Question:medium
A steel girder of length 10 m has one end rigidly attached to a ship's bulkhead and the other end free. It is subjected to a torsional moment of 30 kNm about its longitudinal axis at the free end. The shear modulus of steel is 80 GPa and the polar moment of inertia of the section is \(6 \times 10^{-4}\) m\(^4\).
The total twist at the free end is ______ \(\times 10^{-4}\) rad (rounded off to one decimal place).
A steel girder of length 10 m has one end rigidly attached to a ship's bulkhead and the other end free. It is subjected to a torsional moment of 30 kNm about its longitudinal axis at the free end. The shear modulus of steel is 80 GPa and the polar moment of inertia of the section is \(6 \times 10^{-4}\) m\(^4\).
The total twist at the free end is ______ \(\times 10^{-4}\) rad (rounded off to one decimal place).
The total twist at the free end is ______ \(\times 10^{-4}\) rad (rounded off to one decimal place).
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Use the torsion equation theta equals T L over G J with the full length carrying the applied torque.
Updated On: Jul 28, 2026
Show Solution
Correct Answer: 62.5
Solution and Explanation
Step 1: Compute the torsional stiffness GJ first.
$GJ = 80 \times 10^9 \times 6 \times 10^{-4} = 4.8 \times 10^7$ N.m$^2$. This single number tells us how much torque is needed per unit twist per unit length.
Step 2: Divide the torque-length product by this stiffness.
A cantilevered girder with torque applied only at the free end carries a constant internal torque of $T = 30000$ Nm along its whole $L = 10$ m span, so the total twist is $\theta = TL/(GJ) = (30000 \times 10) / (4.8 \times 10^7)$.
Step 3: Carry out the division.
$\theta = 300000 / (4.8 \times 10^7) = 0.00625$ rad. Writing this in the requested form, $0.00625 = 62.5 \times 10^{-4}$ rad.
Final Answer:
The free end twists by $62.5 \times 10^{-4}$ rad, right in the middle of the 62.4 to 62.6 key band. \[ \boxed{\theta \approx 62.5 \times 10^{-4} \text{ rad}} \]
$GJ = 80 \times 10^9 \times 6 \times 10^{-4} = 4.8 \times 10^7$ N.m$^2$. This single number tells us how much torque is needed per unit twist per unit length.
Step 2: Divide the torque-length product by this stiffness.
A cantilevered girder with torque applied only at the free end carries a constant internal torque of $T = 30000$ Nm along its whole $L = 10$ m span, so the total twist is $\theta = TL/(GJ) = (30000 \times 10) / (4.8 \times 10^7)$.
Step 3: Carry out the division.
$\theta = 300000 / (4.8 \times 10^7) = 0.00625$ rad. Writing this in the requested form, $0.00625 = 62.5 \times 10^{-4}$ rad.
Final Answer:
The free end twists by $62.5 \times 10^{-4}$ rad, right in the middle of the 62.4 to 62.6 key band. \[ \boxed{\theta \approx 62.5 \times 10^{-4} \text{ rad}} \]
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