Question

A 1.0-in.-diameter hole is drilled on the centerline of a long, flat steel bar that is 1 2 thick and 4 in. wide. The bar is subjected to a tensile load of 30,000 lb. Calculate the average stress in the plane of the reduced cross section and the maximum tensile stress immediately adjacent to the hole.

Answer 1

Answer:

The answers are

The average stress = 20000 lb/in²

The maximum tensile stress immediately adjacent to the hole

= 31076.92 lb/in²

Explanation:

To solve the question we have

Weight of tensile load = 30,000 lb

Width of steel bar = 4 in

Thickness of steel bar = 1/2 in

Average Stress = Force/Area  

Size of hole drilled = 1.0 in diameter

Available width at cross section where the 1.00 in diameter hole is drilled =

(4 - 1) in = 3 inches

Cross sectional area at the point of reduced cross section due to the drilled hole = Width × Thickness (Since the item is a flat bar)

= 3 in × 1/2 in = 1.5 in²

Therefore Stress = (30000 lb)/(1.5 in²) = 20000 lb/in²

the maximum tensile stress immediately adjacent to the hole.

Bending stress = $\sigma_B= \frac{M_y}{I}$ where $I = \frac{(0.5^2 + 4^2)}{12}$

0.5*30000/I = 11076.92 lb/in²

Max stress = 31076.92 lb/in²

Answer 2

Answer:

A: The average stress of in the plane of reduced cross section is 20000 PSI.

B:  The maximum tensile stress is immediately adjacent to the hole and its value is : 47000 PSI.

Explanation:

See attachment for detailed answer step wise.

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