Question 1.6: A semicircular bracket assembly is used to support a steel s...
A semicircular bracket assembly is used to support a steel staircase in an office building. Steel rods are attached to each of two brackets using a clevis and pin; the upper end of the rod is attached to a cross beam near the roof. Photos of the bracket attachment and hanger rod supports are shown in Fig. 1-24. The weight of the staircase, and any building occupants who are using the staircase, is estimated to result in a force of 4800 N on each hanger rod.
(a) Obtain a formula for the maximum stress \sigma_{\max } in the rod, taking into account the weight of the rod itself.
(b) Calculate the maximum stress in the rod in MPa using numerical proper-ties L_{r} = 12 m, d_{r} = 20 mm, F_{r} = 4800 N (note that the weight density \gamma_{r} of steel is 77.0 kN/m^{3} [from Table H-1 in Appendix H])
Table H-1 Weights and Mass Densities
| Material | Weight density γ | Mass density ρ |
| kN/m³ | kg/m³ | |
| Aluminum alloys
2014-T6, 7075-T6 6061-T6 |
26–28
28 26 |
2,600–2,800 2,800 2,700 |
| Brass | 82–85 | 8,400–8,600 |
| Bronze | 80–86 | 8,200–8,800 |
| Cast iron | 68 – 72 | 7,000 – 7,400 |
| Concrete
Plain Reinforced Lightweight |
23 24 11-18 |
2,300 2,400 1,100 – 1,800 |
| Copper | 87 | 8,900 |
| Glass | 24–28 | 2,400–2,800 |
| Magnesium alloys | 17–18 | 1,760–1,830 |
| Monel (67% Ni, 30% Cu) | 87 | 8,800 |
| Nickel | 87 | 8,800 |
| Plastics
Nylon Polyethylene |
8.6-11 9.4-14 |
880–1,100 960–1,400 |
|
Rock Granite, marble, quartz Limestone, sandstone |
26–28 20–28 |
2,600–2,900 2,000–2,900 |
| Rubber | 9–13 | 960–1,300 |
| Sand, soil, gravel | 12–21 | 1,200–2,200 |
| Steel | 77 | 7,850 |
| Titanium | 44 | 4,500 |
| Tungsten | 190 | 1,900 |
| Water, fresh
sea |
9.81
10.0 |
1,000
1,020 |
| Wood (air dry)
Douglas fir Oak Southern pine |
4.7–5.5 6.3–7.1 5.5–6.3 |
480–560 640–720 560–640 |
Numerical data:
L_{r}=12 m d_{r}=20 mm \gamma_{r}=77 kN/m³ F_{r}=4800 N
(a) Obtain a formula for the maximum stress σ_{\max} in the rod, taking into account the weight of the rod itself.
The maximum axial force F_{\max } in the rod occurs at the upper end and is equal to the force F_{r } in the rod due to the combined staircase and occu-pants’ weights plus the weight W_{r} of the rod itself. The latter is equal to the weight density \gamma_{r} of the steel times the volume V_{r } of the rod, or
W_{r}=\gamma_{r}\left(A_{r} L_{r}\right) (1-10)
in which Ar is the cross-sectional area of the rod. Therefore, the formula for the maximum stress [from Eq. (1-6) \sigma=\frac{P}{A}] becomes
\sigma_{\max }=\frac{F_{\max }}{A_{r}} \text { or } \sigma_{\max }=\frac{F_{r}+W_{r}}{A_{r}}=\frac{F_{r}}{A_{r}}+\gamma_{r} L_{r} (1-11)
(b) Calculate the maximum stress in the rod in MPa using numerical properties.
To calculate the maximum stress, we substitute numerical values into the preceding equation. The cross-sectional area A_{r}=\pi d_{r}^{2} / 4 , where d_{r } = 20 mm, and the weight density \gamma_{r} of steel is 77.0 kN/m^{3} (from Table H-1 in Appendix H).
Thus,
A_{r}=\frac{\pi d_{r}^{2}}{4}=314 mm^{2} \quad F_{r}=4800 {N}The normal stress in the rod due to the weight of the staircase is
\sigma_{\text {stair }}=\frac{F_{r}}{A_{r}}=15.3 {MPa}and the additional normal stress at the top of the rod due to the weight of the rod itself is
\sigma_{\text {rod }}=\gamma_{r} L_{r}=0.924 {MPa}So the maximum normal stress at the top of the rod is the sum of these two normal stresses:
\sigma_{\max }=\sigma_{\text {stair }}+\sigma_{\text {rod }} \quad \sigma_{\max }=16.2 MPa (1-12)
Note that
\frac{\sigma_{\text {rod }}}{\sigma_{\text {stair }}}=6.05 \%In this example, the weight of the rod contributes about 6% to the maxi-mum stress and should not be disregarded .