CHAPTER 1
(SIMPLE STRESSES)
, NORMAL STRESS
Problem 104
1. A hollow steel tube with an inside diameter of 100 mm must carry a tensile load
of 400 kN. Determine the outside diameter of the tube if the stress is limited to
120 MN/m.
SOLUTION:
𝑃 = 𝜎𝐴
𝐺𝑖𝑣𝑒𝑛:
1000 𝑁
𝑃 = 400 𝑘𝑁 ( ) = 400 000 𝑁
1 𝑘𝑁
𝜎 = 120𝑀𝑃𝑎 (𝑁𝑜𝑡𝑒: 1 𝑀𝑁/𝑚 = 1 𝑀𝑃𝑎)
1
𝐴= 𝜋 (𝐷 2 − (100𝑚𝑚)2 )
4
𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛:
𝑃 = 𝜎𝐴
1
400 000 𝑁 = 120𝑀𝑃𝑎 𝜋 (𝐷 2 − (100𝑚𝑚)2 )
4
400 000 𝑁 = 30 𝑀𝑃𝑎 𝜋 (𝐷 2 − (10 000 𝑚𝑚 2 )
400 000 𝑁 = 30 𝑀𝑃𝑎 𝜋 𝐷 2 − 300 000 𝑚𝑚 2 𝜋 )
400 000 𝑁 + 300 000 𝑀𝑃𝑎 𝑚𝑚 2 𝜋
𝐷2 =
30 𝑀𝑃𝑎 𝜋
2
√𝐷 = √14244. 131815 𝑚𝑚 2 (𝑆𝑞𝑢𝑎𝑟𝑒 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠 𝑡𝑜 𝑔𝑒𝑡 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝐷)
𝑫 = 𝟏𝟏𝟗. 𝟑𝟓 𝒎𝒎𝟐
Problem 106
2. The homogeneous bar shown in Fig. P-106 is supported by a smooth pin at C
and a cable that runs from A to B around the smooth peg at D. Find the stress in
the cable if its diameter is 0.6 inch and the bar weighs 6000 lb.
SOLUTION:
𝛴 𝑀ʗ = 0
3
5𝑓𝑡 𝑇 + 10𝑓𝑡 ( 𝑇) = 5𝑓𝑡 (6000𝑙𝑏)
√34
10.14496𝑇 = 30 000𝑓𝑡 • 𝑙𝑏
𝑇 = 2957.13𝑙𝑏
,𝐺𝑖𝑣𝑒𝑛:
𝑇 = 2957.13 𝑙𝑏
𝐷 = 0.6 𝑖𝑛
𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑: 𝜎 =?
𝑇 = 𝜎𝐴
1
2957.13 𝑙𝑏 = 𝜎 𝜋 (𝐷2 )
4
1
2957.13 𝑙𝑏 = 𝜎 𝜋 (0.6 𝑖𝑛2 )
4
𝜎 = 10 522.38 𝑙𝑏/ 𝑖𝑛2
Thus, 10 522.38 lb/ 𝒊𝒏𝟐 = 10 522.38 psi
Problem 107
3. A rod is composed of an aluminum section rigidly attached between steel and
bronze sections, as shown in Fig. P-107. Axial loads are applied at the positions
indicated. If P = 3000 lb and the cross-sectional area of the rod is 0.5 𝑖𝑛2 ,
determine the stress in each section.
