A bar has a constant cross sectional area A. The bar is devided in 2 segments segment 1 (1/3 of the bar) and segment 2 (2/3 of the bar). Segment 1 is made of a material with mass density ρ and Young's modulus E. Segment 2 is made of a different material, with density 0,5*ρ and Young's modulus 2E. The bar is fixed at the floor at point C, at x=L, and at the ceiling, B, at x=0. Total bar has a length of L.
Note 1: The bar deforms under its own weight. For a material with density ρ, gravity results in a load per unit volume ρ*g.
Note 2: The gravity load on both segments of the bar as densities are different but comparable.
Note 3: No other load is applied.
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Question 1:
Find a symbolic expression for the distributed load per unit length due to gravity (use the terms of ρ, g & A)
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Hint:
You have to find 2 expressions(!)
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My comment:
If i'm not mistaken the first expression is for 0≤x<L/3 and second for L/3<x≤L
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Question 2:
Find a symbolic expression for the reaction RCx at the support C (use the terms ρ, g, A & L)
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Question 3:
Find a symbolic expression for the axial force in the bar N(x) (use the terms ρ, g, A, L & x)
Question 1:
To find the distributed load per unit length due to gravity on segment 1 (0 ≤ x < L/3), we can use the equation:
Load per unit volume = ρ * g
Since segment 1 has a mass density of ρ, the load per unit volume on segment 1 is ρ * g.
Since the cross-sectional area of the bar is constant at A, we can multiply the load per unit volume by A to get the distributed load per unit length on segment 1:
Distributed load per unit length on segment 1 = ρ * g * A
For segment 2 (L/3 < x ≤ L), the material has a density of 0.5 * ρ, so the load per unit volume on segment 2 is 0.5 * ρ * g. Multiplying this by the cross-sectional area A gives us the distributed load per unit length on segment 2:
Distributed load per unit length on segment 2 = 0.5 * ρ * g * A
Therefore, the symbolic expressions for the distributed load per unit length due to gravity on segment 1 and segment 2 are:
For 0 ≤ x < L/3: ρ * g * A
For L/3 < x ≤ L: 0.5 * ρ * g * A
Question 2:
To find the reaction RCx at the support C, we need to consider the equilibrium of forces acting on the bar at support C.
The only vertical force acting on the bar at support C is the reaction RCy. The horizontal forces at support C are balanced (since the bar is fixed at C), so we don't need to consider them.
Since the bar deforms under its own weight, we can consider the weight of each segment as a distributed load per unit length.
For segment 1 (0 ≤ x < L/3), the distributed load per unit length due to gravity is ρ * g * A, which acts in the downward direction.
For segment 2 (L/3 < x ≤ L), the distributed load per unit length due to gravity is 0.5 * ρ * g * A, which also acts in the downward direction.
To find the reaction RCy, we need to integrate the distributed load per unit length over the length of each segment and balance it with the reaction RCy. However, since there is no horizontal force acting on the bar at support C, the reaction RCx is zero.
Therefore, the symbolic expression for the reaction RCx at the support C is:
RCx = 0
Question 3:
To find the axial force in the bar N(x), we need to consider the equilibrium of forces acting on an infinitesimal segment of the bar at position x.
For 0 ≤ x < L/3, the axial force in segment 1 is equal to the integral of the distributed load per unit length over the length of segment 1. The distributed load per unit length on segment 1 is ρ * g * A, and the length of segment 1 is L/3.
Therefore, for 0 ≤ x < L/3, the axial force in segment 1 is:
N(x) = ∫[0 to x] (ρ * g * A) * (L/3) dx
For L/3 < x ≤ L, the axial force in segment 2 is equal to the integral of the distributed load per unit length over the length of segment 2. The distributed load per unit length on segment 2 is 0.5 * ρ * g * A, and the length of segment 2 is 2L/3.
Therefore, for L/3 < x ≤ L, the axial force in segment 2 is:
N(x) = ∫[L/3 to x] (0.5 * ρ * g * A) * (2L/3) dx
Note that the limits of integration change depending on the segment of the bar you are considering.
Therefore, the symbolic expression for the axial force in the bar N(x) is:
For 0 ≤ x < L/3: N(x) = ∫[0 to x] (ρ * g * A) * (L/3) dx
For L/3 < x ≤ L: N(x) = ∫[L/3 to x] (0.5 * ρ * g * A) * (2L/3) dx
To find the expressions for the distributed load per unit length due to gravity for the two segments of the bar, we need to consider the gravitational load acting on each segment.
Question 1:
Expression for the distributed load per unit length due to gravity in Segment 1 (0 ≤ x < L/3):
For Segment 1, the density is ρ and the gravitational load per unit volume is ρ * g. Since the cross-sectional area is constant, the distributed load per unit length can be calculated by multiplying the gravitational load per unit volume by the cross-sectional area, giving us:
Load1(x) = ρ * g * A
Expression for the distributed load per unit length due to gravity in Segment 2 (L/3 < x ≤ L):
For Segment 2, the density is 0.5 * ρ and the gravitational load per unit volume is (0.5 * ρ) * g. Again, multiplying it by the cross-sectional area, we get:
Load2(x) = (0.5 * ρ) * g * A
Question 2:
To find the reaction RCx at the support C, we need to consider the equilibrium of forces. Since there is no applied load, the reaction force at C must balance the gravitational loads on the bar.
Considering Segment 1 (0 ≤ x < L/3), the reaction RCx can be calculated by integrating the distributed load Load1(x) over the length of Segment 1:
RCx = ∫ [0 to L/3] {ρ * g * A} dx
RCx = ρ * g * A * ∫ [0 to L/3] dx
RCx = ρ * g * A * [x] [0 to L/3]
RCx = ρ * g * A * (L/3 - 0)
RCx = ρ * g * A * (L/3)
Question 3:
To find the axial force in the bar N(x), we need to consider the deformation of the bar under its own weight. The axial force is related to the strain and Young's modulus of the material.
Considering Segment 1 (0 ≤ x < L/3), the axial force N(x) can be calculated using Hooke's Law:
N(x) = E * A * ε(x)
ε(x) = (δL(x) / L)
δL(x) = Displacement of Segment 1 at x
Therefore, δL(x) is equal to the elongation of Segment 1 due to gravity. Integrating the distributed load Load1(x) over the length of Segment 1, we get:
δL(x) = ∫ [0 to x] {Load1(x) / (E * A)} dx
δL(x) = ∫ [0 to x] {(ρ * g * A) / (E * A)} dx
δL(x) = (ρ * g) / E * ∫ [0 to x] dx
δL(x) = (ρ * g * x) / E
Now, substituting δL(x) into the equation for ε(x):
ε(x) = (ρ * g * x) / (E * L)
Therefore, the symbolic expression for the axial force in Segment 1 (0 ≤ x < L/3) is:
N(x) = E * A * (ρ * g * x) / (E * L)
N(x) = (ρ * g * A * x) / L
For Segment 2 (L/3 < x ≤ L), the axial force N(x) can be calculated in a similar way. However, since the Young's modulus is 2E, the expression becomes:
N(x) = (2E * A * (0.5 * ρ * g * x)) / L
N(x) = (E * A * ρ * g * x) / L
Note that these expressions for N(x) represent the axial force at any given point x along the bar.