The maximum safe load of a rectangle beam varies jointly as its width and the square of its depth and inversely as the length of the beam. If a beam 8.00 in. wide, 11.5 in. Deep, and 16.0 ft long can safely support 15.000 lb, find the safe load for a beam 6.50 in. Wide, 13.4 in. Deep, and 21.0 ft long made of the same material.
SL=15 * (6.5/8) * (13.4/11.5)^2*(16/21)= 12.61 Lbs.
To find the safe load for the second beam, we can use the concept of joint variation and inverse variation.
Given the following information:
Width of the first beam (w₁) = 8.00 in.
Depth of the first beam (d₁) = 11.5 in.
Length of the first beam (l₁) = 16.0 ft = 192.0 in.
Safe load of the first beam (load₁) = 15,000 lb.
We can set up the following proportion using joint variation:
(load₁) / (w₁ * d₁² * l₁) = (load₂) / (w₂ * d₂² * l₂)
where:
w₂ = width of the second beam = 6.50 in.
d₂ = depth of the second beam = 13.4 in.
l₂ = length of the second beam = 21.0 ft = 252.0 in.
load₂ = safe load for the second beam (to be determined).
Now, we can substitute the given values into the proportion and solve for load₂.
(15,000 lb) / (8.00 in. * (11.5 in.)² * 192.0 in.) = (load₂) / (6.50 in. * (13.4 in.)² * 252.0 in.)
Simplifying the equation:
15,000 lb / (2196.00 in³) = (load₂) / (9447.24 in³)
Cross-multiplying:
(15,000 lb) * (9447.24 in³) = (load₂) * (2196.00 in³)
(load₂) = (15,000 lb * 9447.24 in³) / (2196.00 in³)
Calculating the value:
(load₂) = 64,620.60 lb
Therefore, the safe load for the second beam is approximately 64,620.60 lb.
To solve this problem, we can use the concept of joint variation, which states that when a quantity varies jointly with two or more other quantities, it is proportional to the product of those quantities.
Let's represent the maximum safe load as "L", width as "w", depth as "d", and length as "l". Therefore, we have the following proportional equation:
L ∝ w * d^2 / l
To find the constant of variation, we need to use the given data. We know that a beam 8.00 in. wide, 11.5 in. deep, and 16.0 ft long can safely support 15,000 lb. Plugging in these values into our equation, we have:
15,000 = 8 * (11.5)^2 / 16
Now, we can calculate the constant of variation:
k = 15,000 * 16 / (8 * 11.5^2)
k ≈ 62.609
With the constant of variation, we can find the safe load for the second beam with the given dimensions: 6.50 in. wide, 13.4 in. deep, and 21.0 ft long.
L = k * w * d^2 / l
L = 62.609 * 6.50 * (13.4)^2 / 21.0
Calculating this expression, we find:
L ≈ 16,668.61 lb
Therefore, the safe load for a beam 6.50 in. wide, 13.4 in. deep, and 21.0 ft long is approximately 16,668.61 lb.