Question

Problem 2. Determine the smallest allowable diameter of the shaft which is subjected to the two concentrated forces. The journal bearings at A and B only support vertical forces. The allowable bending stress is $\sigma_{allow}=150$ MPa. A 0.5 m 0.4 m 0.6 m 12 kN M(kN·m) 20 kN B 1.20 0.5 0 0.9 +X(m) 15 -6.00

Name: problem 2 determine the smallest allowable diameter of the shaft which is subjected to the two concentrated forces the journal bearings at a and b only support vertical forces the allowable 76876
Uploaded: 2022-06-13T05:00:23-08:00
Duration: 3 min 59 s
Channel: Prachita Kush
Description: problem 2 determine the smallest allowable diameter of the shaft which is subjected to the two concentrated forces the journal bearings at a and b only support vertical forces the allowable 76876

          Problem 2.
Determine the smallest allowable diameter of the
shaft which is subjected to the two concentrated
forces. The journal bearings at A and B only
support vertical forces. The allowable bending
stress is $\sigma_{allow}=150$ MPa.
A
0.5 m
0.4 m
0.6 m
12 kN
M(kN·m)
20 kN
B
1.20
0.5
0
0.9
+X(m)
15
-6.00

problem 2 determine the smallest allowable diameter of the shaft which is subjected to the two concentrated forces the journal bearings at a and b only support vertical forces the allowable 76876

Added by Marissa M.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Prachita Kush

06/13/2022

Step 1

From the bending moment diagram, the maximum absolute bending moment is $M_{max} = |-6.00 \text{ kN} \cdot \text{m}| = 6.00 \text{ kN} \cdot \text{m}$. Convert this to $\text{N} \cdot \text{mm}$: $M_{max} = 6.00 \text{ kN} \cdot \text{m} = 6.00 \times 10^3 \text{ Show more…

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Problem 2. Determine the smallest allowable diameter of the shaft which is subjected to the two concentrated forces. The journal bearings at A and B only support vertical forces. The allowable bending stress is $\sigma_{allow}=150$ MPa. A 0.5 m 0.4 m 0.6 m 12 kN M(kN·m) 20 kN B 1.20 0.5 0 0.9 +X(m) 15 -6.00