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3 years ago

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A small pad subjected to a shearing force is deformed at the top of the pad 0.08 in. The height of the pad is 1.38 in. What is t

he shearing strain (rad) on the pad?

1 answer:

Aleksandr-060686 [28]3 years ago

7 0

Answer:

The shear strain is 0.05797 rad.

Explanation:

Shear strain is the ratio of change in dimension along the shearing load direction to the height of the plate under application of shear load. Width of the plate remains same. Length of the plate slides under shear load.

Step1

Given:

Height of the pad is 1.38 in.

Deformation at the top of the pad is 0.08 in.

Calculation:

Step2

Shear strain is calculated as follows:

$tan\phi=\frac{\bigtriangleup l}{h}$

$tan\phi=\frac{0.08}{1.38}$

$tan\phi= 0.05797$

For small angle of $\phi$ , $tan\phi$ can take as $\phi$ .

$\phi = 0.05797 rad.$

Thus, the shear strain is 0.05797 rad.

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A ductile hot-rolled steel bar has a minimum yield strength in tension and compression of 350 MPa. Using the distortion-energy a

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Answer:

Explanation:

From the given question:

Using the distortion energy theory to determine the factors of safety FOS can be expressed by the relation:

$\dfrac{Syt}{FOS}= \sqrt{ \sigma x^2+\sigma y^2-\sigma x \sigma y+3 \tau_{xy^2}}$

where; syt = strength in tension and compression = 350 MPa

The maximum shear stress theory can be expressed as:

$\tau_{max} = \dfrac{Syt}{2FOS}$

where;

$\tau_{max} =\sqrt{ (\dfrac{\sigma x-\sigma y}{2})^2+ \tau _{xy^2$

a. Using distortion - energy theory formula:

$\dfrac{350}{FOS}= \sqrt{94^2+0^2-94*0+3 (-75)^2}}$

$\dfrac{350}{FOS}=160.35$

${FOS}=\dfrac{350}{160.35}$

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USing the maximum-shear stress theory;

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${ FOS} =\dfrac{350}{2 \times 88.51}$

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b. σx = 110 MPa, σy = 100 MPa

Using distortion - energy theory formula:

$\dfrac{350}{FOS}= \sqrt{ 110^2+100^2-110*100+3(0)^2}$

$\dfrac{350}{FOS}= \sqrt{ 12100+10000-11000$

$\dfrac{350}{FOS}=105.3565$

$FOS=\dfrac{350}{105.3565}$

FOS =3.322

USing the maximum-shear stress theory;

$\dfrac{350}{2 FOS} =\sqrt{ (\dfrac{110-100}{2})^2+ (0)^2$

$\dfrac{350}{2 FOS} ={ (\dfrac{110-100}{2})^2$

$\dfrac{350}{2 FOS} =25$

FOS = 350/2×25

FOS = 350/50

FOS = 70

c. σx = 90 MPa, σy = 20 MPa, τxy =−20 MPa

Using distortion- energy theory formula:

$\dfrac{350}{FOS}= \sqrt{ 90^2+20^2-90*20+3(-20)^2}$

$\dfrac{350}{FOS}= \sqrt{ 8100+400-1800+1200}$

$\dfrac{350}{FOS}= 88.88$

FOS = 350/88.88

FOS = 3.939

USing the maximum-shear stress theory;

$\dfrac{350}{2 FOS} =\sqrt{ (\dfrac{90-20}{2})^2+ (-20)^2$

$\dfrac{350}{2 FOS} =\sqrt{ (35)^2+ (-20)^2$

$\dfrac{350}{2 FOS} =\sqrt{ 1225+ 400$

$\dfrac{350}{2 FOS} =40.31$

$FOS} =\dfrac{350}{2*40.31}$

FOS = 4.341

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3 years ago