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Question

bl hs ontebi  by EA RY > VOY Rb Sat Ry Sn Blige since iy = Whit gg ~ WW.
hg
Te
Pa i)
La de Ve
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Fon te
me en i do oh 8 te et 8A md ne ee
pe
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VIPSolutions · Accepted Answer

Let's break down the problem step by step.

### **Given Data**
- Torques: $ T_B = 1400 $ Nm, $ T_C = 800 $ Nm
- Lengths: $ L_{AB} = 700 $ mm, $ L_{BC} = 800 $ mm
- Allowable shear stress: $ \tau_{allow} = 55 $ MPa
- Shear modulus: $ G = 41 $ GPa
- Neglect stress concentrations

---

### **Step 1: Find Torques in Each Segment**

#### **Segment AB:**
- To the left of B, only $ T_B $ is applied.
- To the right of B, both $ T_B $ and $ T_C $ are applied, but let's assume equilibrium at each segment.
- **At AB:** The internal torque is the sum due to both torques acting in opposite directions:
  $$
  T_{AB} = T_B + T_C = 1400 + 800 = 2200 \text{ Nm}
  $$
  (assuming both applied in the same rotational sense)

#### **Segment BC:**
- Only $ T_C $ acts:
  $$
  T_{BC} = T_C = 800 \text{ Nm}
  $$

---

### **Step 2: Calculate Minimum Diameters**

The maximum shear stress in a solid shaft:
$$
\tau_{max} = \frac{T \cdot r}{J} = \frac{16T}{\pi d^3}
$$
Set $\tau_{max} = \tau_{allow}$:

$$
\tau_{allow} = \frac{16T}{\pi d^3}
$$
$$
d = \left(\frac{16T}{\pi \tau_{allow}}\right)^{1/3}
$$

#### **For Segment AB:**
$$
d_{AB} = \left(\frac{16 \times 2200 \times 10^3}{\pi \times 55 \times 10^6}\right)^{1/3}
$$

#### **For Segment BC:**
$$
d_{BC} = \left(\frac{16 \times 800 \times 10^3}{\pi \times 55 \times 10^6}\right)^{1/3}
$$

---

### **Step 3: Plug in Values**

#### **AB:**
$$
d_{AB} = \left(\frac{16 \times 2200 \times 10^3}{\pi \times 55 \times 10^6}\right)^{1/3}
$$
$$
= \left(\frac{35,200,000}{172,787,595}\right)^{1/3}
$$
$$
= (0.2037)^{1/3}
$$
$$
\approx 0.588 \text{ m} \approx 58.8 \text{ mm}
$$

#### **BC:**
$$
d_{BC} = \left(\frac{16 \times 800 \times 10^3}{\pi \times 55 \times 10^6}\right)^{1/3}
$$
$$
= \left(\frac{12,800,000}{172,787,595}\right)^{1/3}
$$
$$
= (0.07409)^{1/3}
$$
$$
\approx 0.420 \text{ m} \approx 42.0 \text{ mm}
$$

*Round each to the nearest integer:*

- **d_AB = 59 mm**
- **d_BC = 42 mm**

---

### **Step 4: Calculate Total Twist Angle**

The angle of twist for a segment:
$$
\theta = \frac{T L}{J G}
$$
where $ J = \frac{\pi d^4}{32} $.

Calculate for each segment and sum:

#### **Segment AB:**
$$
\theta_{AB} = \frac{T_{AB} L_{AB}}{J_{AB} G}
$$

#### **Segment BC:**
$$
\theta_{BC} = \frac{T_{BC} L_{BC}}{J_{BC} G}
$$

#### **Total:**
$$
\theta_{total} = \theta_{AB} + \theta_{BC}
$$

Plug in:
- $ L_{AB} = 700 $ mm = 0.7 m
- $ L_{BC} = 800 $ mm = 0.8 m
- $ G = 41 $ GPa = $ 41 \times 10^9 $ Pa

**Calculate $ J_{AB} $ and $ J_{BC} $:**
$$
J = \frac{\pi d^4}{32}
$$

- $ d_{AB} = 0.059 $ m
- $ d_{BC} = 0.042 $ m

$$
J_{AB} = \frac{\pi (0.059)^4}{32} \approx 1.88 \times 10^{-6} \text{ m}^4
$$
$$
J_{BC} = \frac{\pi (0.042)^4}{32} \approx 3.85 \times 10^{-7} \text{ m}^4
$$

Now:

$$
\theta_{AB} = \frac{2200 \times 0.7}{1.88 \times 10^{-6} \times 41 \times 10^9}
= \frac{1540}{77,080}
= 0.01998 \text{ radians}
$$

$$
\theta_{BC} = \frac{800 \times 0.8}{3.85 \times 10^{-7} \times 41 \times 10^9}
= \frac{640}{15,785}
= 0.0406 \text{ radians}
$$

$$
\theta_{total} = 0.01998 + 0.0406 = 0.0606 \text{ radians}
$$

---

### **Summary of Answers**

- **Smallest diameters:**
  - $ d_{AB} = 59 $ mm
  - $ d_{BC} = 42 $ mm

- **Total angle of twist:**  
  - $ \theta_{total} = 0.061 $ radians (rounded to three decimal places)

---

If you need help plugging these into the answer boxes or further explanation, let me know!

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