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IRISSAK [1]
3 years ago
9

Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 280

MPa, and a factor of safety of 3.8 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P = 450 kN.
Engineering
1 answer:
stiks02 [169]3 years ago
8 0

Answer:

Minimum Allowable diameter for each of the 5 bolts = 0.0394 m = 3.94 cm

Explanation:

Maximum Working Stress = Ultimate Shear Stress/factor of safety

Maximum Working Stress = 280 MPa/3.8 = 73.68 MPa

Working Stress = Applied load/Minimum allowable Area = L/A

Minimum Allowable Area = Applied Load/Maximum Working Stress

A = 450000/73680000 = 0.00611 m²

This area is supplied by 5 bolts, so each bolt supplies A/5 = 0.0061/5 = 0.00122 m²

Cross sectional Area of bolts = πD²/4

0.00122 = πD²/4

D² = 4 × 0.00122/π = 0.00155

D = √0.00155 = 0.0394 m = 3.94 cm

Each of the five bolt can have a minimum diameter of 3.94 cm

Hope this Helps!!!

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Explanation:

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Since

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π2 = h•b^a•v^b•ρ^c.

Since

h = L, b = L, v = LT^-1 and ρ= FL^-4T²

Since π is dimensionless then,

π = F^0•L^0•T^0

(L)•(L^a)•(LT^-1)^b•(FL^-4T²)^c = F^0•L^0•T^0

Rearranging

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