Bar BC in the figure has length L, constant cross sectional area A, and is composed of a homogeneous material with modulus E. The bar is fixed between walls at B (x=0) and C (x = L). The bar is subjected to a variable distributed load per unit length, oriented in the direction indicated in the figure. The magnitude of the distributed load, p(x)=bx, linearly increases from B to C, with a known value for the constant parameter b. Note that b has dimensions [Nm2], so that p(x) has the desired dimensions [Nm].

You will use the force method, taking the wall at B as the redundant support, to solve this SI problem.

The KNOWN problem parameters are L[m], A[m2], E[Pa], and b[Nm2].

1) Obtain a symbolic expression for the axial force resultant along the bar, N(x), in terms of x, b, and of the unknown redundant reaction RxB (enter this as R_B):

N(x)=

2) Obtain a symbolic expression for the axial strain in the bar, ϵa(x) in terms of x, b, E, A, and of the unknown redundant reaction RxB (enter this as R_B):

ϵa(x)=

3) Obtain the redundant reaction, RxB in terms of b and L:

RxB=

4) Obtain a symbolic expression for the axial strain in the bar, ϵa(x) in terms of x, and of the known problem parameters b, E, L and A:

ϵa(x)=

5) Obtain expressions for the axial strain at the two ends of the bar (in terms of the known problem parameters b, E, L and A), and for the position x0 along the bar where the axial strain goes to zero (ϵa(x0)=0), (in terms of L):

ϵa(x=0)=
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ϵa(x=L)=
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ϵa(x0)=0atx0=

6) Obtain a symbolic expression for the displacement field, ux(x), in terms of x, and of the known problem parameters b, E, L and A:

ux(x)=

7) Obtain a symbolic expression for the position where the magnitude of the displacement ux is maximum (in terms of L). You might want to use the strain field along the bar for guidance.

xmax=

8) Obtain a symbolic expression for the maximum magnitude (absolute value) of displacement along the bar, umax=|ux|max, in terms of the known problem parameters b, E, L and A:

umax=

To solve this problem using the force method, let's analyze each step one by one:

1) The axial force resultant along the bar, N(x), can be obtained by integrating the distributed load p(x) over the length of the bar:

N(x) = ∫(0 to x) p(x') dx'

Since the distributed load is given as p(x) = bx, we can substitute this into the equation:

N(x) = ∫(0 to x) bx dx'

Taking the integral, we get:

N(x) = (1/2)bx^2 + constant

2) The axial strain in the bar, ϵa(x), can be obtained by dividing the axial force resultant N(x) by the product of modulus E and the constant cross-sectional area A:

ϵa(x) = N(x) / (E * A)

Substituting the expression for N(x) obtained in Step 1:

ϵa(x) = (1/2)bx^2 / (E * A) + constant / (E * A)

3) To obtain the unknown redundant reaction RxB, we need to analyze the equilibrium of forces at the fixed support B. Considering only the axial forces, we have:

RxB = -N(0)

Substituting the expression for N(x) obtained in Step 1:

RxB = -(1/2)b(0)^2 - constant = 0

Simplifying the equation:

RxB = 0

4) To obtain the axial strain in terms of only the known problem parameters, we can substitute the expression for N(x) obtained in Step 1 into the equation for ϵa(x):

ϵa(x) = (1/2)bx^2 / (E * A) + constant / (E * A)

Simplifying:

ϵa(x) = (1/2)bx^2 / (E * A)

5) To obtain the axial strain at the two ends of the bar and the position x0 where ϵa(x0) = 0, we substitute the corresponding values of x:

ϵa(x=0) = (1/2)b(0)^2 / (E * A) = 0

ϵa(x=L) = (1/2)b(L)^2 / (E * A)

To find x0 where ϵa(x0) = 0, we set the expression for ϵa(x) equal to zero:

(1/2)b(x0)^2 / (E * A) = 0

Solving for x0:

x0 = 0

6) The displacement field, ux(x), can be obtained by integrating the axial strain ϵa(x) with respect to x:

ux(x) = ∫(0 to x) ϵa(x') dx'

Substituting the expression for ϵa(x) obtained in Step 4:

ux(x) = ∫(0 to x) (1/2)bx'^2 / (E * A) dx'

Taking the integral, we get:

ux(x) = (1/6)bx^3 / (E * A) + constant

7) To find the position where the magnitude of the displacement ux is maximum, we can analyze the strain field along the bar. From the expression for ϵa(x) obtained in Step 4, we can see that the axial strain increases quadratically with x. Therefore, the maximum magnitude of the displacement ux will occur at the end of the bar, x = L.

xmax = L

8) The maximum magnitude of displacement along the bar, umax = |ux|max, can be obtained by substituting x = L into the expression for ux(x) obtained in Step 6:

umax = |ux|max = (1/6)bL^3 / (E * A) + constant