Let's solve the given problem step-by-step. **Given:** - Simply supported beam: W 12x14 - Point Load: 20 kips at 4 ft from left support (point B) - Beam length: 8 ft - Locations to analyze (at \(x = 2'\)): 1. Top of beam 2. Top of web 3. Neutral axis **Required:** Find the **principal stresses** and **maximum shear stress** at the specified locations. --- ## 1. **Support Reactions** Let’s first find the reactions at supports A and C. Summing moments about A: \[ \sum M_A = 0 = R_C \times 8 - 20 \times 4 \implies R_C = \frac{80}{8} = 10 \text{ k} \] Vertical equilibrium: \[ R_A + R_C = 20 \implies R_A = 10 \text{ k} \] --- ## 2. **Shear Force & Bending Moment at \(x = 2'\)** - **Shear just left of \(x=2'\):** \[ V = R_A = 10 \text{ k} \] - **Moment at \(x = 2'\):** \[ M = R_A \times 2' = 10 \times 2 = 20 \text{ k-ft} \] --- ## 3. **Section Properties (W12x14):** From steel tables: - **Area (A):** 4.12 in² - **Depth (h):** 12.22 in - **Flange width (bf):** 4.02 in - **Web thickness (tw):** 0.230 in - **Flange thickness (tf):** 0.345 in - **Moment of Inertia (I):** 98.1 in⁴ --- ## 4. **Normal Stress due to Bending (\(\sigma\))** \[ \sigma = \frac{My}{I} \] Where \(y\) is the distance from the neutral axis: - **At the top of beam:** \(y = +h/2 = 12.22/2 = 6.11 \text{ in}\) - **At neutral axis:** \(y = 0\) - **At top of web:** \(y = h/2 - tf = 6.11 - 0.345 = 5.765 \text{ in}\) Convert moment to in-lb: \[ M = 20 \text{ k-ft} = 20 \times 1000 \times 12 = 240,000 \text{ in-lb} \] So, - **At top of beam:** \[ \sigma_{top} = \frac{240,000 \times 6.11}{98.1} = \frac{1,466,400}{98.1} \approx 14,954 \text{ psi} = 14.95 \text{ ksi (compression)} \] - **At top of web:** \[ \sigma_{web} = \frac{240,000 \times 5.765}{98.1} \approx \frac{1,383,600}{98.1} \approx 14,107 \text{ psi} = 14.1 \text{ ksi (compression)} \] - **At neutral axis:** \[ \sigma_{NA} = 0 \] --- ## 5. **Shear Stress (\(\tau\))** \[ \tau = \frac{VQ}{Ib} \] Where: - \(V = 10,000 \text{ lb}\) - \(I = 98.1 \text{ in}^4\) - \(b =\) width at location **At different locations:** **A. Top of beam (in flange):** - Shear stress is negligible (most shear is carried by web). **B. Top of web:** - \(b = t_{web} = 0.230\) in Calculate Q (area above point times distance to N.A.): - Area of flange: \(A_f = bf \times tf = 4.02 \times 0.345 = 1.387 \text{ in}^2\) - Distance from N.A. to centroid of flange: \(y' = 6.11 - 0.345/2 = 5.9375 \text{ in}\) \[ Q = A_f \times y' = 1.387 \times 5.9375 = 8.23 \text{ in}^3 \] \[ \tau_{web,top} = \frac{10,000 \times 8.23}{98.1 \times 0.230} = \frac{82,300}{22.063} \approx 3,731 \text{ psi} = 3.73 \text{ ksi} \] **C. At N.A. (max shear, web):** - Area above N.A.: Flange + half web - Area of half web: \(0.230 \times 6.11 = 1.405 \text{ in}^2\) - Total area above N.A.: \(1.387 + 1.405 = 2.792 \text{ in}^2\) - Centroid distance to N.A.: - Flange: \(5.9375\) - Web: centroid is \(6.11/2 = 3.055\) above N.A. - Q = \(1.387 \times 5.9375 + 1.405 \times 3.055 = 8.23 + 4.29 = 12.52\) \[ \tau_{NA} = \frac{10,000 \times 12.52}{98.1 \times 0.230} = \frac{125,200}{22.063} \approx 5,678 \text{ psi} = 5.68 \text{ ksi} \] --- ## 6. **Principal Stresses** \[ \sigma_1, \sigma_2 = \frac{\sigma}{2} \pm \sqrt{\left(\frac{\sigma}{2}\right)^2 + \tau^2} \] ### a) **At Top of Beam** (\(\sigma = -14.95 \text{ ksi}\), \(\tau \approx 0\)): \[ \sigma_1 = -14.95 \text{ ksi},\ \sigma_2 = 0 \] Max shear: \[ \tau_{max} = \frac{|-14.95|}{2} = 7.48 \text{ ksi} \] ### b) **At Top of Web** (\(\sigma = -14.1 \text{ ksi}\), \(\tau = 3.73 \text{ ksi}\)): \[ \sigma_1, \sigma_2 = \frac{-14.1}{2} \pm \sqrt{\left(\frac{-14.1}{2}\right)^2 + (3.73)^2} \] \[ = -7.05 \pm \sqrt{49.7 + 13.9} = -7.05 \pm \sqrt{63.6} = -7.05 \pm 7.98 \] \[ \sigma_1 = 0.93 \text{ ksi},\ \sigma_2 = -15.03 \text{ ksi} \] Max shear: \[ \tau_{max} = \frac{0.93 - (-15.03)}{2} = 7.98 \text{ ksi} \] ### c) **At N.A.** (\(\sigma = 0\), \(\tau = 5.68 \text{ ksi}\)): \[ \sigma_1, \sigma_2 = 0 \pm 5.68 \text{ ksi} \] \[ \sigma_1 = 5.68 \text{ ksi},\ \sigma_2 = -5.68 \text{ ksi} \] Max shear: \[ \tau_{max} = 5.68 \text{ ksi} \] --- ## **Final Answers Table:** | Location | Normal Stress \(\sigma\) (ksi) | Shear Stress \(\tau\) (ksi) | Principal Stresses (ksi) | Max Shear Stress (ksi) | |---------------|-------------------------------|-----------------------------|--------------------------|------------------------| | Top of Beam | -14.95 | ~0 | -14.95, 0 | 7.48 | | Top of Web | -14.1 | 3.73 | 0.93, -15.03 | 7.98 | | Neutral Axis | 0 | 5.68 | 5.68, -5.68 | 5.68 | --- **Let me know if you need a more detailed calculation or explanation for any step!**