Mechanics of Materials Lecture 22 PDF

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Engr. Danish Saeed Mechanics of Materials[Fall 2019]

1/6/2020

Mechanics of Materials CIEN 2101 Lecture # 22

By Engr. Danish Saeed Lecturer, CED, KFUE&IT, RYK 1

Torsion • Torsion is the twisting of an object due to an applied torque.

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Khwaja Fareed University of Engineering & IT, RYK

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Engr. Danish Saeed Mechanics of Materials[Fall 2019]

1/6/2020

Torsional Deformation of a Circular Shaft • Torque tends to twist a member about its longitudinal axis. • Consider a shaft made of highly deformable material such as rubber as shown in figure in next slide. • When the torque is applied, the circles and longitudinal grid lines originally marked on the shaft tend to distort into the pattern shown in Figure. 3

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Khwaja Fareed University of Engineering & IT, RYK

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Engr. Danish Saeed Mechanics of Materials[Fall 2019]

1/6/2020

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Angle of twist • If the shaft is fixed at one end and a torque is applied to its other end, the dark green shaded plane in Figure on next will distort into a skewed form as shown in figure • Here a radial line located on the cross section at a distance x from the fixed end of the shaft will rotate through an angle Φ(x). • The angle Φ(x), so defined, is called the angle of twist. • It depends on the position x and will vary along the shaft as shown. 6

Khwaja Fareed University of Engineering & IT, RYK

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Engr. Danish Saeed Mechanics of Materials[Fall 2019]

1/6/2020

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Angle of twist

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Khwaja Fareed University of Engineering & IT, RYK

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Engr. Danish Saeed Mechanics of Materials[Fall 2019]

1/6/2020

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• Consider a small element located at a radial distance ρ (rho) from the axis of the shaft as shown in figure. • Due to the deformation, the front and rear faces of the element will undergo a rotation. • Back face will rotate by Φ(x), and the front face by Φ(x) +ΔΦ. • As a result, the difference in these rotations, ΔΦ, causes the element to be subjected to a shear strain 10

Khwaja Fareed University of Engineering & IT, RYK

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Engr. Danish Saeed Mechanics of Materials[Fall 2019]

1/6/2020

ρ c

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Khwaja Fareed University of Engineering & IT, RYK

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Engr. Danish Saeed Mechanics of Materials[Fall 2019]

1/6/2020

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• Before deformation the angle between the edges AB and AC is 90°. • After deformation, however, the edges of the element are AD and AC and the angle between them is θ′. • From the definition of shear strain, we have

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Khwaja Fareed University of Engineering & IT, RYK

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Engr. Danish Saeed Mechanics of Materials[Fall 2019]

1/6/2020

Cont… • Considering the length of arc BD, we have ρ

∆Φ ∆

= 

• The above equation shows that shear strain varies linearly with radial distance ρ because

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• In other words, the shear strain within the shaft varies linearly along any radial line, from zero at the axis of the shaft to a maximum γmax at its outer boundary. •

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Khwaja Fareed University of Engineering & IT, RYK

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Engr. Danish Saeed Mechanics of Materials[Fall 2019]

1/6/2020

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Shear Modulus • Shear Modulus (Modulus of Rigidity) is the elasticity coefficient for shearing or torsion. • It can be defined as “The ratio of shear stress to shear strain”. • According to hooks law • Shear stress = Shear modulus × Shear strain • τ=G×γ

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Khwaja Fareed University of Engineering & IT, RYK

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Engr. Danish Saeed Mechanics of Materials[Fall 2019]

1/6/2020

The Torsion Formula • When an external torque is applied to a shaft, it creates a corresponding internal torque within the shaft. • If the material is linear-elastic, then Hooke’s law applies, τ = G × γ • Linear variation in shear strain leads to a corresponding linear variation in shear stress along any radial line on the cross section.

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Cont…

• Hence, τ will vary from zero at the shaft’s longitudinal axis to a maximum value, τmax, at its outer surface. • This variation is shown in Figure on next slide on the front faces of a selected number of elements, located at an intermediate radial position ρ and at the outer radius c. • Due to the proportionality, we can write.

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Engr. Danish Saeed Mechanics of Materials[Fall 2019]

1/6/2020

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• Shear force at each element of area is given as • dF = τ × dA • The torque produced by this force is • dT = (τ × dA)× ρ • And for whole cross section, we have

• ∫A dT= • For equilibrium of shaft, this torque must be equal to internal resultant torque T. 22

Khwaja Fareed University of Engineering & IT, RYK

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Engr. Danish Saeed Mechanics of Materials[Fall 2019]

1/6/2020

• So we have

• The integral ∫   represents the polar moment of inertia and denoted by “J”. • So the above equation may be written as: •

 =

 

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Cont…

• Here •  = the maximum shear stress in the shaft, which occurs at the outer surface. • T = the resultant internal torque acting at the cross section. Its value is determined from the method of sections. • J = the polar moment of inertia of the crosssectional area c = the outer radius of the shaft.

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Khwaja Fareed University of Engineering & IT, RYK

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Engr. Danish Saeed Mechanics of Materials[Fall 2019]

1/6/2020

Polar moment of inertia • Polar moment of inertia “J” is the moment of inertia of a cross section about an axis that is perpendicular to plane of area. •  = ∫   ×  •  = ∫   ×  •  = ∫  × 

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Cont….

• Polar moment of inertia of a cross section describes its ability to resist torsion. Polar moment of inertia of solid shaft:

Where c is the radius of shaft Polar moment of inertia of tubular shaft:

Where co is outer radius and ci is inner radius 26

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