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A steel plate is to be attached to a support with three bolts. The plate is 50 mm wide and 11.2 mm thick and the yield strength of the steel is 250 MPa. The ultimate bearing strength of the bolts is 540 MPa. A factor of safety of 1.67 with respect to yield is required for the plate. A factor of safety of 4.0 with respect to the ultimate bearing strength is required for the bolts. Determine the minimum bolt diameter required to develop the full strength of the plate. (Note: Consider only the gross cross-sectional area of the plate)

Respuesta :

Answer:

d = 16.20 mm

Explanation:

Given data:

palte dimension = 50mm wide and 11.2 mm thick

yield strength of steel 250 MPa = 250 N?mm^2

ultimate bearing strength of bolt = 540 MPa

factor of safety with respect to yield for plate is 1.67

factor of safety with respect to yield for bolts is 4.0

cross sectional area = 50* 11.2 = 560 mm^2

plate strength [tex]= \frac{560 \times 250}{1.67} = 83832 N[/tex]

Number of bolts required is 3

shear force exerted by single bolt [tex]= \frac{83832}{3} = 27844 N[/tex]

Allowable shear stress [tex]= \frac{540}{4} = 135 N/mm^2[/tex]

we know that

stress = force/area

[tex]135 \times \frac{\pi\times d^2}{4} = 27844 [/tex]

solving for d

d = 16.20 mm

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