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Give step-by-step solution with explanation and final answer:Q3. A thin walled cylindrical pipe with open ends is subjected to an internal pressure p at room

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Act like a helpful tutor and exlain me :
Give step-by-step solution with explanation and final answer:Q3. A thin walled cylindrical pipe with open ends is subjected to an internal pressure p at room temperature.
The tube material is isotropic whose properties are listed below.
ZT Geometric properties
74 WN Diameter : 6 inches
- Thickness : 0.06 inches
Ne NL Material properties
~ §
8 = E=10 Msi
SENT v=03
, p A Y=20 ksi
2 2 a=16x10"® (in/in)/°F
Lay K=80 Btu/hr ft? °F/ft
bd
X
a) While holding the temperature constant, if we keep increasing the internal pressure p, what will be its
value when the tube begins to yield according to Tresca criterion. (20 points)

VIPSolutions · Accepted Answer

Let's solve the problem step by step.

---

# Step 1: **Identify What is Asked**

- **Given:** Thin-walled cylindrical pipe, open ends, internal pressure $ p $, isotropic material, properties and dimensions given.
- **Asked:** At constant temperature, **what is the value of $ p $ when the tube begins to yield** according to the **Tresca criterion**?

---

# Step 2: **List the Given Data**

**Geometric Properties:**
- Diameter ($ D $) = 6 in
- Thickness ($ t $) = .06 in

**Material Properties:**
- Yield strength ($ Y $) = 20 ksi

---

# Step 3: **Determine Stresses in Thin-Walled Cylinder**

For a thin-walled cylinder under internal pressure ($ p $):
- **Hoop (circumferential) stress ($ \sigma_{\theta} $):**
  $$
  \sigma_{\theta} = \frac{p D}{2t}
  $$
- **Longitudinal (axial) stress ($ \sigma_x $):**
  $$
  \sigma_x = \frac{p D}{4t}
  $$
- **Radial stress ($ \sigma_r $):**
  - At the inner wall: $ \sigma_r = -p $
  - At the outer wall: $ \sigma_r \approx  $ (thin wall assumption)

But in thin-walled approximation, radial stress is much smaller and often neglected for yield.

---

# Step 4: **Apply Tresca Yield Criterion**

**Tresca criterion:** Yield occurs when the maximum shear stress equals half the yield strength.

- **Principal stresses:** $ \sigma_1 = \sigma_{\theta} $, $ \sigma_2 = \sigma_x $, $ \sigma_3 =  $ (neglecting radial stress, since it's small compared to the other two)

Tresca criterion:
$$
\text{Max} \left| \sigma_i - \sigma_j \right| = Y
$$

Calculate differences:
$$
\sigma_{\theta} - \sigma_x = \frac{pD}{2t} - \frac{pD}{4t} = \frac{pD}{4t}
$$
$$
\sigma_{\theta} -  = \frac{pD}{2t}
$$
$$
\sigma_x -  = \frac{pD}{4t}
$$

The **maximum** is $ \frac{pD}{2t} $.

Set this equal to $ Y $:
$$
\frac{pD}{2t} = Y
$$

---

# Step 5: **Solve for Critical Pressure $ p $**

Plug in the values:
- $ D = 6 $ in
- $ t = .06 $ in
- $ Y = 20 $ ksi

$$
\frac{p \times 6}{2 \times .06} = 20,000 \text{ psi}
$$
$$
\frac{p \times 6}{.12} = 20,000
$$
$$
p \times 50 = 20,000
$$
$$
p = \frac{20,000}{50}
$$
$$
p = 400 \text{ psi}
$$

---

# Step 6: **Final Answer**

## **Final Answer**
**The tube will begin to yield according to the Tresca criterion at:**

$$
\boxed{p = 400 \text{ psi}}
$$

---

### **Summary Table**

| Step | Calculation | Result |
|---|---|---|
| 1 | Hoop stress formula | $ \sigma_{\theta} = \frac{pD}{2t} $ |
| 2 | Set equal to yield (Tresca) | $ \frac{pD}{2t} = Y $ |
| 3 | Substitute values, solve for $ p $ | $ p = 400 $ psi |

Let me know if you'd like further explanation or a check for the von Mises criterion!

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