The second moment of area of a rectangle through its centroid is given by:
bL3/12
Using the approximation for binomial series valid for small x (+ve or –ve):
(1+x)^n=1+nx/1!+(n(n-1) x^2)/2!+⋯ 1+nx
Determine the approximate change in the second moment of area if b is increased by 1.5% and L is reduced by 2.0%
please help, this is sample question of assessment which I have tomorrow.
Thanks
If M = bL^3/12
replace b by 1.015b
and L by 0.98L
and you have
(1.015b)(0.98L)^3/12
= (1.015*0.98^3)bL^3/12
= 0.955M
so, M has decreased by 4.5%
I see no reason to resort to binomial approximation. But, if you want to try it, and compare results, feel free.
To find the approximate change in the second moment of area, we can use the binomial approximation formula provided. Let's break down the given formula first:
The second moment of area of a rectangle through its centroid is given by:
I = (b * L^3) / 12
where b is the width of the rectangle and L is the length.
Now, let's use the binomial approximation formula to calculate the change in I when b is increased by 1.5% and L is reduced by 2.0%.
For the change in b:
Let x be the change in b. Since the change is given as a percentage, we have x = 0.015b (1.5% = 0.015).
Similarly, for the change in L:
Let y be the change in L. Since the change is given as a percentage, we have y = -0.02L (-2.0% = -0.02).
Now, let's substitute these values back into the binomial approximation formula:
(I + ΔI) = ((b + x)(L + y)^3) / 12
≈ (bL^3/12) + [(3bL^2y) / 12] + [(bL^2x) / 12]
≈ I + (0.25bL^2x) + (0.25bL^2y)
Substitute the values for x and y:
≈ I + (0.25bL^2 * 0.015) + (0.25bL^2 * -0.02)
≈ I + (0.00375bL^2) - 0.005bL^2
Therefore, the approximate change in the second moment of area, ΔI, is:
ΔI = (0.00375bL^2) - 0.005bL^2
You can now substitute the values of b and L into the equation to find the numerical value of ΔI.
To determine the approximate change in the second moment of area, we can use the given formula and the approximation for the binomial series.
The second moment of area of a rectangle through its centroid is given by "bL^3/12". Let's call this value "I", for simplicity.
Now, we need to find the change in I when b is increased by 1.5% and L is reduced by 2.0%.
Let's start by finding the value of I when b and L are their original values. Let's call this value "I_0".
I_0 = bL^3/12
Now, let's find the value of I when b is increased by 1.5%. The new value of b can be calculated as follows:
b_new = b + (1.5/100) * b
Similarly, let's find the value of I when L is reduced by 2.0%. The new value of L can be calculated as follows:
L_new = L - (2.0/100) * L
Now, we can calculate the new value of I, let's call it "I_new", using the modified values of b and L:
I_new = b_new * L_new^3/12
= (b + (1.5/100) * b ) * (L - (2.0/100) * L)^3/12
To calculate this approximation using the binomial series, let's rewrite it as:
I_new = (1 + (1.5/100)) * (L^3/12) * (1 - (2.0/100))^3
Now, we can apply the approximation for the binomial series to the expression (1 - (2.0/100))^3:
(1 - (2.0/100))^3 ≈ 1 + 3 * (-2.0/100) + (3 * 2 * (-2.0/100)^2)/2
Substituting this approximation back into the expression for I_new, we get:
I_new ≈ (1 + (1.5/100)) * (L^3/12) * (1 + 3 * (-2.0/100) + (3 * 2 * (-2.0/100)^2)/2)
This expression represents the approximate change in the second moment of area.
To get the numerical value, you need to substitute the actual values of b, L, and evaluate this expression using a calculator or algebra tools.
Note: The approximation for the binomial series becomes more accurate as the value of x becomes smaller. In this case, the approximation is valid since the changes in b and L are relatively small.