Question 3.17: Refer to Figure 3–13. The square steel tube carries 130 kN o...
Refer to Figure 3–13. The square steel tube carries 130 kN of axial compressive force. Compute the compressive stress in the tube and the bearing stress between each mating surface. Consider the weight of the concrete pier to be 1500 N.
Given Load F = 130 000 N compression. Weight of pier = 1500 N. Geometry of members shown in Figure 3–13.
Analysis Tube: Use direct compressive stress formula. Bearing stresses: Use Equation (3–20) for each pair of mating surfaces.
\sigma_{b} = \frac{Applied load}{Bearing area} = \frac{F}{A_{b}} (3-20)
Results Compressive stress in the tube: (area = A = 1536 mm²)
σ = F/A = 130 000 N/1536 mm² = 84.6 N/mm²= 84.6 MPa
Bearing stress between tube and square plate: This will be equal in magnitude to the compressive stress in the tube because the cross-sectional area of the tube is the smallest area in contact with the plate. Then,
\sigma_{b} = 84.6 MPa
Bearing stress between plate and top of concrete pier: The bearing area is that of the square plate because it is the smallest area at the surface.
\sigma_{b} = F/A_{b} = 130 000N/(150 mm)² = 5.8 MPa
Bearing stress between the pier and the gravel: The bearing area is that of a square, 600 mm on a side. Add 1500 N for the weight of the pier.
\sigma_{b} = F/A_{b} = 131500/(600 mm)² = 0.37 N/mm² = 0.37 MPa = 370 kPa
Comment Allowable bearing stresses are discussed later.