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].
Find:
1: expression for the axial force resultant along the bar, N(x), in terms of x, b, and of the unknown redundant reaction R_xB
2. expression for the axial strain in the bar, ϵa(x) in terms of x, b, E, A, and of the unknown redundant reaction R_xB
3. redundant reaction, R_xB in terms of b and L
4. 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
5. 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 (ϵ_a(x0)=0), (in terms of L):
ϵ_a(x=0)=
ϵ_a(x=L)=
ϵ_a(x_0=0) at x_0=0 =
6. expression for the displacement field, u(x), in terms of x, and of the known problem parameters b, E, L and A
7. expression for the position where the magnitude of the displacement ux is maximum (in terms of L).
8. 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:
To find the expressions for the various quantities, we can use basic principles of mechanics, specifically axial force and strain equations.
1. The axial force resultant along the bar, N(x), can be found by integrating the distributed load function p(x) over the length of the bar. Since p(x) = bx, the integration can be written as:
N(x) = ∫p(x) dx = ∫(bx) dx = (b/2)x^2 + C1
where C1 is the constant of integration. Since the bar is fixed at B (x=0), the redundant reaction R_xB provides an additional force at that point. Therefore, we can write N(x=0) = -R_xB.
2. The axial strain in the bar, ϵa(x), can be found by dividing the axial deformation (change in length) by the original length of the bar. We can relate the axial deformation to the axial force using Hooke's law, which states that stress (σ) is equal to the product of Young's modulus (E) and strain (ϵ). Therefore, we can write:
ϵa(x) = N(x)/(A*E)
Substituting the expression for N(x) from step 1, we get:
ϵa(x) = [(b/2)x^2 + C1]/(A*E)
3. The redundant reaction R_xB can be found by solving for N(x=0) = -R_xB using the expression obtained in step 1.
4. The expression for the axial strain in the bar, ϵa(x), in terms of x, b, E, L, and A can be obtained by substituting the known problem parameters into the equation derived in step 2.
5. To find the axial strain at the two ends of the bar, ϵa(x=0) and ϵa(x=L), we can substitute the corresponding values of x into the expression obtained in step 4. Additionally, to find the position x0 along the bar where the axial strain goes to zero (ϵ_a(x0)=0), we can set ϵa(x) = 0 in the expression obtained in step 4 and solve for x0.
6. The displacement field, u(x), can be found by integrating the expression for the axial strain ϵa(x) over the length of the bar. Since ϵa(x) is a function of x, integrating with respect to x will give us the displacement as a function of x. Therefore, we have:
u(x) = ∫ϵa(x) dx
Substituting the expression for ϵa(x) derived in step 4, we get the displacement field in terms of x, b, E, L, and A.
7. To find the position where the magnitude of the displacement ux is maximum, we can differentiate the expression for u(x) from step 6 with respect to x and set it equal to zero. Then, solve for x to find the position where the derivative is zero.
8. The maximum magnitude of displacement along the bar, umax=|ux|max, can be found by substituting the value of x obtained in step 7 into the expression for u(x) from step 6 and taking the absolute value.