Question 3.15: For the support shown in Figure 3–8, the pipe is a standard ...
For the support shown in Figure 3–8, the pipe is a standard DN150 steel pipe completely filled with concrete. If the load F is 690 kN, compute the stress in the concrete and the steel. For steel, use E = 207 × 10^{9} N/m². For concrete, use E = 22.7 × 10^{9} N/m² for a rated strength of s_{c} = 20.7 MPa (see Section 2–10).
Objective Compute the stress in the concrete and the steel.
Given Load = F = 690 kN; E_{s} = 207 \times 10^{9} N/m^{2}; E_{c} = 22.7 \times 10^{9} N/m^{2} .
From Appendix A–9(b), for a DN150 pipe:
A_{s} = 3601 mm²; inside diameter = d= 154.1 mm.
Analysis Use Equation (3–15) to compute the stress in the concrete, σ_{c} . Then, use Equation (3–13) to compute σs. All data are known except A_{c} . But
\sigma_{c} = \frac{FE_{c}}{A_{s}E{s} + A_{c}E_{c}} (3-13)
\sigma_{s} = \frac{ \sigma_{c}E_{s}}{E{c}} (3-15)
A_{c} = \frac{\pi d^{2}}{4} = \frac{ \pi (154.1 mm)^{2}}{4} = 18 651 mm²
Results Then, in Equation (3–15),
\sigma_{c} = \frac{(690 000 N)(22.7 \times 10^{9} N/m²)}{(3601 mm²)(207 \times 10^{9} N/m²)+(18651 mm²)(22.7 \times 10^{9} N/m²} = 13.4 MPa
Using Equation (3–13) gives
\sigma_{s} = \frac{ \sigma_{c}E_{s}}{E_{c}} = \frac{(134 MPa)(207 \times 10^{9} N/m²)}{22.7 \times 10^{9} N/m²} = 122.2 MPa
Comment These stresses are fairly high. If it were desired to have at least a design factor of 2.0 based on the yield strength of the steel and 4.0 on the rated strength of the concrete, the required strengths would be
Steel: s_{y} = 2(122.2 MPa) = 244.4 MPa
Concrete: Rated \sigma_{c} =4(13.4 MPa) = 53.6 MPa
The steel could be similar to SAE 1020 annealed or any stronger condition. A rated strength of 20.7 MPa for the concrete would not be satisfactory.