Download 50mm 100mm 100mm Centroid Neutral Axis Center of Curvature

Problem 4.21
Determine the minimum diameter of a solid shaft used to transmit 500 kW of power from a 2000rpm motor so that the shear stress does not exceed 50 MPa. Ans. d=62.4 mm.
Notes: The torque is related to the power transmitted by Eq. (4.42). The maximum shear stress
is given by Eq. (4.34). This should be expressed as a function of the diameter, set equal to the
allowable stress of 50 MPa, and then solved for the diameter.
Solution: The angular velocity of the shaft is ω = 2000 rpm = 209 rad/s. Therefore, the torque
transmitted is given by Eq. (4.42) as:
From Table 4.1, the polar moment of inertia for a round cross section is J = πd4/32. c from Eq.
(4.34) is d/2. Therefore, the diameter is obtained form the maximum shear stress from Eq. (4.34):
or d = 0.0624 m = 62.4 mm.
Problem 4.27
A curved bar has a rectangular cross section with height h = ro −ri = 50 mm and width b = 100
mm. Its inner radius is 100 mm. Find the distance between the neutral axis and the centroid.
Ans. e=1.68 mm.
Notes: This problem merely requires application of Eq. (4.58).
Solution: It’s best to sketch this to avoid making errors.
100mm
50mm
100mm
Centroid
Neutral
Axis
Center of Curvature
Note that ro = ri + h = 150 mm. The eccentricity is given by Eq. (4.58) as:
The neutral radius itself is at 123.32 mm.
Problem 4.28
The curved bar in Problem 4.27 is loaded with a bending moment of 3000 N-m. Find the stress at
the innermost and outermost radii. Ans. σi=-83.29 MPa, σo = 63.52 MPa.
Notes: These stresses are obtained from Eqs. (4.65) and (4.66). The results from Prob. 4.27
are used to obtain this solution. Here the assumption is that the moment is tending to curve the
bar more, rather than straighten it.
Solution: As derived in Prob. 4.27, e = 1.68 mm = 0.00168 m. With a height of 50 mm and a
width of 100 mm, the cross sectional area is A = 0.005 m2. Therefore, the stress at the inner
surface is, from Eq. (4.65),
The stress at the outer surface is, from Eq. (4.66),
The distances from the neutral axis to the inner and outer surfaces are ci = 23.32 mm and co =
26.68mm, respectively.
Problem 4.32
A cantilevered beam with a circular-tube cross section has an outer diameter of 130 mm and a
wall thickness of 10 mm. The load perpendicular to the beam is 20,000 N, and the beam is 1.3 m
long from the point of force to the wall where the beam is fastened. Calculate the maximum
bending and shear stresses. Ans. σmax = 247.3 MPa, σmax = 10.61 MPa.
Notes: The approach is similar to Prob. 4.31. Equation (4.48) gives the maximum bending
stress, and Table 4.3 gives the maximum shear stress.
Solution: Start with a good sketch.
20,000 N
1.3m
The maximum moment occurs at the wall and equals M = PL = (20, 000 N)(1.3 m) = 26kNm. The
maximum shear force is V = P = 20 kN. The moment of inertia of the beam is (see Table
4.1):
From Table 4.3 on page 179, the maximum transverse shear stress is:
From Eq. (4.48), the maximum bending stress is:
The bending stress maximum is located on the top of the beam at the wall. The transverse shear
stress maximum is located on the neutral axis of the beam, along the whole length of the tube.
Because these two stresses are located at different positions, we do not combine them.