Mechanics of Solids

This document provides study material and solved assignments for the subject of Mechanics of Solids. It includes solutions for questions on free body diagrams, stress, shear stress, and more. The document also covers topics such as bearing stresses and deflection of bars. The subject code and course name are not mentioned.

Running Head: MECHANICS OF SOLIDS 1
MECHANICS OF SOLIDS
Student’s Name
Professor
Date

MECHANICS OF SOLIDS 2
Mechanics of Solids
Question 1
Free body diagram of figure 1.
RD
420 k N
300
RB RA
0.96 mm 0.09mm 1.05 mm
RD sin 30 – 420 + RB- RA………………………………………………. (i)
Moment at A.
RD sin 30(2.1) + RB (1.14)- 420(1.05)
2.1 RD sin 30 + 1.14(420+RA- RD sin 30) – 441= 0
2.1 RD sin 30 + 478.8 +1.14 RA- 1.14 RD sin 30-441=0
0.96 * 50 sin 30 + 37.8 +1.14 RA=0
RA = -54.21 k N
Substuting RA in equation (i)
2.1 RD sin 30 + 1.14 RB- 441= 0
RB= 420+RA -RD sin 30
RB= 340.8 k N

MECHANICS OF SOLIDS 3
Stress in the strut BC.
𝜕= 𝐹
𝐴 == 340.8
0.2
= 1703.95 N.mm
Shear stress on bolt at A.
𝜏 = 16𝑉
3𝜋𝑑2
𝜏 =16 ∗ −54.21
3𝜋(0.05)2
𝜏 = −36.807 𝑀𝑝𝑎
Bearing stresses on flange
3𝑉
2𝑏𝑑
3∗ −54.21
2∗0.15∗0.05 = -10.842 Mpa
Bearing stress on platform
3∗ −54.21
2∗0.2∗0.05
= -8.131 Mpa

MECHANICS OF SOLIDS 4
Question 2
Considering Static equilibrium
(∑M) A=0
(-Fy * 504.5) +(Fx *86.4) + Ry* (815.9 + 504.5) + (Rx *531.3)
=0
(-68*5530*504.5) +(68sin30 * 86.6) + Rcos 20(1320.6) + Rsin 20* 531.3=0
68[-393.709] + R [1422.485] =0
R= 68∗393.709
1422.985
R = 1822 k N
(∑Fy) =0
Ay – Fy + Ry = 0
Ay- 68cos 30 +18.82cos 20=0
Ay= 41.20 k N
(∑Fx) = 0
Ax – Fx + Rx = 0
Ax= 68sin 30-18.82sin 20
Ax= 27.56 k N
Resultant force at A.
FA= √((𝐴𝑥)2 + ((Ay)2) = √((27.56)2 + ((41.204)2)
FA = 49.57 k N
Both bolts are under single shear
𝜏A =
𝐹𝐴
(𝐴𝑟𝑒𝑎)𝐴
= 49.57∗103
(𝜋
4∗252)𝐴
= 100.99 Mpa
𝜏B =
𝑅
(𝐴𝑟𝑒𝑎)𝐵
= 18.82∗103
(𝜋
4∗202)𝐴
= 59.

MECHANICS OF SOLIDS 5
Question 3 FD
Assume axis to be: W P
Y- axis FA
X – axis
∑𝐹𝑦= 0
FA + FD = W+ P
FA + FD= 5000 N……………………………………………………(i)
Moment about A.
MA = 0
0 = W *AG cos 𝜃 + P* AN cos 𝜃 = FD* PD cos 𝜃
800* 0.9 + 3000 * 1.26 = FD * 1.8 FA
FD = 3100 N
From equation (i): FA = 1900 N C
Considering bar ABC:
𝛿𝐴𝐵𝐶= 𝛿𝐴𝐵+ 𝛿𝐵𝐶
𝐹𝐴∗ 𝐿𝐴𝐵
𝐴𝐴𝐵+𝐸𝐴𝐵
+ 𝐹𝐴𝐷∗ 𝐿𝐵𝐶
𝐴𝐵𝐶+𝐸𝐵𝐶
𝐸𝐴𝐵= 𝐸𝑆𝑡𝑒𝑒𝑙= 210 Gpa A
𝐸𝐵𝐶= 𝐸𝐴𝑙 = 80 Gpa
FA

