(1 point) 4-61 Rigid YokesThe rigid yokes B and C shown are securely fastened to the 2.1-in. square steel (E=31000 ksi) bar AD. The forces applied to the bar are P1𝑃1 = 83 kip, P2𝑃2 = 47 kip, P3𝑃3 = 28.5 kip and P4𝑃4 = 46 kip. If the lengths of the bar are a = 9 ft, b = 6 ft, and c = 3.5 ft, determine:(a) The maximum normal stress in the bar.(b) The change in length of segment AB.(c) The change in length of segment BC.(d) The change in length of the complete bar.(a)σmax𝜎𝑚𝑎𝑥 = ksi(b)δAB𝛿𝐴𝐵 = in.(c)δBC𝛿𝐵𝐶 = in.(d)δtotal𝛿𝑡𝑜𝑡𝑎𝑙 = in.
Question
(1 point) 4-61 Rigid YokesThe rigid yokes B and C shown are securely fastened to the 2.1-in. square steel (E=31000 ksi) bar AD. The forces applied to the bar are P1𝑃1 = 83 kip, P2𝑃2 = 47 kip, P3𝑃3 = 28.5 kip and P4𝑃4 = 46 kip. If the lengths of the bar are a = 9 ft, b = 6 ft, and c = 3.5 ft, determine:(a) The maximum normal stress in the bar.(b) The change in length of segment AB.(c) The change in length of segment BC.(d) The change in length of the complete bar.(a)σmax𝜎𝑚𝑎𝑥 = ksi(b)δAB𝛿𝐴𝐵 = in.(c)δBC𝛿𝐵𝐶 = in.(d)δtotal𝛿𝑡𝑜𝑡𝑎𝑙 = in.
Solution
To solve this problem, we need to use the formula for normal stress, which is σ = P/A, where P is the force and A is the cross-sectional area. We also need to use the formula for deformation, which is δ = PL/AE, where L is the length, E is the modulus of elasticity, and A is the cross-sectional area.
(a) The maximum normal stress in the bar:
The maximum normal stress will occur where the force is greatest. The total force on the bar is P1 + P2 + P3 + P4 = 83 kip + 47 kip + 28.5 kip + 46 kip = 204.5 kip.
The cross-sectional area of the bar is A = (2.1 in)^2 = 4.41 in^2.
So, the maximum normal stress is σ_max = P/A = 204.5 kip / 4.41 in^2 = 46.37 ksi.
(b) The change in length of segment AB:
The force on segment AB is P1 = 83 kip.
The length of segment AB is a = 9 ft = 108 in.
So, the change in length of segment AB is δ_AB = PL/AE = 83 kip * 108 in / (4.41 in^2 * 31000 ksi) = 0.065 in.
(c) The change in length of segment BC:
The force on segment BC is P1 + P2 = 83 kip + 47 kip = 130 kip.
The length of segment BC is b = 6 ft = 72 in.
So, the change in length of segment BC is δ_BC = PL/AE = 130 kip * 72 in / (4.41 in^2 * 31000 ksi) = 0.068 in.
(d) The change in length of the complete bar:
The total length of the bar is a + b + c = 9 ft + 6 ft + 3.5 ft = 18.5 ft = 222 in.
So, the change in length of the complete bar is δ_total = PL/AE = 204.5 kip * 222 in / (4.41 in^2 * 31000 ksi) = 0.342 in.
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