s i s y l a n A s s e r t S : 2 r e t p a h C
n t a a r g s n o i r h a K S n i i n B B i i d r u m S a d H h . o M M
4 1 0 2
Mechanical engineering department Politeknik Merlimau
[email protected],
[email protected]
This topic explains the principles of stress, deformation analysis for several kinds of stresses such as direct stress, shear stress, torsional shear stress and formulate torque, power and rotational speed.
JJ513 Engineering Design
Stress Analysis
THE BIG PICTURE
2-1
2-2
2-3
2-4
• STRESS
• STRAIN
• SHEAR STRESS • RELATIONSHIP AMONG TORQUE, POWER AND TRANSFER ROTATIONAL SPEED
OBJECTIVES OF THIS CHAPTER
After completing this chapter, you will be able to: 1. Describe the direct stresses 2. Explain the deformation under direct axial loading 3. Describe the shear stress
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JJ513 Engineering Design
Stress Analysis
INTRODUCTION
In our daily life, we see that when we pull the rubber band, it elongates. The amount, by which the rubber band elongates, depends upon the amount of load and the nature as well as cross sectional area of the rubber band material. It has been experimentally found that the cohesive force, between molecules of rubber band, offers resistance against the deformation, and the force of resistance increase with the deformation. It has also been observed that the process of deformation stops when the force of resistance is equal to the external force. In the succeeding pages, we shall discuss the effects produced by the application of loads, on the materials.
2-1
STRESS
Every material is elastic in nature. This is why, whenever some external system of forces acts on a body, it undergoes some deformation. As the body undergoes deformation, its molecules set up some resistance to deformation. This resistance per unit area to deformation is known as stress. Mathematically stress may be defined as the force per unit area.
P A
Where, P = load or force acting on the body A = cross – sectional area of the body 2 Unit = N / m
2-2
STRAIN
Whenever a single force acts on a body, it undergoes some deformation. This deformation per unit length is known as strain. Mathematically strain may be defined as the deformation per unit length.
l
l
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JJ513 Engineering Design
Stress Analysis
Where, = deformation of the body = original length of the body
2-3
SHEAR STRESS
When a section is subjected to two equal and opposite forces, acting tangentially across the resisting section, as a result of which the body tends to shear off across the section as shown in figure 2.1 below, the stress induced is called shear stress.
Before the force is subject
P P After the force is subjected Figure 2.1 shearing stress
2-4
RELATIONSHIP AMONG TORQUE, POWER AND POWER TRANSFER ROTATIONAL SPEED
Power is the rate of doing work; Where; P = force, N U = velocity, m/s T = torque, Nm = rotational speed, rads
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JJ513 Engineering Design
2-5
Stress Analysis
SOLVED PROBLEM
PROBLEM 1
A steel rod 1m long & 20mm x 20mm in cross-section is subjected to tensile force of 40KN. Determine the elongation of the rod, if modulus of elasticity for the rod material is 200 Gpa. Given: = 1 m -6
A = 20m x 20m = 400 x 10 m² P= 40 KN = 40 x 103 N E= 200 Gpa = 200 x 10⁹ N/m²
Solution: l
Pl AE
=
(40 103 )(1) (400 10 6 )(200 10 9 )
= 0.5 mm.
PROBLEM 2
A hollow cylinder 2m long has an outside diameter of 50mm & inside diameter of 30mm. If the cylinder is carrying a load of 25 KN, find the stress in the cylinder. Also find the deformation of the cylinder , if the value of the mudulus of elasticityfor the cylinder materials is 100 Gpa. Given: l
= 2m
D = 50mm D = 30mm P = 25 KN
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JJ513 Engineering Design
Stress Analysis
Solution:
A =
4
(50 103 ) 2 (30 103 ) 2 1.257mm 2 4
( D 2 d 2 )
P
=
25 103
2
A 1.257 10
p AE
6
= 19.9 MN/m .
(25 103 )(2) 1.257 10 6 (100 109 )
= 0.4mm PROBLEM 3
A load of 5 KN is to be raised with the help of a steel wire. Find the minimum diameter of the steel wire if the stress is not exceeding 100 MPa. Given: P = 5 kN, 100 MN / m
2
.
Solution:
P d 2
P
d
4 4 P
4 P 2
d
d
A
2
4 P
4(5 103) (100 10
6
)
7.99mm @8mm
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JJ513 Engineering Design
Stress Analysis
PROBLEM 4
In an experiment, a steel specimen at 13mm diameter was found to elongate 0.2mm in a 200mm gauge length when it was subjected to a tensile force of 26.8kN. If the specimen was tested within the elastic range, what is the value of young’s modulus? Given: d = 13mm l = 0.2mm
l 200mm P = 26.8kN
Solution: A l
E
d2
4 pl
4
(13 103 )2 1.327 104 m2
AE pl A l
26.8 103 200 103 1.327 104 0.2 103
202
GN m2
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JJ513 Engineering Design
Stress Analysis
PROBLEM 5
A hollow steel tube 3.5m long has external diameter of 120mm. In order to determine the internal diameter, the tube was subjected to a tensile load of 400kN and extension was measured to be 2mm. If the modulus of elasticity for the tube material is 200Gpa, determine the internal diameter of the tube. Given:
3.5m D 120 103 m P 400 103 N
2 103 m
E 200GPa 200 109
N m2
Solution:
P AE P
(120 103 ) 2 d 2 E 4
4 P (120 10
3 2
) d 2 E
(120 103 ) 2 d 2 (120 103 ) 2
4 P (
4 P (
) E
d (120 103 ) 2 3 2
d (120 10 )
) E
d 2
4 P (
) E
4(400 103 )3.5 (2 10
3
)(200 109 )
d 99.72mm
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JJ513 Engineering Design
Stress Analysis
REFERENCES
Bernard J.Hamrock, S. R. (2005). fundamentals of machine elements 2nd edition. kuala lumpur: The McGraw Hill Companies. Khurmi, R. S. (2007). Strength of Materials. New Delhi: S. Chand.
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