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3 years ago

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A solid copper bar of circular cross section has length L 5 1.25 m and shear modulus of elasticity G 5 45 GPa. The bar is design

ed to carry a 250 N m? torque acting at the ends. If the allowable shear stress is 30 MPa and the allowable angle of twist between the ends is 2.58, what is the minimum required diameter?

1 answer:

Harman [31]3 years ago

3 0

Answer:

See explanations

Explanation:

solid copper bar of circular cross section has length L 5 1.25 m and shear modulus of elasticity G 5 45 GPa. The bar is designed to carry a 250 N m? torque acting at the ends. If the allowable shear stress is 30 MPa and the allowable angle of twist between the ends is 2.58

Answers: dmin = 35.7mm

7 7 L=1.25 m

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We can find the change in the enthalpy through the tables A5 for Saturated water, pressure table.

For 1bar=1000kPa:

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$Q=21.4378*2666.17$

$Q=57157.036kW$

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A HSS152.4 × 101.6 × 6.4 structural steel [E = 200 GPa] section (see Appendix B for crosssectional properties) is used as a colu

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Answer:

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Explanation:

Solution

Given

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The elasticity modules, E = 200 GPa

The factor of safety with respect to failing buckling . F.S =2

Geometric properties of structural steel shapes for, A HSS152.4 × 101.6 × 6.4

the moment of inertial about y axis Iy =4 .62 * 10^ 6 mm ^4

For

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The effective length factor, K =1

The critical buckling load , Pcr = π²EI/(KL)²

Pcr = π² * 200 * 10 ^3 * 4.62 * 10 ^6/( 1* 6* 10 ^3)

= 253319.85N

= 253.32N

The maximum safe load , Pallow = 253.32 /2 = 126.66kN

hence, the maximum safe for the column is 126.66kN

(b)If the end condition is fixed free-free

the effective length factor, K= 2

The critical buckling load , Pcr = π²EI/(KL)²

Pcr = π² * 200 * 10 ^3 * 4.62 * 10 ^6/( 2 * 6 * 10 ^3)²

= 63329.96N

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The maximum safe load, P allow = Pcr/F.S

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(c) If the end condition is fixed- pinned

The effective length factor K =0.7

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The effective factor, K =0.5

The critical buckling load , Pcr = π²EI/(KL)²

Pcr = π² * 200 * 10 ^3 * 4.62 * 10 ^6/( 0.5 * 6 * 10 ^3)²

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P allow = 506.64 kN

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Answer:

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Ec(l) = (EmEp)/(VmEp + VpEm)

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Ec(l) = (68 x 380)/(0.37 x 380) + (0.63 x 68)

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