*4-60. The assembly consists of two posts AD and CF made of A-36 steel and having a cross-sectional area of 1000 mm2, and a 2014-T6 aluminum post BE having a cross- sectional area of 1500 mm2. If a central load of 400 kN is applied to the rigid cap, determine the normal stress in each post. There is a small gap of 0.1 mm between the post BE and the rigid member ABC. 400 kN -0.5 m -0.5 m 0.4 m D

Structural Analysis
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ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
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Why do we add 0.1 to sBE instead of deducting 0.1 ?
anics of X
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*4-60. The assembly consists of two posts AD and CF
made of A-36 steel and having a cross-sectional area of
1000 mm2, and a 2014-T6 aluminum post BE having a cross-
sectional area of 1500 mm2. If a central load of 400 kN is
applied to the rigid cap, determine the normal stress in each
post. There is a small gap of 0.1 mm between the post BE
and the rigid member ABC.
400 kN
-0.5 m
-0.5 m
0.4 m
Equation of Equilibrium. Due to symmetry, FAD FCF = F. Referring to the
FBD of the rigid cap, Fig. a,
+1EF, = 0;
F BE + 2F
- 400(10)
= 0
(1)
Compatibility Equation. Referring to the initial and final position of rods AD (CF)
and BE, Fig. b,
8 = 0.1 + 8BE
F(400)
FBE (399.9)
1(10-) 200(10°)]
= 0.1 +
1.5(10-3) 73.1(10°)]
Transcribed Image Text:anics of X Chapter 04.pdf Chapter 04.pdf Chapter 05.pdf C:/Users/Alia/Desktop/Chapter%2004.pdf O D Page view A Read aloud| ▼ Draw E Highlight O 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *4-60. The assembly consists of two posts AD and CF made of A-36 steel and having a cross-sectional area of 1000 mm2, and a 2014-T6 aluminum post BE having a cross- sectional area of 1500 mm2. If a central load of 400 kN is applied to the rigid cap, determine the normal stress in each post. There is a small gap of 0.1 mm between the post BE and the rigid member ABC. 400 kN -0.5 m -0.5 m 0.4 m Equation of Equilibrium. Due to symmetry, FAD FCF = F. Referring to the FBD of the rigid cap, Fig. a, +1EF, = 0; F BE + 2F - 400(10) = 0 (1) Compatibility Equation. Referring to the initial and final position of rods AD (CF) and BE, Fig. b, 8 = 0.1 + 8BE F(400) FBE (399.9) 1(10-) 200(10°)] = 0.1 + 1.5(10-3) 73.1(10°)]
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