the left side of 1-1.

So, for an equilibrium of the right portion of 1-1

$\sum_{}^{} F_{V} = 0 = > R_{G Y} - (F_{H E} \times \frac{3}{5}) = 0 = > F_{H E} = \frac{5}{3} R_{G Y} W h e n t h e m o v i n g l o a d o f 1 K N i s o n t h e r i g h t s i d e o f 1 - 1 {\sum F_{V} = 0 (f o r t h e l e f t p o r t i o n)}_{}^{} = > R_{A Y} + F_{E H} \times \frac{3}{5} = 0 = > F_{E H} = - \frac{5}{3} R_{A Y}$

(b) For member JE

When the moving load is on the left of section 2-2

$\sum_{}^{} F_{V} = 0 (f o r t h e r i g h t s i d e o f s e c t i o n 2 - 2) R_{G Y} + (F_{E J} \times \frac{3}{5}) = 0 F_{E J} = - \frac{5}{3} R_{G Y}$

When the moving load is on the right of section 2-2;

$\sum_{}^{} F_{V} = 0 (F o r t h e l e f t s i d e o f s e c t i o n 2 - 2) R_{A Y} - F_{J E} \times \frac{3}{5} = 0 F_{J E} = \frac{5}{3} R_{A Y}$

Step 3: For member EH;

For member JI

When the moving load is on the left of section 2-2;

$\sum_{}^{} M_{E} = 0 (f o r t h e r i g h t s i d e o f s e c t i o n 2 - 2) (R_{G Y} \times 8) + (F_{I J} \times 3) = 0 T h e r e f o r e, F_{I J} = - \frac{8}{3} R_{G Y}$

When the moving load is on the right of section 2-2;

$\sum_{}^{} M_{E} = 0 (F o r t h e l e f t s i d e o f s e c t i o n 2 - 2) (R_{A Y} \times 16) + (F_{J I} \times 3) = 0 T h e r e f o r e; F_{J I} = - \frac{16}{3} R_{A Y}$

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