A hollow cylinder with outer radius r0 and inner radius ri is loaded by a uniformly distributed load q across its length L. The beam is supported by a roller at its left end and a pin L5 from its right end.
What is the maximum bending moment in the beam? Express your answer in terms of q and L.
Mmax
At what point along the beam does the maximum bending moment occur? Express your answer in terms of the fraction of the length L, from the left hand support, A.
x:
Assume that the length of the beam is 5m and the uniformly distributed load is 1kN/m. Draw the shear force and bending moment diagrams, then indicate the value of the shear force and bending moment at x=0m, x=1m, x=2m, x=3m, x=4m and x=5m.
Shear forces:
V(x=0m) in (kN):
V(x=1m) in (kN):
V(x=2m) in (kN):
V(x=3m) in (kN):
V just before the support B, at (x=4m) in (kN):
V just after the support B, at (x=4m) in (kN):
V(x=5m) in (kN):
Bending moments:
M(x=0m) in (kNm):
M(x=1m) in (kNm):
M(x=2m) in (kNm):
M(x=3m) in (kNm):
M(x=4m) in (kNm):
M(x=5m) in (kNm):
Assume that the inner radius of the cylinder ri is 10cm, and the thickness of the cylinder wall is 2mm, and, as before, the length L is 5m and the distributed load q is 1kN/m. What is the magnitude of the maximum stress in the cylinder? (Please remember that common moments of inertia can be easily looked up.)
Magnitude of σmax (in MPa):
M_max=(9/128)*q*L^2
x=3/8 * L
V(x=0m) in (kN):1.88
V(x=1m) in (kN):0.875
V(x=2m) in (kN):-0.125
V(x=3m) in (kN):-1.13
V just before the support B, at (x=4m) in (kN):-2.13
V just after the support B, at (x=4m) in (kN):1
V(x=5m) in (kN):0
Bending moments:
M(x=0m) in (kNm):0
M(x=1m) in (kNm):1.38
M(x=2m) in (kNm):1.76
M(x=3m) in (kNm):1.13
M(x=4m) in (kNm):-0.50
M(x=5m) in (kNm):0
To find the maximum bending moment in the beam, we can use the formula for bending moment, which is given by:
M = (q * L^2) / 8
where q is the uniformly distributed load and L is the length of the beam.
For this problem, q = 1 kN/m and L = 5 m. Plugging these values into the formula, we can calculate the maximum bending moment (Mmax):
Mmax = (1 kN/m * (5 m)^2) / 8
Mmax = (5 kN.m^2) / 8
Mmax = 0.625 kN.m^2
To find the point along the beam where the maximum bending moment occurs, we need to determine the fraction of the length L from the left-hand support (A). In this case, the maximum bending moment occurs at the right end of the beam, which is at x = L. So the fraction of the length from the left-hand support (A) is 1.
The maximum bending moment occurs at x = 1.
To draw the shear force and bending moment diagrams, we need to consider the different sections of the beam and determine the forces acting on them.
First, let's consider the left-hand section (from x = 0 to x = 1):
At x = 0: The shear force (V) and bending moment (M) are both equal to 0.
At x = 1: The shear force (V) can be calculated using the formula:
V = q * x = 1 kN/m * 1 m = 1 kN
The bending moment (M) at x = 1 can be obtained by integrating the shear force from x = 0 to x = 1:
M = Integral(q * x, 0, 1) = Integral(1 kN/m * x, 0, 1) = (1/2) kN.m
Next, let's consider the middle section (from x = 1 to x = 4):
At x = 1: The shear force (V) remains 1 kN.
At x = 2: The shear force (V) can be calculated using the same formula as before:
V = q * x = 1 kN/m * 2 m = 2 kN
The bending moment (M) at x = 2 can be obtained by integrating the shear force from x = 1 to x = 2:
M = Integral(q * x, 1, 2) = Integral(1 kN/m * x, 1, 2) = (3/2) kN.m
Similarly, we can calculate the shear forces and bending moments at x = 3 and x = 4 using the same process:
At x = 3: V = 3 kN, M = (4/2) kN.m
At x = 4: V = 4 kN, M = (5/2) kN.m
Finally, at x = 5 (the right end of the beam):
The shear force (V) just before the support B is 4 kN.
The shear force (V) just after the support B is also 4 kN.
The bending moment (M) is (5/2) kN.m.
Now, let's consider the case where the beam is a hollow cylinder. The inner radius (ri) is given as 10 cm, which is equivalent to 0.1 m. The thickness of the cylinder wall is given as 2 mm, which is equivalent to 0.002 m.
To find the maximum stress in the cylinder, we can use the formula for the maximum stress in a hollow cylinder subjected to bending, which is given as:
σmax = (Mmax * (ri + ro)) / (I * t)
where Mmax is the maximum bending moment, ri is the inner radius, ro is the outer radius, I is the moment of inertia of the cross-section, and t is the thickness of the cylinder wall.
We have already calculated Mmax as 0.625 kN.m. The outer radius (r0) can be obtained by adding the thickness of the cylinder wall to the inner radius:
r0 = ri + t = 0.1 m + 0.002 m = 0.102 m
The moment of inertia (I) for a hollow cylinder can be looked up in common engineering references. For simplicity, let's assume the moment of inertia is 0.23 m^4.
Now, we can plug these values into the formula to find the magnitude of the maximum stress (σmax):
σmax = (0.625 kN.m * (0.1 m + 0.102 m)) / (0.23 m^4 * 0.002 m)
σmax = (0.625 kN.m * 0.202 m) / (0.00046 m^5)
σmax = 0.03175 kN/m^2
Since 1 MPa (Megapascal) is equivalent to 1 kN/m^2, the magnitude of the maximum stress in the cylinder is:
Magnitude of σmax = 0.03175 MPa