composite shaft AC is made by shrink-fitting a titanium-alloy sleeve of length L3 = 4 ft over segment BC of a 6-ft-long aluminum-alloy rod. This makes a three-component shaft, which is fixed against rotation at ends A and C. The shear moduli are G1 = G2 = 6 X 103 ksi and G3 = 4 X 103 ksi, respectively, and the diameters and d1 = d2 = (di)3 = 2 in., and (do)3 = 2.5 in. (a) Using the Displacement Method, determine the angle of ro- tation at B, og. produced by an applied torque Tg = 50 kip . in. (b) Determine the maximum shear stress (Tmax)1 in the alu- minum segment AB, (Tma)2 in the aluminum core segment BC, and (Tmax), in the titanium sleeve BC.

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a composite shaft AC is made by shrink-fitting a titanium-alloy sleeve of length L3 = 4 ft over segment BC of a 6-ft-long aluminum-alloy rod. This makes a three-component shaft, which is fixed against rotation at ends A and C. The shear moduli are G1 = G2 = 6 X 103 ksi and G3 = 4 X 103 ksi, respectively, and the diameters and d1 = d2 = (di)3 = 2 in., and (do)3 = 2.5 in.

(a) Using the Displacement Method, determine the angle of ro- tation at B, og. produced by an applied torque Tg = 50 kip . in.

(b) Determine the maximum shear stress (Tmax)1 in the alu- minum segment AB, (Tma)2 in the aluminum core segment BC, and (Tmax), in the titanium sleeve BC.

The image presents a diagram of a cylindrical shaft with varying diameters and sections. Here's a transcription and detailed explanation:

**Diagram Description:**

1. **Sections**:
   - Section (1): A solid cylindrical section on the left side with diameter \( d_1 \) and length \( L_1 \). This section is anchored at point A.
   - Section (3): A hollow cylindrical section with an outer diameter \( d_2 \) and length \( L_2 = L_3 \). It is fixed at point C on the right.

2. **Points**:
   - **A**: The leftmost fixed support for section (1).
   - **B**: Between sections (1) and (3), where a torque \( T_B \) is applied.
   - **C**: The rightmost fixed support for section (3).

3. **Dimensions**:
   - \( d_1 \): Diameter of section (1) (solid section).
   - \( d_2 \): Outer diameter of section (3) (hollow section).
   - \( L_1 \): Length of the solid section.
   - \( L_2 = L_3 \): Length of the hollow section.

**Additional Details**:

- The diagram shows a torque \( T_B \) applied at the junction point B, indicated by an arrow curling around the shaft.
- The shaft appears to have a transition at point B from the solid section to the hollow section.

This diagram is likely used to illustrate concepts such as torsional mechanics, analyzing the effects of applied torque on shafts with different properties across their sections.
Transcribed Image Text:The image presents a diagram of a cylindrical shaft with varying diameters and sections. Here's a transcription and detailed explanation: **Diagram Description:** 1. **Sections**: - Section (1): A solid cylindrical section on the left side with diameter \( d_1 \) and length \( L_1 \). This section is anchored at point A. - Section (3): A hollow cylindrical section with an outer diameter \( d_2 \) and length \( L_2 = L_3 \). It is fixed at point C on the right. 2. **Points**: - **A**: The leftmost fixed support for section (1). - **B**: Between sections (1) and (3), where a torque \( T_B \) is applied. - **C**: The rightmost fixed support for section (3). 3. **Dimensions**: - \( d_1 \): Diameter of section (1) (solid section). - \( d_2 \): Outer diameter of section (3) (hollow section). - \( L_1 \): Length of the solid section. - \( L_2 = L_3 \): Length of the hollow section. **Additional Details**: - The diagram shows a torque \( T_B \) applied at the junction point B, indicated by an arrow curling around the shaft. - The shaft appears to have a transition at point B from the solid section to the hollow section. This diagram is likely used to illustrate concepts such as torsional mechanics, analyzing the effects of applied torque on shafts with different properties across their sections.
Expert Solution
Step 1: Angle of rotation:

(a) Angle of Rotation at B (θB):

Using the Displacement Method, we can calculate the angular displacement at point B due to the applied torque Tg.


θB = Tg / (G1 * L1) + Tg / (G2 * L2)


Where:

  • Tg = Applied torque = 50 kip.in
  • G1 = Shear modulus of aluminum alloy in segment AB = 6 x 10^3 ksi
  • G2 = Shear modulus of aluminum alloy in segment BC = 6 x 10^3 ksi
  • L1 = Length of segment AB = 2 ft (given)
  • L2 = Length of segment BC = 4 ft (given)


Substitute the values to calculate θB,

θB = (50 kip.in) / (6 x 10^3 ksi * 2 ft) + (50 kip.in) / (6 x 10^3 ksi * 4 ft)

θB = 0.0017 radians

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