MECHANICS OF DEFORMABLE
BODIES
PROBLEM SET NO.1
Submitted by:
Submitted to:
Average Normal Stress
Problem 1.1
A shop crane consists of a boom AC that’s supported by a pin at A and by
a rectangular tension bar BD, as shown in Fig.1. Details of the pin joints at A and
B are shown in views a-a and b-b, respectively. The tension bar BD has a width
w=1.5in. and a thickness t=0.5in. If the vertical load at C is P=5kips, what is the
average tensile stress in the bar BD?
Given:
1. w=1.5in.
2. t=0.5in
3. P=5kips
Required:
1. The average tensile stress in the bar BD.
Solution:
+↑ ∑ 𝐹𝑦 = 0
𝑇𝑠𝑖𝑛𝜃 − 𝑃 = 0 eq.1
+→ ∑ 𝐹𝑥 = 0
𝑇𝑐𝑜𝑠𝜃 = 0 eq.2
LBD= √(6𝑓𝑡)2 + (4𝑓𝑡)2 = √36 + 16 = √52 = 7.21𝑓𝑡
𝜃 = tan−1 (
𝑜𝑝𝑝
4𝑓𝑡
) = 𝑡𝑎𝑛−1 (
) = 33.69°
𝑎𝑑𝑗
6𝑓𝑡
subs. eq.1:
𝑇𝑠𝑖𝑛(33.69°) − 𝑃 = 0
𝑇=
𝑃
5𝑘𝑖𝑝𝑠
=
0.555 0.555
𝑇 = 9.01𝑘𝑖𝑝𝑠
subs eq.2:
𝑇𝑐𝑜𝑠𝜃 = 0
9.01𝑘𝑖𝑝𝑠 × 0.832 = 7.49𝑘𝑖𝑝𝑠
𝑇 = 7.49𝑘𝑖𝑝𝑠
Solved Problems in Mechanics of Deformable Bodies
1
Average Normal Stress
𝑇
𝜎=𝐴
𝐴 = 𝑤𝑡 = 1.5𝑖𝑛 × 0.5𝑖𝑛 = 0.75𝑖𝑛2
𝜎=
9.01𝑘𝑖𝑝𝑠
0.75𝑖𝑛2
𝝈 = 𝟏𝟐. 𝟎𝟏𝟑𝟑𝒌𝒔𝒊
Solved Problems in Mechanics of Deformable Bodies
1
Average Normal Stress
Problem 1.2
Two solid circular rods are welded to a plate at B to form a single rod, as shown
in Fig.1. Consider the 30kN force at B to be uniformly distributed around the
circumference of the collar at B and the 10kN load at C to be applied at the centroid
of the end cross section. Determine the axial stress in each portion of the rod.
Given:
1. Pb=30kN
2. Pc=10kN
3. d₁=20mm
4. d₂=15mm
Required:
1. the axial stress in each
portion of the rod.
Solution:
𝐅
𝐀
FAB= 𝟑𝟎𝒌𝑵(Tensile) = 30,000N
FBC= 𝟏𝟎𝒌𝑵(Compressive) = 10,000N
𝛔=
For AB:
𝑨𝟏 =
𝛔=
𝝅𝑫𝟐 𝝅(𝟐𝟎𝒎𝒎)𝟐
=
= 𝟑𝟏𝟒. 𝟏𝟔𝒎𝒎𝟐
𝟒
𝟒
𝐅
𝟑𝟎, 𝟎𝟎𝟎𝐍
=
= 𝟗𝟓. 𝟒𝟗𝟐𝟕𝐌𝐏𝐚
𝐀 𝟑𝟏𝟒. 𝟏𝟔𝐦𝐦𝟐
For BC:
𝛑(𝟏𝟓𝐦𝐦)𝟐
𝟐
𝐀 =
= 𝟏𝟕𝟔. 𝟕𝟏𝐦𝐦𝟐
𝟒
𝛔=
𝟏𝟎, 𝟎𝟎𝟎𝐍
= 𝟓𝟔. 𝟓𝟖𝟗𝟗𝐌𝐏𝐚
𝟏𝟕𝟔. 𝟕𝟏𝐦𝐦𝟐
Solved Problems in Mechanics of Deformable Bodies
1
Average Normal Stress
Problem 1.3
The plate has a width of 0.5m. if the stress distribution at the support varies as
shown. Determine the force P applied to the plate and the distance d to where it is
applied.
Given:
1. w=0.5m
2. L=4m
3. δ=(15x½) MPa
4. δ=30MPa
Required:
1. The force P applied to the
plate and the distance d to
where it is applied.
