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Give step-by-step solution with explanation and final answer:A single 3/8-inch diameter bolt is used to connect five 1/4-inch metal plates as shown below. Draw a free-body diagram of the bolt showing shear forces at all plate interfaces, and: (a) Calculate the largest shear stress in the bolt. (b) Calculate the largest bearing stresses on the plates. 300 Ib RE : 400 lb Son 300 Ib ee 500 1b

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Give step-by-step solution with explanation and final answer:Uploaded ImageA single 3/8-inch diameter bolt is used to connect five 1/4-inch metal plates as shown below. Draw a free-body diagram of the bolt showing shear forces at all plate interfaces, and: (a) Calculate the largest shear stress in the bolt. (b) Calculate the largest bearing stresses on the plates. 300 Ib RE : 400 lb Son 300 Ib ee 500 1b

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Asked by: Veera manikanta
Created at: 2025-09-03 11:31:26

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Let's solve the problem step by step: --- ### **Step 1: Free-Body Diagram of the Bolt** There are 5 plates and 4 interfaces where shear occurs (between each pair of plates). The forces on the interfaces from top to bottom are: - **1st interface (between Plate 1 and Plate 2):** 300 lb to the left, 500 lb to the right **Shear force = 300 lb (left)** - **2nd interface (between Plate 2 and Plate 3):** (300 lb left + 400 lb left = 700 lb left), (500 lb right) **Shear force = 400 lb (left)** [since Plate 2 takes 400 lb load] - **3rd interface (between Plate 3 and Plate 4):** (700 lb left), (500 lb right + 500 lb right = 1000 lb right) **Shear force = 300 lb (right)** [since Plate 3 takes 300 lb load] - **4th interface (between Plate 4 and Plate 5):** (1000 lb right), (300 lb left) **Shear force = 0 lb** [since Plate 4 takes no load] **Summary of Shear Forces at Interfaces:** - 1st: 300 lb - 2nd: 400 lb - 3rd: 300 lb - 4th: 0 lb --- ### **Step 2: Calculate the Largest Shear Stress in the Bolt** **Shear Stress Formula:** \[ \tau = \frac{V}{A} \] where \(V\) is the shear force, and \(A\) is the cross-sectional area. #### **Bolt Area:** Diameter \(d = 3/8\) in \(= 0.375\) in \[ A = \frac{\pi}{4} d^2 = \frac{\pi}{4} (0.375)^2 = 0.1104 \text{ in}^2 \] #### **Largest Shear Force:** From above, the largest shear force is **400 lb** (at the second interface). \[ \tau_{\text{max}} = \frac{400}{0.1104} = 3624.6 \text{ psi} \] --- ### **Step 3: Calculate the Largest Bearing Stress on the Plates** **Bearing Stress Formula:** \[ \sigma_b = \frac{P}{t \cdot d} \] where \(P\) is the force, \(t\) is the plate thickness, and \(d\) is the bolt diameter. - Plate thickness \(t = 1/4\) in \(= 0.25\) in - Bolt diameter \(d = 0.375\) in #### **Largest Force on a Plate:** Look at the plates: - Plate 2: 400 lb - Plate 4: 500 lb **Largest is 500 lb** \[ \sigma_{b,\text{max}} = \frac{500}{0.25 \times 0.375} = \frac{500}{0.09375} = 5333.3 \text{ psi} \] --- ## **Final Answers:** ### (a) Largest Shear Stress in the Bolt: \[ \boxed{3620~\text{psi}} \] ### (b) Largest Bearing Stress on the Plates: \[ \boxed{5330~\text{psi}} \] *Rounded to 2 significant digits for clarity.* --- **If you need the free-body diagram sketched out, let me know!**

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