A cantilever beam, shown below, is subjected to two random loads, S_1 and S_2, t

Question

A cantilever beam, shown below, is subjected to two random loads, S_1 and S_2, that are statistically independent with respective means and standard deviations of mu_1, sigma_1 and mu_2 sigma_2. The shear force Q and bending moment M at the fixed support of the beam are both functions of the two loads, respectively, as follows: Q = S_1 + S_2 M = aS_1 + 2 aS_2 a. Q and M are also random variables. Write expressions for their means and variances. b. Find E[QM] in terms of mu_1, sigma_1, mu_2, and sigma_2. c. Find the covariance and corresponding correlation coefficient for Q and M.

Explanation / Answer

Solution

Back-up Theory

Let X and Y be two random variables and let E, V, SD and C respectively stand for Expected Value or Expectation i.e., mean (average), Variance, Standard Deviation and Covariance.

E(aX) = aE(X); E(bY) = bE(Y); E(aX + bY) = aE(X) + bE(Y) …………………………(1)

In particular, E(X + Y) = E(X) + E(Y) ………………………..…………………………(2)

V(aX) = a2V(X); V(bY) = b2V(Y); V(aX + bY) = a2V(X) + b2V(Y) + 2abC(X,Y)……..(3) In particular, V(X + Y) = V(X) + V(Y) + 2C(X,Y)……..………………….……………(4)

                      V(X - Y) = V(X) + V(Y) - 2C(X,Y)……..………………….……………(5)

V(X) = E(X2) – {E(X)}2 …………………………………………………………………(6)

C(X, Y) = E(XY) – {E(X).E(Y)} ………………………..………………………………(7)

Special Case: When X and Y are Statistically Independent (or can also be said as simply Independent),

E(XY) = E(X)E(Y); and C(X, Y) = 0 ……………………………………………………(8)

and hence      V(aX + bY) = a2V(X) + b2V(Y) ………………………………………… (9)

In particular, V(X + Y) = V(X) + V(Y) ……..…………….……………….……………(10)

                      V(X - Y) = V(X) + V(Y) ……..…………………………….……………(11)

Now, to work out the solution,

Given E(S1) = µ1, E(S2) = µ2, V(S1) = 12 , V(S2) = 22 ; S1 and S2 are statistically independent; Q = S1 + S2 and M = aS1 + 2aS2

Part (a)

E(Q) = E(S1 + S2) = E(S1) + E(S2) = µ1 + µ2 [vide (2) under Back-up Theory]

V(Q) = V(S1 + S2) = V(S1) + V(S2) = 12 + 22 [vide (10) under Back-up Theory]

E(M) = E(aS1 + 2aS2) = aE(S1) + 2aE(S2) = aµ1 + 2aµ2 [vide (1) under Back-up Theory]

V(M) = V(aS1 + 2aS2) = a2V(S1) + 4a2V(S2) = a212 + 4a222 [vide (9) under Back-up Theory]

DONE

Part (b)

E(QM) = E{(S1 + S2)(aS1 + 2aS2)} = E(aS12 + 2aS1S2 + aS1S2 + 2aS22)

= aE(S12) + 3aE(S1S2) + 2aE(S22)

= aE(S12) + 3aE(S1)E(S2)+ 2aE(S22) [vide (8) under Back-up Theory]

= a[V(S1) + {E(S1)}2] + 3aE(S1)E(S2)+ 2a[V(S2) + {E(S2)}2]

= a(12 + µ12) + 3aµ1µ2 + 2a(22 + µ22) = a12 + 2a22 + aµ12 + 2aµ22 + 3aµ1µ2.

DONE

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