The statically-indeterminate composite beam AB of length 2L is fixed at A (x=0) and is simply supported at B (x=2L). The beam is composed of a core of modulus E0 of constant square cross section of dimensions 2h0×2h0 bonded inside a sleeve of modulus 2E0 and constant square cross section of outer dimensions 4h0×4h0, as indicated in the figure. The beam is loaded by a downward concentrated load of magnitude P applied at the midspan of the beam (x=L) as indicated.
The KNOWN quantities in this problem are P[N], L[m], h0[m], and E0[Pa]. In symbolic expressions, enter P, L, h0 and E0 as P, L, h_0 and E_0, respectively.
1)Consider the SD companion problem obtained by selecting the roller at B as the redundant support, and obtain the bending moment resultant M(x) in terms of x, L, the unknown redundant reaction RBy (enter this as R_y^B), and P.
for 0≤x≤L,M(x)=
for L≤x≤2L,M(x)=
2)Obtain a symbolic expression for the redundant reaction RBy in terms of P.
RBy=
3)Obtain symbolic expressions for the curvature of the beam at the fixed support A, (1ρ(x=0)), and at the midspan of the beam, (1ρ(x=L)), in terms of P, L, h0, and E0.
1ρ(x=0)=
1ρ(x=L)=
4)Obtain symbolic expressions in terms of the known quantities for the maximum tensile stresses in the core and in the sleeve on the x=0 cross section, and indicate at what y each of them occurs. Express your answers in terms of P, L, and h0.
σmaxcore=
at y= -
σmaxsleeve=
at y=
hamare samajh me angreji nhi aati
Looks like a course 2 MITx problem to me. You are on your own.
To answer these questions, we need to analyze the beam and apply the appropriate equations and principles. Here are the step-by-step explanations to find the solutions:
1) To obtain the bending moment resultant M(x) in terms of x, L, RBy, and P in the given range,
a) For 0 ≤ x ≤ L, the beam is loaded with a concentrated load at midspan (x = L), resulting in a symmetric distribution. The corresponding bending moment can be obtained using the equation for simply supported beams. The moment at any point x in this range is M(x) = (P/2)(L-x).
b) For L ≤ x ≤ 2L, the beam is still subjected to the concentrated load at midspan (x = L), but the beam behavior changes due to the fixed support at A. The bending moment at any point x in this range can be obtained by using the equation for a fixed-fixed beam. The moment equation in this range is M(x) = (P/2)(x-L).
2) To obtain a symbolic expression for the redundant reaction RBy in terms of P,
a) Since the roller at B is selected as the redundant support, it implies that its vertical reaction, RBy, is zero. Therefore, RBy = 0.
3) To obtain symbolic expressions for the curvature of the beam at the fixed support A (x = 0) and at the midspan (x = L) in terms of P, L, h0, and E0,
a) The curvature, ρ, is related to the bending moment, M, by the equation ρ = M / (EI), where E is the modulus of elasticity and I is the moment of inertia of the cross-section.
b) At x = 0, the bending moment can be obtained by substituting x = 0 in the equation in step 1a: M(x=0) = (P/2)(L-0) = (P/2)L.
The moment of inertia for the composite section can be calculated using the parallel axis theorem: I = Icore + Isleeve, where Icore = (h0^3)/12 and Isleeve = (4h0^3)/12.
Substituting the values, we can find 1ρ(x=0) = (M(x=0)) / (EI) = ((P/2)L) / (E0*((h0^3)/12 + (4h0^3)/12)).
c) At x = L, the bending moment can be obtained by substituting x = L in the equation in step 1a: M(x=L) = (P/2)(L-L) = 0.
The moment of inertia remains the same as in step 3b: I = Icore + Isleeve.
Substituting the values, we can find 1ρ(x=L) = (M(x=L)) / (EI) = 0.
4) To obtain symbolic expressions for the maximum tensile stresses in the core and in the sleeve on the x = 0 cross-section,
a) At x = 0 cross-section, the maximum tensile stress in the core occurs at the outermost fiber (y = h0) and is given by the formula σmaxcore = (My * h0) / (Icore), where My = M(0) = (P/2)L. Substitute the values to find σmaxcore.
b) At x = 0 cross-section, the maximum tensile stress in the sleeve occurs at the outermost fiber (y = 2h0) and is given by the formula σmaxsleeve = (My * h0) / (Isleeve), where My = M(0) = (P/2)L. Substitute the values to find σmaxsleeve.
Note: It may be helpful to check the figure provided in the problem to ensure the correct application of equations and values.