A solid steel shaft ABCDE turns freely in bearings at points A and E. The shaft is driven by the gear at C, which applies a torq
1 answer:
Answer:
Hello your question is poorly written below is the well written question and the diagram attached to your question is attached below as w ell
A solid steel shaft ABCDE turns freely in bearings at points A and E. The shaft is driven by the gear at C, which applies a torque T2 325 lb-ft. Gears at B and D are driven by the shaft and have resisting torques T1 200 lb-ft and T3 125 lb-ft, respectively. Segments BC and CD have lengths LBC 20 in. and LCD 15 in. and the shear modulus G 11,600 ksi. Determine the minimum required diameter (d) of the shaft if the allowable shear stress t a 56 ksi. Also calculate the angle of twist between gears B and D.
answer :
a) d ≈ 12.7 inches
b) ∅bc = 1.62 * 10^-3 rad
∅cd = 7.59 * 10^-4 rad
Explanation:
Given data :
G = 11600 ksi
Ta11 = 6 ksi
<u>a) determine the minimum required diameter </u>
Apply the relation below
Toc / Ip = Ta11 / R = G*∅ / L
∴ Ip = Tbc * R / Ta11
π / 32 * (d)^4 = [ ( 200 * 12 ) * ( d/2 )] / 6
therefore : d = 12.7 inches
<u>b) determine angle of twist between gears B and D </u>
attached below is the detailed solution
) ∅bc = 1.62 * 10^-3 rad
∅cd = 7.59 * 10^-4 rad
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Answer:
shear strength = 2682.31 Ib/ft^2
Explanation:
major principal stress = 100 Ib / in2
minor principal stress = 20 Ib/in2
Normal stress = 3000 Ib/ft2
<u>Determine the shear strength when direct shear test is performed </u>
To resolve this we will apply the coulomb failure criteria relationship between major and minor principal stress a
for direct shear test
use Mohr Coulomb criteria relation between normal stress and shear stress
Shear strength when normal strength is 3000 Ib/ft = 2682.31 Ib/ft^2
attached below is the detailed solution
