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lilavasa [31]
3 years ago
14

Two steel plates are to be held together by means of 16-mm-diameter high-strength steel bolts fitting snugly inside cylindrical

brass spacers. Knowing that the average normal stress must not exceed 210 MPa in the bolts and 145 MPa in the spacers, determine the outer diameter of the spacers that yields the most economical and safe design.
Engineering
1 answer:
dusya [7]3 years ago
5 0

Answer:

The outer diameter of the spacers that yields the most economical and safe design is 25.03 mm

Explanation:

For steel bolt

Stress = 210 MPa or 210 N/mm2

Pressure = Stress* Area

Pbolt = 210 N/mm2 * 16^2 *(pi)/4

Pbolt = 210 N/mm2 * 200.96 mm^2 = 42201.6  N

For Brass spacer

Pressure = 42201.6  N

Area of Brass spacer = Pressure/Stress

Area of Brass spacer = 42201.6  N/145 N/mm^2 = 291.044 mm^2

Area of Brass spacer = (pi) (d^2 - 16^2)/4 =  291.044 mm^2

d^2 - 16^2 = 291.044 mm^2* 4/(pi) = 370.758

d^2 =  370.758 + 16^2

d^2 =   626.758

d = 25.03 mm

The outer diameter of the spacers that yields the most economical and safe design is 25.03 mm

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Answer:

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Explanation:

Given:

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Find the torque T_max

Solution:

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                             T = τ*pi*d^3 / 16

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                             T_c = 12*1000*pi*1^3 / 16

                             T_c = 2356.2 lb.in

- Design Torque T for Steel rod:

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                             T_s = 15*1000*pi*0.25^3 / 16

                             T_s = 46.02 lb.in

- The design torque must conform to the allowable shear stress for both copper and steel. The maximum allowable would be:

                             T = min ( 2356.2 , 46.02 )

                             T = 46.02 lb-in

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Answer:

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4 years ago
Determine the design moment strength for a W21x73 steel beam with a simple span of 18 ft when lateral bracing for the compressio
SVETLANKA909090 [29]

This question is incomplete, the complete question is;

Determine the design moment strength (ϕMn) for a W21x73 steel beam with a simple span of 18 ft when lateral bracing for the compression flange is provided at the ends only (i.e., Lb = 18 ft). Report the result in kip-ft.

Use Fy=50 ksi and assume Cb=1.0 (if needed).

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Explanation:

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section  W 21 x 73 steel beam;

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also given that; fy = 50 ksi and Cb = 1.0

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QMn = 0.9 × 50 × 151

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3 years ago
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