Exam (elaborations)

Solutions for Understanding Structures, 1st Edition Sozen (All Chapters included)

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Course
SE - Structural Engineer

Institution
SE - Structural Engineer

Complete Solutions Manual for Understanding Structures, 1st Edition by Mete A. Sozen, Toshikatsu Ichinose ; ISBN13: 9781420068610. (Full Chapters included Chapter 1 to 6)....1. Trusses 2. Moments and Deflections in Cantilever Beams 3. Moment and Deflections in Simply Supported Beams 4. Bending a...

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Uploaded on January 24, 2024
Number of pages 40
Written in 2009/2010
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Institution SE - Structural Engineer
Course SE - Structural Engineer

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SOLUTIONS MANUAL FOR
Understanding Structures:
An Introduction
to Structural
Analysis

by
Mete A. Sozen
Toshikatsu Ichinose

** Immediate Download
** Swift Response
** All Chapters included

75039_FM.indd 1 10/25/07 1:10:31

, 2007/10/25

Answers to Problems

P AB = 5 3
1-1. Equilibrium of forces at node A is shown graphically in Fig. F = 10

1-1. Because 5 3 ≈ 8.7 , No. 5 is correct.
P AC = 5

Fig. 1-1

1-2. Noting that (stress) = (axial force)/(cross-sectional area), the stresses of members AB and AC

are

PAB 5 3 PAC 5
σ AB = = (kN/mm 2 ) and σ AC = = (kN/mm 2 )
A 100 A 100

Because (strain) = (stress)/(Young’s modulus), strains in members AB and AC are

σ AB 5 3 σ AC 5
ε AB = = (No unit) and ε AC = = (No unit)
E 100 × 200 E 100 × 200

Recalling (deformation) = (strain)x(length), we have

5 3 5 3
eAB = ε AB × LAB = × 10, 000 = ≈ 4.3 ( mm )
100 × 200 2
5 5 3
eAC = ε AC × LAC = × 10, 000 3 = ≈ 4.3 ( mm )
100 × 200 2

No. 3 is correct.

1-3. Because the strength of the material is 200 N/mm2 and its cross-sectional area is 100 mm2, the

maximum possible axial force is 200 x 100 = 20 kN. On the other hand, the axial forces in members

AB and AC caused by an external force F are as follows (see Fig. 1-1).

394

, 2007/10/25

3 1
PAB = F PAC = F
2 2

Noting that PAB is larger than PAC, we can assume that the truss fails if PAB = 20 kN.

3 2
PAB = F = 20 leads to F = × 20 ≈ 23.1 (kN )
2 3

No. 4 is correct.

1-4. Equilibrium of the moments around support A leads to a reaction at support E expressed as

5
RE = × 20 = 5 ( N) .
20

Equilibrium of vertical forces leads to

RA = 20 − RE = 15 ( N) .

Equilibrium at joint A shown in Fig. 1-2a yields PAC = 15 kN, while equilibrium at joint E shown in

Fig. 1-2b yields PCE = 5 kN.

No. 1 is correct.

PAB PDE

PAB PDE
A PAC 15 kN 5 kN
E
PAC PCE PCE
15 kN 5 kN
(a) Equilibrium at node A (b) Equilibrium at node E

Fig. 1-2

1-5. The elongations of members AC and CE are calculated as below.

395

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Exam (elaborations)

Solutions for Understanding Structures, 1st Edition Sozen (All Chapters included)

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