Solution:
+↑ ∑ 𝐹𝑦 = 0
𝐿
𝑃 = ∫ 𝜎(𝑥) ∙ 𝑤𝑑𝑥
0
4
1
𝑃 = ∫ 15𝑥 2 × 0.5𝑑𝑥
0
4
1
𝑃 = 7.5 ∫ 𝑥 2 𝑑𝑥
0
2 3 4
2
𝑃 = 7.5 [ 𝑥 2 ] = 7.5 ( × 8) = 40𝑘𝑁
3
0
3
+↺ ∑ 𝑀𝑥 = 0
𝐿
𝑀 = ∫ 𝑥 ∙ 𝜎(𝑥) ∙ 𝑤𝑑𝑥
0
Solved Problems in Mechanics of Deformable Bodies
1
Average Normal Stress
4
3
𝑀 = 7.5 ∫ 𝑥 2 𝑑𝑥
0
2 5 4
2
𝑀 = 7.5 [ 𝑥 2 ] = 7.5 ( × 32) = 96𝑘𝑁 ∙ 𝑚
5
0
5
𝑀𝑝 = 𝑃𝑑; 𝑀𝑝 = 𝑀
𝑃𝑑 = 𝑀
40𝑘𝑁 × 𝑑 = 96𝑘𝑁 ∙ 𝑚
𝑑=
96𝑘𝑁 ∙ 𝑚
= 2.4𝑚
40𝑘𝑁
Solved Problems in Mechanics of Deformable Bodies
1
Average Normal Stress
Problem 1.4
The frame shown consists of four wooden members, ABC, DEF, BE, and CF.
Knowing that each member has a 2x4in. rectangular cross section and that each pin
has a ½in. diameter, determine the maximum value of the average normal stress (a) in
member BE, (b) in member CF.
Given:
1. W=480lb
2. d=½in
3. A=2x4in
Required:
1. The maximum
value of the
average normal
stress (a) in
member BE, (b)
in member CF.
Solution:
At Joint C:
+↑ ∑ 𝐹𝑦 = 0
𝐹 CF 𝑠𝑖𝑛𝜃CF−𝑊 = 0
𝐹 CF 𝑠𝑖𝑛𝜃CF= 480𝑙𝑏
The angle for 𝜃CF:
𝜃 = tan−1 (
15𝑖𝑛
) = 26.57°
30𝑖𝑛
Solved Problems in Mechanics of Deformable Bodies
1
Average Normal Stress
480𝑙𝑏
𝐹 CF=
sin(26.57°)
=
480
0.447
= 1074.2𝑙𝑏
+→ ∑ 𝐹𝑥 = 0
𝐹 CF 𝑐𝑜𝑠𝜃CF−𝐹 BC 𝑐𝑜𝑠𝜃BC= 0
At Joint B:
+↑ ∑ 𝐹𝑦 = 0
𝐹 BE 𝑠𝑖𝑛𝜃BE−𝐹 BC 𝑠𝑖𝑛𝜃BC= 0
45𝑖𝑛
𝜃BE= tan−1 (30𝑖𝑛) = 56.31
+→ ∑ 𝐹𝑥 = 0
𝐹 BE 𝑐𝑜𝑠𝜃BE−𝐹 AB 𝑐𝑜𝑠𝜃AB= 0
The angle for 𝜃BE:
45𝑖𝑛
𝜃BE= tan−1 (30𝑖𝑛) = 56.31
The angle for 𝜃BC:
45𝑖𝑛
𝜃BC= tan−1 (40𝑖𝑛) = 48.37°
𝐹 BE=
𝐹𝐵𝐶𝑠𝑖𝑛𝜃𝐵𝐶
𝑠𝑖𝑛𝜃𝐵𝐸
Assuming 𝐹 BC= 480𝑙𝑏
𝑭BE
𝟒𝟖𝟎×𝒔𝒊𝒏(𝟒𝟖.𝟑𝟕°)
𝝈=
𝒔𝒊𝒏(𝟓𝟔.𝟑𝟏°)
=
𝟒𝟖𝟎×𝟎.𝟕𝟒𝟖
𝟎.𝟖𝟑𝟐
= 𝟒𝟑𝟏. 𝟓𝒍𝒃
𝑭
𝑨
Solved Problems in Mechanics of Deformable Bodies
1
Average Normal Stress
𝑨 = 𝟐𝒊𝒏 × 𝟒𝒊𝒏 = 𝟖𝒊𝒏𝟐
For member BE:
𝜎=
𝐹 431.5𝑙𝑏
=
= 53.9375𝑝𝑠𝑖
𝐴
8𝑖𝑛2
For member CF:
𝜎=
𝐹 1074.2𝑙𝑏
=
= 134.275𝑝𝑠𝑖
𝐴
8𝑖𝑛2
Solved Problems in Mechanics of Deformable Bodies
1
Average Normal Stress
Problem 1.5
An aircraft tow bar is positioned by means of a single hydraulic cylinder
connected by a 25mm diameter steel rod to two identical arm and wheel units DEF. The
mass of the entire tow bar is 200 kg, and its center of gravity is located at G. For the
position shown, determine the normal stress in the rod.
Given:
1. D=25mm
2. m=200kg
Required:
1. The normal
stress in the
rod.
Solution:
𝜎=
𝐹
𝐴
𝐹 = 𝑚𝑔 = 200𝑘𝑔 ×
9.81𝑚
= 1962𝑁
𝑠2
𝐴=
𝜋𝐷2 𝜋(25𝑚𝑚)2
=
= 490.87𝑚𝑚2
4
4
𝜎=
𝐹
1962𝑁
=
𝐴 490.87𝑚𝑚2
𝝈 = 𝟒𝑴𝑷𝒂
Solved Problems in Mechanics of Deformable Bodies
1