Strength of Materials Test 4

Result

Strength of Materials Test 4

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Rank
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Weekly Quiz Competition

Question 1
1 / -0

A cantilever beam having a rectangular cross section of width 60 mm and depth 100 mm, is made of aluminum alloy. The material mechanical properties are: Young’s modulus, E = 73 GPa and ultimate stress, σ_u = 480 MPa. Assuming a factor of safety of 4, the maximum bending moment (in kN-m) that can be applied on the beam is _____.

A
12-12

B
124-121

C
127-126

D
129-122

Solution

Permissible bending stress:
\({\sigma _b} = \frac{{{\sigma _{ut}}}}{{FOS}} = \frac{{480}}{4} = 120\;MPa\)
\(\sigma = \frac{M}{I}y \Rightarrow M = \sigma \times \frac{I}{y}\)
\(I = \frac{{b{d^3}}}{{12}}\)
\(\frac{I}{y} = \frac{{b{d^3}}}{{12}} \times \frac{2}{d} = \frac{{b{d^2}}}{6}\)
\(M = \sigma \times \frac{{b{d^2}}}{6} = \frac{{120 \times 60 \times 100 \times 100}}{6}\)
M = 12 × 10⁶ N.mm
M = 12 kN.m
Question 2
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A solid shaft is subjected to a bending moment of 400 Nm and a torque of 600 Nm. If the Permissible normal stress for the shaft material is limited to 80 MPa. What is the minimum diameter (in mm, rounded up to nearest integer) of the shaft required?

A
41-42

B
414-421

C
417-426

D
419-422

Solution

Concept:
For combined loading:
Equivalent bending moment:
\({{\rm{M}}_{\rm{E}}} = \frac{{{\rm{M}} + \sqrt {{{\rm{M}}^2} + {{\rm{T}}^2}} }}{2}\)
Equivalent torque:
\({{\rm{T}}_{\rm{E}}} = \sqrt {{{\rm{M}}^2} + {{\rm{T}}^2}} \)
Given:
M = 400 Nm, T = 600 Nm, τ_b(max) = 80 MPa
Calculation:
The shaft is under the action of combined bending and torsion.
\({{\rm{M}}_{\rm{E}}} = \frac{{{\rm{M}} + \sqrt {{{\rm{M}}^2} + {{\rm{T}}^2}} }}{2} = \frac{{400 + \sqrt {{{400}^2} + {{600}^2}} }}{2} = 560.5\;Nm\)
Form bending equation:
\(\therefore \frac{{{{\rm{f}}_{{{\rm{b}}_{{\rm{max}}}}}}}}{{\rm{R}}} = \frac{{{{\rm{M}}_{\rm{E}}}}}{{\rm{I}}}\)
\(\therefore \frac{{\frac{{80}}{{\rm{D}}}}}{2} = \frac{{560.5 \times {{10}^3}}}{{\frac{{{\rm{\pi }}{{\rm{D}}^4}}}{{64}}}}\)
∴ D = 41.48 mm ≈ 42 mm