Answered: 20 kN F E D 3 m C A - 4 m- 4 m- 10 kN…

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

Determine (approximately) the forces in the members of the truss. The diagonals are to be designed to support both tensile and compressive forces, and therefore each is assumed to carry half the panel shear. The support reactions have been computed.

20 kN
F
E
D
3 m
C
A
- 4 m-
4 m-
10 kN
20 kN
10 kN
(а)
Fig. 7–2

Expert Solution

Step 1

Shear panels are ABEF and BCDE

From A to B ,

Shear force=-10+20

Shear force=10kN

Therefore member AE and BF carry half shear force of ABEF shear panel

$F_{AE} = F_{BE} F_{AE} = \frac{10}{2} F_{AE} = 5 kN F_{BE} = 5 kN (Tension)$

Similarly,

From B to C,

shear force=10-20

Shear force=-10kN

$F_{BD} = F_{EC} F_{BD} = \frac{- 10}{2} F_{BD} = - 5 kN F_{EC} = - 5 kN$

Step 2

Draw the free body diagram of joint A

Consider the summation of forces in the vertical direction

$\begin{array}{rcl} 0 & = & F_{AF} + F_{AE} sinθ - 10 \\ F_{AF} & = & 10 - 5 \times \frac{3}{5} \\ F_{AF} & = & 7 kN (Tension) \end{array}$

Consider the summation of forces in the horizontal direction

$\begin{array}{rcl} 0 & = & F_{AB} + F_{AE} cosθ \\ F_{AB} & = & - (5 \times \frac{4}{5}) \\ F_{AB} & = & - 4 kN (Compressive) \end{array}$

Therefore the forces in member AF and AB is 7kN and -4kN respectively

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