F13Q1_1: COMPOSITE BAR UNDER DISTRIBUTED LOADING
The composite bar AB has length L and is fixed at A(x=0).
The bar has a constant cross-section and is composed of two steel plates, of modulus Es=2E0 sandwiching a copper core of modulus Ec=E0.
The steel plates and the copper core have equal width w and equal height h.
The bar is subjected to a distributed load per unit length of the bar fx(x)=p0(xL)13.
The given known quantities are:
L=2m
h=3cm
w=2cm
E0=100GPa
p0=1000kN/m
Helpful hint:
∫xndx=xn+1n+1+Constant
Obtain a symbolic expression for the axial force resultant, N(x), in terms of L, x and p0 (enter this as p_0):
N(x)=
Obtain the numerical value of the axial force resultant at A, N(x=0), in kN.
N(x=0)=
Obtain a symbolic expression for the elongation of bar AB, δAB, in terms of L, h, w, E0, and p0.
δAB=
Obtain the numerical value of the elongation of bar AB, δAB, in mm.
δAB=
No (former MIT professor), I do not do MITx problems in Course 8 or Course 2. You are on your own. The 8.01x crowd swamped this board the last time around.
Gotcha, I got the other problems fine this ones really troubling me so thought I'd take a shot , atleast try to get closer to what I'm doing wrong. Thanks for the response and I completely understand !:)
To obtain the symbolic expression for the axial force resultant, N(x), in terms of L, x, and p0:
1. Start with the formula for the axial force resultant in a composite bar under distributed loading:
N(x) = ∫[0,x] (fx(x) * A(x)) dx
where fx(x) is the distributed load per unit length of the bar, and A(x) is the cross-sectional area of the bar at position x.
2. Substitute the given values for fx(x) and A(x):
N(x) = ∫[0,x] (p0(x/L)^13 * A(x)) dx
3. Determine the cross-sectional area of the composite bar at position x:
The cross-sectional area of each steel plate is w * h, and the cross-sectional area of the copper core is w * h. Since they are sandwiched together, the total cross-sectional area at position x is:
A(x) = 2 * (w * h) + (w * h) = 4 * (w * h)
4. Substitute the expression for A(x) into the integral:
N(x) = ∫[0,x] (p0(x/L)^13 * 4 * (w * h)) dx
Now, simplify this expression using integration techniques, and you will obtain the symbolic expression for N(x).
To obtain the numerical value of the axial force resultant at A, N(x=0), in kN:
Plug x=0 into the symbolic expression for N(x) and evaluate the result in kN.
To obtain the symbolic expression for the elongation of bar AB, δAB, in terms of L, h, w, E0, and p0:
1. Start with Hooke's law for linear elasticity:
δAB = (N(x) * L) / (A(x) * E),
where N(x) is the axial force resultant at position x, L is the length of the bar, A(x) is the cross-sectional area at position x, and E is the modulus of elasticity.
2. Substitute the expressions for N(x) and A(x) derived earlier:
δAB = [(p0(x/L)^13 * 4 * (w * h)) * L] / [(4 * (w * h)) * E0],
3. Simplify the expression:
δAB = (p0 * L^14 * (x/L)^13) / (E0)
To obtain the numerical value of the elongation of bar AB, δAB, in mm:
Plug in the given numerical values of L, h, w, E0, and p0 into the symbolic expression for δAB.