Answer:
See explanation for step by step procedure to get answer.
Explanation:
Given that:
The conveyor belt is moving downward at 4 m>s. If the coefficient of static friction between the conveyor and the 15-kg package B is ms = 0.8, determine the shortest time the belt can stop so that the package does not slide on the belt.
See the attachments for complete steps to get answer.
Answer: Attached below is the missing detail and Mohr's circle.
i) б1 = 9.6 Ksi
б2 = -10.7 ksi
ii) 10.2 Ksi
iii) -0.51Ksi
Explanation:
First step :
direct compressive stress on shaft
бd = P / π/4 * d^2
= -20 / 0.785 * 5^2 = -1.09 Ksi
shear stress at the outer surface due to torsion
ζ = 16*T / πd^3
= (16 * 250 ) / π * 5^3 = 010.19 Ksi
<u>Calculate the Principal stress, maximum in-plane shear stress and average normal stress</u>
Using Mohr's circle ( attached below )
<u>i) principal stresses:</u>
б1 = 4.8 cm * 2 = 9.6 Ksi
б2 = -5.35 cm * 2 = -10.7 ksi
<u>ii) maximum in-plane shear stress</u>
ζ = radius of Mohr's circle
= 5.1 cm = 10.2 Ksi ( Given that ; 1 cm = 2Ksi )
<u>iii) average normal stress</u>
= 9.6 + ( - 10.7 ) / 2
= -0.51Ksi
