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 = [tex]\sigma_B= \frac{M_y}{I}[/tex] where [tex]I = \frac{(0.5^2 + 4^2)}{12}[/tex]
0.5*30000/I = 11076.92 lb/in²
Max stress = 31076.92 lb/in²
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.