SOLUTION:
Given: STEEL: ALUMINUM: BROZE:
P = 3000 lb 𝛔_𝑠𝑡 𝐴_𝑠𝑡 = 𝑃𝑠𝑡
P = 3000 lb 𝛔𝑠𝑡 (𝐀𝑠𝑡 ) = 𝐏𝑠𝑡 𝛔𝑎𝑙 (𝐀𝑎𝑙 ) = 𝐏𝑎𝑙 𝛔𝑏𝑟 (𝐀𝑏𝑟 ) = 𝐏𝑏𝑟
𝐴𝑠𝑡 = 0.5 𝑖𝑛2 𝛔𝑠𝑡 (0.5 𝑖𝑛2 ) = 12 𝛔𝑎𝑙 (0.5 𝑖𝑛 2 ) = 12 𝛔𝑏𝑟 (0.5 𝑖𝑛 2 ) = 12
𝐴𝑏𝑟 = 0.5 𝑖𝑛2 𝛔𝒔𝒕 = 𝟐𝟒 𝒌𝒔𝒊 𝛔𝒂𝒍 = 𝟒 𝒌𝒔𝒊 𝛔𝒃𝒓 = 𝟏𝟖 𝒌𝒔𝒊
, SHEARING STRESS
Problem 115
1. What force is required to punch a 20-mm-diameter hole in a plate that is 25 mm
thick? The shear strength is 350 MN/𝑚 2 .
SOLUTION:
𝑮𝒊𝒗𝒆𝒏:
Required diameter of hole = 20
mm Thickness of plate = 25 mm
Shear strength of plate = 350 MN/m2
𝑹𝒆𝒒𝒖𝒊𝒓𝒆𝒅: 𝐹𝑜𝑟𝑐𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑡𝑜 𝑝𝑢𝑛𝑐ℎ 𝑎 20 − 𝑚𝑚 − 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 ℎ𝑜𝑙𝑒
V = τA = P
P = 350 N/𝑚 2 [ π (20 mm) (25 mm)]
P = 549 778.7 N (Covert to kN where; 1 N = 0.001 kN)
𝐏 = 𝟓𝟒𝟗. 𝟖 𝐤𝐍
Problem 117
2. Find the smallest diameter bolt that can be used in the clevis shown in Fig. 1-11b
if P = 400 kN. The shearing strength of the bolt is 300 MPa.
SOLUTION:
Given:
Force P = 400 kN
Shear strength of the bolt = 300 MPa
𝑉 = 𝜏𝐴 = 𝑃
1
400 kN (1000) = 300 MPa [ 2 ( 𝜋 𝑑2 )]
4
𝒅 = 𝟐𝟗. 𝟏𝟑 𝒎𝒎
(SIMPLE STRESSES)
, NORMAL STRESS
Problem 104
1. A hollow steel tube with an inside diameter of 100 mm must carry a tensile load
of 400 kN. Determine the outside diameter of the tube if the stress is limited to
120 MN/m.
SOLUTION:
𝑃 = 𝜎𝐴
𝐺𝑖𝑣𝑒𝑛:
1000 𝑁
𝑃 = 400 𝑘𝑁 ( ) = 400 000 𝑁
1 𝑘𝑁
𝜎 = 120𝑀𝑃𝑎 (𝑁𝑜𝑡𝑒: 1 𝑀𝑁/𝑚 = 1 𝑀𝑃𝑎)
1
𝐴= 𝜋 (𝐷 2 − (100𝑚𝑚)2 )
4
𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛:
𝑃 = 𝜎𝐴
1
400 000 𝑁 = 120𝑀𝑃𝑎 𝜋 (𝐷 2 − (100𝑚𝑚)2 )
4
400 000 𝑁 = 30 𝑀𝑃𝑎 𝜋 (𝐷 2 − (10 000 𝑚𝑚 2 )
400 000 𝑁 = 30 𝑀𝑃𝑎 𝜋 𝐷 2 − 300 000 𝑚𝑚 2 𝜋 )
400 000 𝑁 + 300 000 𝑀𝑃𝑎 𝑚𝑚 2 𝜋
𝐷2 =
30 𝑀𝑃𝑎 𝜋
2
√𝐷 = √14244. 131815 𝑚𝑚 2 (𝑆𝑞𝑢𝑎𝑟𝑒 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠 𝑡𝑜 𝑔𝑒𝑡 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝐷)
𝑫 = 𝟏𝟏𝟗. 𝟑𝟓 𝒎𝒎𝟐
Problem 106
2. The homogeneous bar shown in Fig. P-106 is supported by a smooth pin at C
and a cable that runs from A to B around the smooth peg at D. Find the stress in
the cable if its diameter is 0.6 inch and the bar weighs 6000 lb.
