Question

A high strength steel band saw of 90 mm width and 0.5 mm thickness runs over a pulley of 500 mm diameter. Assuming e

Answer 1

200 MPa is the answer

Answer 2

Given:

height = 90 mm

width = 0.5 mm

Diameter = 500

To Find:

maximum flexural stress and minimum diameter pulleys

Explanation:

Flexural stress developed:

M=EI / ρ

f=Mc /I = [(EI/ρ)c]I

f=Ec/ρ=[900000(0.5/2)]250

fb=   900 MPa

Minimum diameter of pulley:

fb=Ecρ

400=900000(0.5/2)ρ

ρ=  0.017mm

Diameter, d =0.017 mm          

Answer = 0.034 mm

Complete question is-

A high strength steel band saw of 90 mm width and 0.5 mm thickness runs over a pulley of 500 mm diameter. What maximum flexural stress is developed? What minimum diameter pulleys can be used without exceeding a flexural stress of 400 MPa? Assume E = 200 GPa.