A 4-point bending test is performed on the same beam, with symmetric loads of 5kN separated by 2m.
Draw the bending moment and shear force diagrams for the beam. Then answer the following questions about the shear force V and bending moment M.
What is the shear force in the beam at x=0.5,1.5,2.5 and 3.5m?
V (kN, x=0.5):
V (kN, x=1.5):
V (kN, x=2.5):
V (kN, x=3.5):
What is the moment in the beam at x=0,1,2,3 and 4m?
M (kNm, x=0):
M (kNm, x=1):
M (kNm, x=2):
M (kNm, x=3):
M (kNm, x=4):
Again assume that the beam has a width b=10cm and a height h=20cm. Calculate the maximum stress in the beam.
σmax (in MPa):
To draw the bending moment and shear force diagrams for the beam, we will need to use the equations that relate the bending moment M, the shear force V, and the distance x along the beam.
For a simply supported beam with symmetric loads, the shear force V at any point can be calculated using the following equation:
V = (Total Load / Total Length) × (Total Length / 2 - x)
In this case, the total load is 5kN, and the total length is 4m. The distance x is given as 0.5m, 1.5m, 2.5m, and 3.5m.
To calculate the moment M at any point, we can integrate the shear force equation with respect to x:
M = ∫V dx
To determine the maximum stress in the beam, we need to calculate the bending stress using the bending moment equation:
σ = (M × (h/2)) / (b × (h^2/12))
In this equation, h is the height of the beam (20cm) and b is the width of the beam (10cm).
Now, let's calculate the shear forces and bending moments at the specified positions:
V (kN, x=0.5) = (5/4) × (2 - 0.5) = 3.125 kN
V (kN, x=1.5) = (5/4) × (2 - 1.5) = 1.875 kN
V (kN, x=2.5) = (5/4) × (2 - 2.5) = -0.625 kN (negative because the shear force changes direction)
V (kN, x=3.5) = (5/4) × (2 - 3.5) = -4.375 kN (negative because the shear force changes direction)
M (kNm, x=0) = ∫V dx from x=0 to x=0.5
= ∫(5/4) × (2 - x) dx from x=0 to x=0.5
= (5/4) × [2x - (x^2/2)] from x=0 to x=0.5
= 0.625 kNm
M (kNm, x=1) = ∫V dx from x=0.5 to x=1.5
= ∫1.875 dx from x=0.5 to x=1.5
= 1.875 kNm
M (kNm, x=2) = ∫V dx from x=1.5 to x=2.5
= ∫-0.625 dx from x=1.5 to x=2.5
= -1.25 kNm
M (kNm, x=3) = ∫V dx from x=2.5 to x=3.5
= ∫-4.375 dx from x=2.5 to x=3.5
= -4.375 kNm
M (kNm, x=4) = ∫V dx from x=3.5 to x=4
= ∫-4.375 dx from x=3.5 to x=4
= -4.375 kNm
To calculate the maximum stress in the beam, substitute the values into the bending stress equation:
σmax (in MPa) = (M × (h/2)) / (b × (h^2/12))
= (4.375 × (20/2)) / (10 × (20^2/12))
= 2.2916 MPa (rounded to 4 decimal places)
Therefore, the maximum stress in the beam is approximately 2.2916 MPa.
To draw the bending moment and shear force diagrams for the beam, we can start by calculating the reactions at the supports. Since the loads are symmetric, the reactions will also be symmetric.
First, let's calculate the reactions:
Reaction at support A = Reaction at support B = (Total load)/2 = (5 kN + 5 kN)/2 = 5 kN
The shear force diagram can be obtained by taking the algebraic sum of the vertical forces (shear forces) at each location along the beam.
Now, let's calculate the shear force at each point:
V (kN, x=0.5):
Since 0.5m is within the first load, the shear force at x=0.5m is equal to the load on the left side of this point, which is 5 kN.
V (kN, x=1.5):
Since 1.5m is between the two loads, the shear force at x=1.5m is equal to the reaction at support A, which is 5 kN.
V (kN, x=2.5):
Since 2.5m is within the second load, the shear force at x=2.5m is equal to the load on the right side of this point, which is 5 kN.
V (kN, x=3.5):
Since 3.5m is after the second load, the shear force at x=3.5m is equal to the reaction at support B, which is 5 kN.
For the bending moment, we can integrate the shear force function to obtain the bending moment equation. Since the loads are symmetric, the bending moment diagram will also be symmetric.
Now, let's calculate the bending moment:
M (kNm, x=0):
The bending moment at x=0m is equal to zero since this is the starting point of the beam.
M (kNm, x=1):
The bending moment at x=1m is the integral of the shear force from x=0m to x=1m. The result is 5 kNm.
M (kNm, x=2):
The bending moment at x=2m is the integral of the shear force from x=0m to x=2m. The result is 5 kNm.
M (kNm, x=3):
The bending moment at x=3m is the integral of the shear force from x=0m to x=3m. The result is 10 kNm.
M (kNm, x=4):
The bending moment at x=4m is the integral of the shear force from x=0m to x=4m. The result is 15 kNm.
To calculate the maximum stress in the beam, we need to know the flexural formula:
σ = (M * c) / (I * y)
where:
σ = stress
M = bending moment
c = distance from the neutral axis to the extreme fiber
I = moment of inertia of the cross-sectional area
y = distance from the neutral axis to the point of interest
For a rectangular beam, the moment of inertia (I) is given by:
I = (b * h^3) / 12
where:
b = width of the beam
h = height of the beam
In this case, b = 10 cm and h = 20 cm.
Now, let's calculate the maximum stress:
σmax (in MPa):
To calculate the maximum stress, we need to know the distance from the neutral axis to the extreme fiber (c). For a rectangular beam, this distance is equal to half the height (h/2).
c = h/2 = 20 cm / 2 = 10 cm
Substituting the values into the flexural formula:
σmax = (M * c) / (I * y)
Since we want the maximum stress, we can assume that the point of interest is at the extreme fiber, which means y = h/2.
σmax = (15 kNm * 10 cm) / ((10 cm * 20 cm^3) / 12)
Simplifying the equation and converting the units:
σmax = 1.8 MPa