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

Solutions Manual for Mechanics of Material in SI Units, 10th Global Edition By Russell Hibbeler




Solutions Manual for
Mechanics of Material in SI
Units, 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
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Solutions Manual for Mechanics of Material in SI Units, 10th Global Edition By Russell Hibbeler

, Solutions Manual for Mechanics of Material in SI Units, 10th Global Edition By 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

Solutions Manual for Mechanics of Material in SI Units, 10th Global Edition By Russell Hibbeler

,1–1. A force of 80 Solutions
N is supported byMechanics
Manual for the bracket as
of Material Chapter
in SI Units, 10th1Global Edition By 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.




Ans:
NA = 77.3 N, VA = 20.7 N, MA = -0.555 N # m

Solutions Manual for Mechanics of Material in SI Units, 10th Global Edition By Russell Hibbeler
1

, 1–2. Solutions Manual for Mechanics of Material in SI Units, 10th Global Edition By 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,
MD = 0.3125 kN # m

Solutions Manual for Mechanics of Material in SI Units, 10th Global Edition By Russell Hibbeler
2