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Chapter 10: Problem 23

Two steel plates are fastened together using four bolts. The bolts each have a shear modulus of \(8.0 \times 10^{10} \mathrm{Pa}\) and a shear strength of $6.0 \times 10^{8} \mathrm{Pa} .\( The radius of each bolt is \)1.0 \mathrm{cm} .$ Normally, the bolts clamp the two plates together and the frictional forces between the plates keep them from sliding. If the bolts are loose, then the frictional forces are small and the bolts themselves would be subject to a large shear stress. What is the maximum shearing force \(F\) on the plates that the four bolts can withstand? (IMAGE NOT COPY)

Short Answer

Expert verified
Answer: The maximum shearing force that the four bolts can withstand is approximately \( 24\pi × 10^{4} Pa.m^2 \).

Step by step solution

01

Finding the Cross-sectional area of a bolt

Since the bolts are cylindrical in shape, we can calculate their cross-sectional area (A) using the formula for the area of a circle with radius r: \( A = \pi r^2 \) Given, radius of each bolt (r) = 1.0 cm which can be converted to meters 1.0 × 10^(-2) m. So, the cross-sectional area of each bolt is: \( A = \pi (1.0 × 10^{-2})^2 = \pi × 10^{-4} m^2 \)
02

Finding the shear stress on each bolt

The shear stress (τ) on each bolt can be calculated using the given shear strength (s) of the bolts: \( τ = s A \) Given, shear strength s = \(6.0 × 10^{8} Pa\). Now we will find the shear stress on each bolt: \( τ = (6.0 × 10^{8} Pa) (\pi × 10^{-4} m^2) = 6.0\pi × 10^{4} Pa.m^2 \)
03

Finding the maximum shearing force on the plates

Each bolt can withstand τ amount of shear stress, and there are 4 bolts. The total maximum shearing force that these bolts can withstand is simply the sum of the shear stress that each bolt can withstand: \( F_{total} = 4 × τ = 4 × 6.0\pi × 10^{4} Pa.m^2 = 24\pi × 10^{4} Pa.m^2 \) Therefore, the maximum shearing force F on the plates that the four bolts can withstand is approximately \( 24\pi × 10^{4} Pa.m^2 \).

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