Question 8.7: A tubular post of square cross section supports a horizontal...
A tubular post of square cross section supports a horizontal platform (Fig. 8-28). The tube has outer dimension b = 6 in. and wall thickness t = 0.5 in. The platform has dimensions 6.75 in. × 24.0 in. and supports a uniformly distributed load of 20 psi acting over its upper surface. The resultant of this distributed load is a vertical force P_{1}:
P_{1} = (20 psi)(6.75 in. × 24.0 in.) = 3240 lb
This force acts at the midpoint of the platform, which is at distance d = 9 in. from the longitudinal axis of the post. A second load P_{2} = 800 lb acts horizontally on the post at height h = 52 in. above the base.
Determine the principal stresses and maximum shear stresses at points A and B at the base of the post due to the loads P_{1} and P_{2}.
Stress resultants. The force P_{1} acting on the platform (Fig. 8-28) is statically equivalent to a force P_{1} and a moment M_{1} = P_{1}d acting at the centroid of the cross section of the post (Fig. 8-29a). The load P_{2} is also shown in this figure.
The stress resultants at the base of the post due to the loads P_{1} and P_{2} and the moment M_{1} are shown in Fig. 8-29b. These stress resultants are the following:
1. An axial compressive force P_{1} = 3240 lb
2. A bending moment M_{1} produced by the force P_{1}:
M_{1}=P_{1}d = (3240 lb)(9 in.) = 29,160 lb-in.
3. A shear force P_{2} = 800 lb
4. A bending moment M_{2} produced by the force P_{2}:
M_{2}=P_{2}h = (800 lb)(52 in.) = 41,600 lb-in.
Examination of these stress resultants (Fig. 8-29b) shows that both M_{1} and M_{2} produce maximum compressive stresses at point A and the shear force produces maximum shear stresses at point B. Therefore, A and B are critical points where the stresses should be determined. (Another critical point is diagonally opposite point A, as explained in the Note at the end of this example.)
Stresses at points A and B. (1) The axial force P_{1} (Fig. 8-29b) produces uniform compressive stresses throughout the post. These stresses are
in which A is the cross-sectional area of the post:
A = b² – (b – 2t)² = 4t(b – t)
= 4(0.5 in.)(6 in. – 0.5 in.) = 11.00 in.²
Therefore, the axial compressive stress is
\sigma_{P_{1}}=\frac{P_{1}}{A}=\frac{3240 lb}{11.00 in.^{2}}=295 psiThe stress \sigma_{P_{1}} is shown acting at points A and B in Fig. 8-29c.
(2) The bending moment M_{1} (Fig. 8-29b) produces compressive stresses \sigma_{M_{1}} at points A and B (Fig. 8-29c). These stresses are obtained from the flexure formula:
in which I is the moment of inertia of the cross-sectional area:
I=\frac{b^{4}}{12}-\frac{(b-2t)^{4}}{12}=\frac{1}{12}\left[(6 in.)^{4}-(5 in.)^{4}\right]=55.92 in.^{4}Thus, the stress \sigma_{M_{1}} is
\sigma_{M_{1}}=\frac{{M_{1}b}}{2I}=\frac{(29,160 lb-in.)(6 in.)}{}=1564 psi(3) The shear force P_{2} (Fig. 8-29b) produces a shear stress at point B but not at point A. From the discussion of shear stresses in the webs of beams with flanges (Section 5.10), we know that an approximate value of the shear stress can be obtained by dividing the shear force by the web area (see Eq. 5-50 in Section 5.10). Thus, the shear stress produced at point B by the force P_{2} is
\tau_{aver} = \frac{V}{th_{1}} (5-50)
\tau_{P_{2}}=\frac{{P_{2}}}{A_{web}}=\frac{P_{2}}{2t(b-2t)}=\frac{800 lb}{}=160 psiThe stress \tau_{P_{2}} acts at point B in the direction shown in Fig. 8-29c.
If desired, we can calculate the shear stress \tau_{P_{2}} from the more accurate formula of Eq. (5-48a) in Section 5.10. The result of that calculation is \tau_{P_{2}} = 163 psi, which shows that the shear stress obtained from the approximate formula is satisfactory.
\tau_{\max}=\frac{V}{8It}(bh^{2}-bh^{2}_{1}+th^{2}_{1}) (5-48a)
(4) The bending moment M_{2} (Fig. 8-29b) produces a compressive stress at point A but no stress at point B. The stress at A is
\sigma_{M_{2}}=\frac{{M_{2}(b/2)}}{I}=\frac{{M_{2}b}}{2I}=\frac{(41,600 lb-in.)(6 in.)}{}=2232 psiThis stress is also shown in Fig. 8-29c.
