## ANSWER:

**Given:**

- Diameter of the shaft (d) = 10 cm
- Tensile force (P) = 10,000 N
- Torque (T) = 5,000 N cm

**Step 1: Calculate the normal stresses**

The normal stress due to the tensile force is:

```
σx = P / A
```

where A is the cross-sectional area of the shaft. The cross-sectional area of a circle is:

```
A = πr²
```

where r is the radius of the shaft. The radius of the shaft is half the diameter, so r = 5 cm. Therefore, the cross-sectional area is:

```
A = π(5 cm)² = 25π cm²
```

Plugging in the values for P and A, we get:

```
σx = 10,000 N / (25π cm²) ≈ 127.4 N/cm²
```

The normal stress due to the torque is:

```
σy = -T / (πr³)
```

Plugging in the values for T and r, we get:

```
σy = -5,000 N cm / (π(5 cm)³) ≈ -63.7 N/cm²
```

**Step 2: Calculate the shear stress**

The shear stress is:

```
τxy = T / (2πr³)
```

Plugging in the values for T and r, we get:

```
τxy = 5,000 N cm / (2π(5 cm)³) ≈ 31.85 N/cm²
```

**Step 3: Calculate the principal stresses**

The principal stresses are the two normal stresses that act perpendicular to each other at a point on the surface of the shaft. The principal stresses can be calculated using the following equation:

```
σ1,2 = (σx + σy) / 2 ± √((σx - σy)² / 4 + τxy²)
```

Plugging in the values for σx, σy, and τxy, we get:

```
σ1 = 31.85 N/cm²
σ2 = -127.4 N/cm²
```

**Step 4: Calculate the octahedral shearing stress**

The octahedral shearing stress is a measure of the intensity of the shearing stresses acting at a point on the surface of the shaft. The octahedral shearing stress can be calculated using the following equation:

```
τoct = √((σ1 - σ2)² / 3 + 3τxy²)
```

Plugging in the values for σ1, σ2, and τxy, we get:

```
τoct = 103.25 N/cm²
```

**Step 5: Calculate the maximum shearing stress**

The maximum shearing stress is the largest of the three shear stresses acting at a point on the surface of the shaft. The maximum shearing stress can be calculated using the following equation:

```
τmax = (σ1 - σ2) / 2
```

Plugging in the values for σ1 and σ2, we get:

```
τmax = 98.55 N/cm²
```

**Therefore, the principal stresses at point A on the surface of the shaft are 31.85 N/cm² and -127.4 N/cm², the octahedral shearing stress is 103.25 N/cm², and the maximum shearing stress is 98.55 N/cm².**