1. A carpenter has to construct a rectangular window frame with dimensions 4 feet by 3 feet. If the frame is to be made from 2-inch by 4-inch wooden planks, how many planks will be required?
Solution:
The perimeter of the rectangular window frame is 2(4+3) = 14 feet. The length of each plank is 4 feet and the width is 2 inches (0.1667 feet).
So, the number of planks required for the length is 14/4 = 3.5, which can be rounded up to 4 planks.
The number of planks required for the width is 2(3)/0.1667 = 36 planks. Therefore, the total number of planks required is 4+36 = 40 planks.
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2. A wooden beam is 12 feet long and 4 inches wide. If it is to support a load of 500 pounds, what is the maximum allowable deflection of the beam if the allowable stress is 1,200 psi?
Solution:
The moment of inertia of the beam is I = (1/12)(0.333)(0.04167)^3 = 1.2218x10^-7 ft^4
The maximum bending moment that the beam can withstand is M = (500 lb)(12 ft) = 6,000 lb-ft.
The maximum allowable stress is σ = 1,200 psi.
The maximum allowable deflection can be calculated as:
δ = (5/384)(6,000 lb-ft)(12 ft)^3 / (1.2218x10^-7 ft^4)(1,200 psi) = 0.084 inches
3. A wooden staircase has 20 steps, each step being 8 inches high and 10 inches deep. If the staircase has to be built to reach a height of 16 feet, what is the total length of the staircase?
Solution:
The height of the staircase is 16x12 = 192 inches. The total height of the 20 steps is 20x8 = 160 inches.
Therefore, the height of the vertical space between the last step and the top of the staircase is 192-160 = 32 inches.
The length of each step is 10 inches, so the total length of the horizontal part of the staircase is 20x10 = 200 inches.
The total length of the staircase is the hypotenuse of a right triangle with height 32 inches and base 200 inches:
L = sqrt((32)^2 + (200)^2) = 202.5 inches
4. A wooden plank is 10 feet long and 6 inches wide. If it is to be used as a shelf to support a load of 100 pounds, what is the maximum allowable stress if the maximum deflection is limited to 0.1 inches?
Solution:
The moment of inertia of the plank is I = (1/12)(0.5)(0.0833)^3 =
1.2237x10^-5 ft^4
The maximum bending moment that the plank can withstand is M = (100 lb)(10 ft) = 1,000 lb-ft.
The maximum allowable deflection is δ = 0.1 inches = 0.00833 feet.
The maximum allowable stress can be calculated as:
σ = (M/I)(c/y) = (1,000 lb-ft)(0.0417 ft) / (1.2237x10^-5 ft^4)
(0.5x0.0833 ft) = 759 psi
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5. A wooden column is 12 feet long and has a cross-sectional area of 30 square inches. If the column is to support a load of 10,000 pounds, what is the maximum allowable stress if the maximum allowable deflection is limited to 0.25 inches?
Solution:
The maximum allowable deflection is δ = 0.25 inches = 0.0208 feet.
The maximum allowable stress can be calculated using the Euler formula for the critical load: Pcr = π^2EI/(KL)^2 where Pcr is the critical load, E is the modulus of elasticity, I is the moment of inertia, K is the effective length factor, and L is the length of the column. Assuming that the column is fixed at both ends, the effective length factor is K = 0.5.
The moment of inertia of the column is I = (1/4)π(0.5)^4 = 0.0491 ft^4.
The modulus of elasticity for wood is typically around 1.5x10^6 psi.
The critical load can be calculated as:
Pcr = π^2(1.5x10^6 psi)(0.0491 ft^4)/(0.5x12 ft)^2 = 3,527 lbs
The maximum allowable stress can be calculated as:
σ = (10,000 lbs)/(0.0208 ft)(30 in^2) = 15,152 psi
Since the critical load is less than the load the column needs to support, the column will buckle. Therefore, the maximum allowable stress is limited by the critical stress, which is:
σ = Pcr/(30 in^2) = 117.6 psi
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