Let's break down the problem step by step as requested: --- # 1. Problem Summary A shaft (200 mm long) is supported by bearings A and B. Belt forces (vertical and horizontal) act on a pulley at the center. The left end is attached to a clutch and transmits **torque**. Nothing is attached at the right end. **Goal:** Calculate the stresses (normal and shear) at the critical cross-section (neglecting stress concentration). --- # (a) **Identify the Critical Cross Section** The shaft is subjected to: - **Vertical Force:** 120 N (down) - **Horizontal Force:** 200 N (to the right) - **Torque:** 250 Nm (counterclockwise, as viewed from the right) The **critical section** is the section where the combined stresses (bending + torsion) are *maximum*. Since the forces and torque are applied at the pulley (center), **the critical cross-section is just to the left or right of the pulley** (i.e., at the center of the shaft, under the pulley). --- # (b) **Sketch & Identify Points of Maximum Stress** **Key Points on Cross-section:** - **Top and Bottom fibers:** Maximum normal (bending) stress due to vertical force. - **Left and Right fibers:** Maximum normal (bending) stress due to horizontal force. - **Maximum Shear stress:** On the surface, at 45° from the neutral axis, due to torsion. ## **Types of Stresses:** - **Bending normal stress:** due to vertical and horizontal forces - **Shear stress:** due to torque (torsion) --- # (c) **Populate the Stress Tensor for Analysis Points** Let’s calculate the stresses at the **outer surface** at a point (for example, at the topmost fiber). ## **Given Data** - Length between bearings: \( L = 200\, \text{mm} = .2\, \text{m} \) - Diameter of shaft: \( d = 30\, \text{mm} = .03\, \text{m} \) - Radius: \( r = \frac{d}{2} = .015\, \text{m} \) - Vertical force: \( F_v = 120\,\text{N} \) - Horizontal force: \( F_h = 200\,\text{N} \) - Torque: \( T = 250\,\text{Nm} \) --- ## **Step 1: Calculate Bending Moments at the Center** Since the forces are applied at the center, reactions at A and B will share the load equally. ### **Vertical Plane:** - Each bearing reaction: \( R_v = \frac{F_v}{2} = 600\,\text{N} \) - Moment at center: \( M_v = R_v \times (L/2) = 600\,\text{N} \times .1\,\text{m} = 60\,\text{Nm} \) ### **Horizontal Plane:** - Each bearing reaction: \( R_h = \frac{F_h}{2} = 100\,\text{N} \) - Moment at center: \( M_h = R_h \times (L/2) = 100\,\text{N} \times .1\,\text{m} = 100\,\text{Nm} \) --- ## **Step 2: Calculate Stresses** ### **Bending Stress, \( \sigma = \frac{M y}{I} \)** Where: - \( y = r = .015\,\text{m} \) - \( I = \frac{\pi d^4}{64} = \frac{\pi (.03)^4}{64} = 3.98 \times 10^{-7} \, \text{m}^4 \) #### **Vertical Bending:** \[ \sigma_{bv} = \frac{M_v r}{I} = \frac{60 \times .015}{3.98 \times 10^{-7}} = 2.26 \times 10^6 \, \text{N/m}^2 = 2.26 \, \text{MPa} \] #### **Horizontal Bending:** \[ \sigma_{bh} = \frac{M_h r}{I} = \frac{100 \times .015}{3.98 \times 10^{-7}} = 3.77 \times 10^6 \, \text{N/m}^2 = 3.77 \, \text{MPa} \] **Total bending stress at outer fiber (adding as vectors):** \[ \sigma_b = \sqrt{(\sigma_{bv})^2 + (\sigma_{bh})^2} = \sqrt{(2.26)^2 + (3.77)^2} = 4.39 \, \text{MPa} \] --- ### **Torsional (Shear) Stress, \( \tau = \frac{T r}{J} \)** Where: - \( J = \frac{\pi d^4}{32} = 2 \times I = 7.96 \times 10^{-7} \, \text{m}^4 \) \[ \tau_{max} = \frac{T r}{J} = \frac{250 \times .015}{7.96 \times 10^{-7}} = 4.71 \times 10^6 \, \text{N/m}^2 = 4.71 \, \text{MPa} \] --- ## **Step 3: Stress Tensor at Surface Point (e.g., Topmost Point)** At the topmost fiber (due to vertical bending and torsion): \[ \boxed{ \begin{bmatrix} \sigma_b & \tau_{max} \\ \tau_{max} & \end{bmatrix} = \begin{bmatrix} 4.39 & 4.71 \\ 4.71 & \end{bmatrix} \, \text{MPa} } \] - At the leftmost or rightmost fiber, the normal stress will be due to horizontal bending, and the shear stress due to torsion. --- # **Summary Table** | Point | Bending Stress (MPa) | Shear Stress (MPa) | |-----------------|---------------------|--------------------| | Top fiber | 4.39 | 4.71 | | Bottom fiber | -4.39 | -4.71 | | Side fibers | 3.77 | ±4.71 | --- # **Final Answers** ### **(a)** *Critical cross-section*: At the center of the shaft, under the pulley. ### **(b)** *Points of max stresses*: - Max normal stress at top/bottom and sides of the shaft cross-section. - Max shear stress at the surface, at 45° to the axes (due to torsion). ### **(c)** *Stress tensor at top fiber*: \[ \boxed{ \begin{bmatrix} 4.39 & 4.71 \\ 4.71 & \end{bmatrix} \, \text{MPa} } \] --- **Let me know if you'd like a diagram or further breakdown!**