For a given loading conditions the state of stress in the wall of a cylinder is expressed as follows: 
(a)  85 MN/m2 tensile 
(b)  25 MN/m2 tensile at right angles to (a) 
(c)  Shear stresses of 60 MN/m2 on the planes 
Angle from plan is 30 degree.
Find various stresses for the given configuration
 solve this numerical 

To solve this problem, we need to find the **principal stresses** and **maximum shear stresses** based on the given state of stress in the cylinder wall. This is a **plane stress** problem, with:

- **σ₁** = 85 MN/m² (Tensile)
- **σ₂** = 25 MN/m² (Tensile, at right angles to σ₁)
- **τ₁₂** = 60 MN/m² (Shear stress on the planes)

We will use **Mohr's Circle** approach to determine the **principal stresses** (σ₁', σ₂') and **maximum shear stress (τ_max)**. Below is the step-by-step solution.

---

### Step 1: Formulae to Use
- **Principal stresses (σ₁', σ₂'):**

\[
σ_{1,2} = \frac{σ₁ + σ₂}{2} \pm \sqrt{\left( \frac{σ₁ - σ₂}{2} \right)^2 + τ_{12}^2}
\]

- **Maximum shear stress (τ_max):**

\[
τ_{max} = \sqrt{\left( \frac{σ₁ - σ₂}{2} \right)^2 + τ_{12}^2}
\]

- **Angle of principal planes (θ):**

\[
\text{tan}(2θ) = \frac{2τ_{12}}{σ₁ - σ₂}
\]

---

### Step 2: Input Values
- **σ₁** = 85 MN/m²
- **σ₂** = 25 MN/m²
- **τ₁₂** = 60 MN/m²

---

### Step 3: Calculate Principal Stresses (σ₁', σ₂')
Using the formula:

\[
\frac{σ₁ + σ₂}{2} = \frac{85 + 25}{2} = \frac{110}{2} = 55 \, \text{MN/m}^2
\]

\[
\frac{σ₁ - σ₂}{2} = \frac{85 - 25}{2} = \frac{60}{2} = 30 \, \text{MN/m}^2
\]

Now, substitute in the principal stress formula:

\[
σ_{1,2} = 55 \pm \sqrt{(30)^2 + (60)^2}
\]

\[
σ_{1,2} = 55 \pm \sqrt{900 + 3600} = 55 \pm \sqrt{4500} = 55 \pm 67.08
\]

So, the principal stresses are:

\[
σ_1' = 55 + 67.08 = 122.08 \, \text{MN/m}^2
\]
\[
σ_2' = 55 - 67.08 = -12.08 \, \text{MN/m}^2
\]

---

### Step 4: Calculate Maximum Shear Stress (τ_max)
Using the formula:

\[
τ_{max} = \sqrt{(30)^2 + (60)^2} = \sqrt{900 + 3600} = \sqrt{4500} = 67.08 \, \text{MN/m}^2
\]

---

### Step 5: Calculate Angle of Principal Plane (θ)
Using the formula:

\[
\text{tan}(2θ) = \frac{2 \cdot 60}{85 - 25} = \frac{120}{60} = 2
\]

\[
2θ = \tan^{-1}(2) \implies 2θ = 63.43^\circ \implies θ = 31.72^\circ
\]

---

### Final Results
- **Principal stresses:**
  - \(σ_1' = 122.08 \, \text{MN/m}^2\)
  - \(σ_2' = -12.08 \, \text{MN/m}^2\)

- **Maximum shear stress:** \(τ_{max} = 67.08 \, \text{MN/m}^2\)

- **Angle of principal planes:** \(θ = 31.72^\circ\)

---

### Interpretation
- One of the principal stresses is **compressive** (\(-12.08 \, \text{MN/m}^2\)), meaning that the material experiences tension in one direction and compression in another.
- The **maximum shear stress** occurs at an angle of \(45^\circ\) from the principal planes.

---

Do you have any further questions, or would you like to explore another aspect of this problem?

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**Related Questions:**
1. What is Mohr's circle and how can it be constructed?
2. How do you find stresses on an arbitrary inclined plane?
3. What is the physical significance of principal stresses in design?
4. How does the presence of shear stress affect the failure of materials?
5. How does this analysis change if there are thermal stresses included?

---

**Tip:** For visualization, try constructing **Mohr’s Circle**—it provides a geometric way to interpret stress states.

For a given loading condition, the state of stress in the wall of a cylinder is expressed as follows: (a) 85 MN/m² tensile, (b) 25 MN/m² tensile at right angles to (a), (c) shear stresses of 60 MN/m² on the planes. The angle from the plane is 30 degrees. Find various stresses for the given configuration.

Learn how to calculate principal and maximum shear stresses in a cylinder wall using Mohr's Circle. This detailed solution includes the calculation of stresses based on tensile and shear stress components with angles, covering essential concepts of plane stress and material mechanics.

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