SOLUTION:
𝛴 𝑀ʗ = 0
3
5𝑓𝑡 𝑇 + 10𝑓𝑡 ( 𝑇) = 5𝑓𝑡 (6000𝑙𝑏)
√34
10.14496𝑇 = 30 000𝑓𝑡 • 𝑙𝑏
𝑇 = 2957.13𝑙𝑏
,𝐺𝑖𝑣𝑒𝑛:
𝑇 = 2957.13 𝑙𝑏
𝐷 = 0.6 𝑖𝑛
𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑: 𝜎 =?
𝑇 = 𝜎𝐴
1
2957.13 𝑙𝑏 = 𝜎 𝜋 (𝐷2 )
4
1
2957.13 𝑙𝑏 = 𝜎 𝜋 (0.6 𝑖𝑛2 )
4
𝜎 = 10 522.38 𝑙𝑏/ 𝑖𝑛2
Thus, 10 522.38 lb/ 𝒊𝒏𝟐 = 10 522.38 psi
Problem 107
3. A rod is composed of an aluminum section rigidly attached between steel and
bronze sections, as shown in Fig. P-107. Axial loads are applied at the positions
indicated. If P = 3000 lb and the cross-sectional area of the rod is 0.5 𝑖𝑛2 ,
determine the stress in each section.
SOLUTION:
Given: STEEL: ALUMINUM: BROZE:
P = 3000 lb 𝛔_𝑠𝑡 𝐴_𝑠𝑡 = 𝑃𝑠𝑡
P = 3000 lb 𝛔𝑠𝑡 (𝐀𝑠𝑡 ) = 𝐏𝑠𝑡 𝛔𝑎𝑙 (𝐀𝑎𝑙 ) = 𝐏𝑎𝑙 𝛔𝑏𝑟 (𝐀𝑏𝑟 ) = 𝐏𝑏𝑟
𝐴𝑠𝑡 = 0.5 𝑖𝑛2 𝛔𝑠𝑡 (0.5 𝑖𝑛2 ) = 12 𝛔𝑎𝑙 (0.5 𝑖𝑛 2 ) = 12 𝛔𝑏𝑟 (0.5 𝑖𝑛 2 ) = 12
𝐴𝑏𝑟 = 0.5 𝑖𝑛2 𝛔𝒔𝒕 = 𝟐𝟒 𝒌𝒔𝒊 𝛔𝒂𝒍 = 𝟒 𝒌𝒔𝒊 𝛔𝒃𝒓 = 𝟏𝟖 𝒌𝒔𝒊
, SHEARING STRESS
Problem 115
1. What force is required to punch a 20-mm-diameter hole in a plate that is 25 mm
thick? The shear strength is 350 MN/𝑚 2 .
SOLUTION:
𝑮𝒊𝒗𝒆𝒏:
Required diameter of hole = 20
mm Thickness of plate = 25 mm
Shear strength of plate = 350 MN/m2
𝑹𝒆𝒒𝒖𝒊𝒓𝒆𝒅: 𝐹𝑜𝑟𝑐𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑡𝑜 𝑝𝑢𝑛𝑐ℎ 𝑎 20 − 𝑚𝑚 − 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 ℎ𝑜𝑙𝑒
V = τA = P
P = 350 N/𝑚 2 [ π (20 mm) (25 mm)]
P = 549 778.7 N (Covert to kN where; 1 N = 0.001 kN)
𝐏 = 𝟓𝟒𝟗. 𝟖 𝐤𝐍
Problem 117
2. Find the smallest diameter bolt that can be used in the clevis shown in Fig. 1-11b
if P = 400 kN. The shearing strength of the bolt is 300 MPa.
SOLUTION:
Given:
Force P = 400 kN
Shear strength of the bolt = 300 MPa
𝑉 = 𝜏𝐴 = 𝑃
1
400 kN (1000) = 300 MPa [ 2 ( 𝜋 𝑑2 )]
4
𝒅 = 𝟐𝟗. 𝟏𝟑 𝒎𝒎