MECHANICS OF SOLIDS 6
FA[ 0.2
𝜆
4∗0.0152∗210∗109 + 0.05
𝜆
4∗0.042∗80∗ 109]
FA [ 5.389 *10−9 + 0.497 ∗ 10−9 ]
𝛿𝐴𝐵𝐶= 1.1103 ∗ 10−5m deflection of ABC
Considering bar DEF
FD
F
E
D
FD
𝛿𝐷𝐸𝐹= Deflection of bar DEF
𝛿𝐷𝐸𝐹= 𝛿𝐷𝐸+ 𝛿𝐵𝐹
𝐹𝐷∗ 𝐿𝐷𝐸
𝐴𝐷𝐸+𝐸𝐷𝐸
+ 𝐹𝐷∗ 𝐿𝐸𝐹
𝐴𝐸𝐹+𝐸𝐸𝐹
𝐸𝐷𝐸= 𝐸𝑆𝑡𝑒𝑒𝑙= 210 Gpa
𝐸𝐸𝐹= 𝐸𝐴𝑙 = 80 Gpa
𝛿𝐷𝐸𝐹= FD*[[ 0.2
𝜆
4∗0.0152∗210∗109 + 0.05
𝜆
4∗0.042∗80∗ 109]
𝛿𝐷𝐸𝐹= 1.824 ∗ 10−5m

MECHANICS OF SOLIDS 7
Question 4
Strut at BC is 40mm
120 k N.mm
B C
120 k N.mm
120 mm
Gbi =120 Gpa
From CD is
30
C D
210 k N.mm
150 mm
GCD= 60 Gpa

MECHANICS OF SOLIDS 8
330 30 50
150 mm
GAB = 80 Gpa
Shear stress in BC
Τ𝐵𝐶
𝜏𝐵𝐶
= 𝜏𝐵𝐶
𝑅
120
𝜋∗404
32 ∗1612 = 𝜏𝐵𝐶
20∗10−3 𝜏𝐵𝐶= 9.6 𝑀𝑝𝑎
Shear stress in CD
Τ𝐶𝐷
𝜏𝐶𝐷
= 𝜏𝐶𝐷
𝑅
240
𝜋∗(304−204)
32 ∗16−12
= 𝜏𝐶𝐷
15∗10−3 𝜏𝐶𝐷= 56.41 𝑀𝑝𝑎
Shear stress in AB
Τ𝐴𝐵
𝜏𝐴𝐵
= 𝜏𝐴𝐵
𝑅
720
𝜋∗(504−304)
32 ∗10−12
= 𝜏𝐴𝐵
25∗10−3 𝜏𝐴𝐵= 33.703𝑀𝑝𝑎
Angle of twist
𝜃𝐷𝐴= 𝜃𝐷𝐶 + 𝜃𝐶𝐵 + 𝜃𝐵𝐴

MECHANICS OF SOLIDS 9
𝜃𝐷𝐴= Τ𝐶𝐷∗𝐿𝐶𝐷
𝐺𝐶𝐷∗𝜏𝐶𝐷
+ Τ𝐶𝐵∗𝑙𝐶𝐵
𝐺𝐶𝐵∗𝜏𝐶𝐵
+Τ𝐵𝐴∗𝐿𝐵𝐴
𝐺𝐴𝐵∗𝜏𝐴𝐵
𝜃𝐷𝐴= ( 240∗0.15
60∗ (304−204) + 120∗0.12
120∗404 + 720∗0.15
80∗(504−304)) 32
𝜋∗ 10−3
𝜃𝐷𝐴= 0.0124 𝜋𝑎𝑑
𝜃𝐷𝐴= 0.7 degrees