Stress elements. The next step is to show the stresses acting on stress elements at points A and B (Figs. 8-29d and e). Each element is oriented so that the y axis is vertical (that is, parallel to the longitudinal axis of the post) and the x axis is horizontal. At point A the only stress is a compressive stress \sigma_{A} in the y direction (Fig. 8-29d):
= 295 psi + 1564 psi + 2232 psi = 4090 psi (compression)
Thus, this element is in uniaxial stress.
At point B the compressive stress in the y direction (Fig. 8-29e) is
\sigma_{B}=\sigma_{P_{1}}+\sigma_{M_{1}} = 295 psi + 1564 psi = 1860 psi (compression)
and the shear stress is
\tau_{P_{2}} = 160 psi
The shear stress acts leftward on the top face of the element and downward on the x face of the element.
Principal stresses and maximum shear stresses at point A. Using the standard notation for an element in plane stress (Fig. 8-30), we write the stresses for element A (Fig. 8-29d) as follows:
\sigma_{x} = 0 \sigma_{y}=-\sigma_{A}=-4090 psi \tau_{xy} = 0
Since the element is in uniaxial stress, the principal stresses are
\sigma_{1} = 0 \sigma_{2} = -4090 psi
and the maximum in-plane shear stress (Eq. 7-26) is
\tau_{\max}=\frac{\sigma_{1}-\sigma_{2}}{2}=\frac{4090 psi}{2}=2050 psiThe maximum out-of-plane shear stress (Eq. 7-28a) has the same magnitude.
(\tau_{\max})_{x_{1}}=\pm\frac{\sigma_{2}}{2} (7-28a)
Principal stresses and maximum shear stresses at point B. Again using the standard notation for plane stress (Fig. 8-30), we see that the stresses at point B (Fig. 8-29e) are
\sigma_{x} = 0 \sigma_{y}=-\sigma_{B}=-1860 psi \tau_{xy}=-\tau_{P_{2}}=-160 psi
To obtain the principal stresses, we use Eq. (7-17), which is repeated here:
\sigma_{1,2}=\frac{\sigma_{x}+\sigma_{y}}{2}\pm\sqrt{\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right)^{2}+\tau^{2}_{xy}} (m)
Substituting for \sigma_{x}, \sigma_{y}, and \tau_{xy}, we get
\sigma_{1,2}=-930 psi\pm944 psi
or
\sigma_{1}=14 psi \sigma_{2}=-1870 psi
The maximum in-plane shear stresses may be obtained from Eq. (7-25):
\tau_{\max}=\sqrt{\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right)^{2}+\tau^{2}_{xy}} (n)
This term was evaluated previously, so we see immediately that
\tau_{\max}=944 psiBecause the principal stresses \sigma_{1} and \sigma_{2} have opposite signs, the maximum in-plane shear stresses are larger than the maximum out-of-plane shear stresses (see Eqs. 7-28a, b, and c and the accompanying discussion). Therefore, the maximum shear stress at point B is 944 psi.
(\tau_{\max})_{y_{1}}=\pm\frac{\sigma_{1}}{2} (\tau_{\max})_{z_{1}}=\pm\frac{\sigma_{1}-\sigma_{2}}{2} (7-28b,c)
Note: If the largest stresses anywhere at the base of the post are needed, then we must also determine the stresses at the critical point diagonally opposite point A (Fig. 8-29c), because at that point each bending moment produces the maximum tensile stress. Thus, the tensile stress acting at that point is
\sigma_{y}=-\sigma_{P_{1}}+\sigma_{M_{1}}+\sigma_{M_{2}} = -295 psi + 1564 psi + 2232 psi = 3500 psi
The stresses acting on a stress element at that point (see Fig. 8-30) are
\sigma_{x} = 0 \sigma_{y}=3500 psi \tau_{xy}=0
and therefore the principal stresses and maximum shear stress are
\sigma_{1} = 3500 psi \sigma_{2}=0 \tau_{\max}= 1750 psi
Thus, the largest tensile stress anywhere at the base of the post is 3500 psi, the largest compressive stress is 4090 psi, and the largest shear stress is 2050 psi. (Keep in mind that only the effects of the loads P_{1} and P_{2} are considered in this analysis. Other loads, such as the weight of the structure, also produce stresses at the base of the post.)
