Paul C. answered • 09/11/21
Bachelor of Mechanical Engineering with Honors
The weight W creates a tension of W in the cable, which exerts a 12kN upward force on the beam.
The pinned joints at each end can exert Forces in the x and y direction, but no concentrated moments.
By applying Newton's first law of motion in the x-direction, the sum of the x-components of the reaction forces must be 0. For beams in pure bending, you can set both of those forces equal to 0.
Thus: a free body diagram on the system would show F_A, the reaction force at A up, 8kN down at C, 12kN up at D, 8kN down at E, and F_B, the reaction force at B up.
Applying Newton's first law in the y-direction gives:
F_A - 8 + 12 - 8 + F_B = 0 => F_A + F_B = +4kN
Summing the moments about point A gives
F_A*(0m) - 8*(1m) + 12*(2m) - 8*(3m) + F_B*(4m) = 0
=>
-8 + 24 - 24 + 4F_B = 0 => 4F_B = 8 => F_B = +2kN
Back to the forces in the y-direction, this means that F_A = +2kN as well.
This beam has 5 concentrated forces on it, no concentrated moments, and no distributed loads. This means that the Shear force diagram will contain only horizontal lines with discontinuities wherever the concentrated loads are applied, with a jump equal to the amount of the applied concentrated force.
The shear force diagram would be:
0 at A
+2kN from A to C
-6kN from C to D
+6kN from D to E
-2kN from E to B
0 at B
Because there are no concentrated moments, the bending moment diagram would be the area under the curve of the shear force diagram from point A. This means that the graph will be a continuous function of lines that change slope sharply at all of the concentrated loads.
The bending moment diagram will be
0 at A
+2 kNm at C
-4 kNm at D
+2 kNm at E
0 at B
with linear functions connecting each of those points.
To determine the maximum shear stress, you must locate the maximum bending moment, which is of magnitude 4kNm at D, which is 4000 Nm.
The peak bending stress in a cross section is σ=M*c/I, where M is the bending moment, c is the largest distance from the neutral axis (the center of area of the section) and I is the second moment of area (sometimes called moment of inertia) of the section, in the direction it is bending.
For standard beams, you may find the quantity I/c listed as the "Section Modulus, 'S'" meaning σ=M/S. The problem tells you that the beam is a W310x23.8. This is a standard beam with properties tabulated by the ASTM. You should be able to find such a table in your textbook. There, the section modulus is given as
S = 280 cm^3 = 280 * 10^{-6} m^3
Therefore, you can calculate the maximum bending stress
σ=M/S = 4000/(280*10^{-6}) = 14.3 MPa
To calculate the shear stress, you need the maximum shear force, which is V = 6 kN = 6000 N. The shear stress comes from τ=VQ/(It) where Q is the maximum first moment of area of the section, and t is the thickness at that location.
This can be complicated to calculate; however, there is a commonly used conservative approximation for the shear stress in a I-beam τ=V/(A_web) where A_web is the cross sectional area of the middle section of the I-beam.
The web area is the thickness of the web multiplied by (the depth minus twice the flange thickness).
A_web = t_web*(d - 2t_flange)
For the given beam, those quantities can be found
A_web = 5.59mm*(305mm - 2*6.73mm) = 1630 mm^2 = 1630*10^{-6} m^2
Now, we can find the shear stress τ=V/(A_web) = 6000/(1630*10^{-6}) = 3.68 MPa
