Solutions Manual Mechanics of Material 10th Global Edition By Russell Hibbeler

Solutions Manual for Mechanics of Material (Global Edition) 10e Russell Hibbeler




Solutions Manual for Mechanics
of Material 10th Global Edition By
Russell Hibbeler
(All Chapters 1-14, 100% Original
Verified, A+ Grade)
This is The Only Original and
Complete Solutions Manual for
10th Global Edition, All Other Files
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Solutions Manual for Mechanics of Material (Global Edition) 10e Russell Hibbeler

, Table of Contents

1. Stress
2. Strain
3. Mechanical Properties of Materials
4. Axial Load
5. Torsion
6. Bending
7. Transverse Shear
8. Combined Loadings
9. Stress Transformation
10. Strain Transformation
11. Design of Beams and Shafts
12. Deflection of Beams and Shafts
13. Buckling of Columns
14. Energy Methods

,1–1. A force of 80 N is supported by the Chapter 1
Solutions Manual for bracket asof Material (Global
Mechanics Edition) 10e Russell Hibbeler
shown. Determine the resultant internal loadings acting on
the section through point A. 0.3 m
A
30�
0.1 m



80 N 45�


Solution
Equations of Equilibrium:
+
Q©Fx¿ = 0; NA - 80 cos 15° = 0

NA = 77.3 N Ans.

a+ ©Fy¿ = 0; VA - 80 sin 15° = 0

VA = 20.7 N Ans.

a+ ©MA = 0; MA + 80 cos 45°(0.3 cos 30°)

- 80 sin 45°(0.1 + 0.3 sin 30°) = 0

MA = - 0.555 N # m Ans.

or

a+ ©MA = 0; MA + 80 sin 15°(0.3 + 0.1 sin 30°)

- 80 cos 15°(0.1 cos 30°) = 0

MA = - 0.555 N # m Ans.

Negative sign indicates that MA acts in the opposite direction to that shown on FBD.




These solutions represent a preliminary version
of the Instructors' Solutions Manual (ISM). It is
possible and even likely that at this preliminary
stage of preparing the ISM there are some
omissions and errors in the draft solutions.
Ans:
Solutions Manual for Mechanics of Material (Global Edition)N10e 77.3 N,
A =Russell VA = 20.7 N, MA = -0.555 N m
Hibbeler #

1

, 1–2.
Solutions Manual for Mechanics of Material (Global Edition) 10e Russell Hibbeler
Determine the resultant internal loadings on the C
cross section at point D.
1m




F 2m

1.25 kN/m



B
Solution A D E
1.5 m
Support Reactions: Member BC is the two force member. 0.5 m 0.5 m 0.5 m
4
a+ ΣMA = 0;   FBC (1.5) - 1.875(0.75) = 0
5
FBC = 1.1719 kN
4
+ c ΣFy = 0;  Ay + (1.1719) - 1.875 = 0
5

Ay = 0.9375 kN

3
+ ΣFx = 0;
S (1.1719) - Ax = 0
5

Ax = 0.7031 kN

Equations of Equilibrium: For point D
+ ΣFx = 0;  ND - 0.7031 = 0
S

ND = 0.703 kN Ans.
+ c ΣFy = 0;  0.9375 - 0.625 - VD = 0

VD = 0.3125 kN Ans.

a+ ΣMD = 0;  MD + 0.625(0.25) - 0.9375(0.5) = 0
MD = 0.3125 kN # m Ans.




Ans:
ND = 0.703 kN,
VD = 0.3125 kN,
Solutions Manual for Mechanics of Material (Global Edition) 10e Russell Hibbeler MD = 0.3125 kN # m



2