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. 8-30). The tube has outer dimension b = 150 mm and wall thickness t = 13 mm . The platform has dimensions 175 mm × 600 mm and sup-ports a uniformly distributed load of 140 kPa acting over its upper sur-face. The resultant of this distributed load is a vertical force P_{1}:
P_{1}=(140 \mathrm{~kPa})(175 \mathrm{~mm} \times 600 \mathrm{~mm})=14.7 \mathrm{~kN}This force acts at the midpoint of the platform, which is at distance d = 225 mm from the longitudinal axis of the post. A second load P_{2} = 3.6 kN acts horizontally on the post at height h = 1.3 m 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} andP_{2}:
Stress resultants. The force P_{1} acting on the platform (Fig. 8-30) 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. 8-31a). 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. 8-31b. These stress resultants are the following:
1. An axial compressive force P_{1} = 14.7 kN
2. A bending moment M_{1} produced by the force P_{1}:
M_{1}=P_{1} d=(14.7 \mathrm{~kN})(225 \mathrm{~mm})=3307.5 \mathrm{~N} \cdot \mathrm{~m}3. A shear force P_{1} = 3.6 kN
4. A bending moment M_{2} produced by the force P_{2}:
M_{2}=P_{2} h=(3.6 \mathrm{~kN})(1.3 \mathrm{~m})=4.68 \mathrm{kN} \cdot \mathrm{~m}Examination of these stress resultants (Fig. 8-31b) 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. 8-31b) produces uniform compressive stresses throughout the post. These stresses
in which A is the cross-sectional area of the post:
\begin{aligned}A &=b^{2}-(b-2 t)^{2}=4 t(b-t) \\&=4(13 \mathrm{~mm})(150 \mathrm{~mm}-13 \mathrm{~mm})=7124 \mathrm{~mm}^{2}\end{aligned}Therefore, the axial compressive stress is
\sigma_{P_{1}}=\frac{P_{1}}{A}=\frac{14.7 \mathrm{~kN}}{7124 \mathrm{~mm}^{2}}=2.06 \mathrm{~MPa}The stress \sigma_{P_{1}} is shown acting at points A and B in Fig. 8-31c.
(2) The bending moment {M_{1}} (Fig. 8-31b) produces compressive stresses \sigma_{M_{1}} at points A and B (Fig. 8-31c). 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-2 t)^{4}}{12}=\frac{1}{12}\left[(150 \mathrm{~mm})^{4}-(124 \mathrm{~mm})^{4}\right]=22.49 \times 10^{-6} \mathrm{~m}^{4}Thus, the stress \sigma_{M_{1}} is
\sigma_{M_{1}}=\frac{M_{1} b}{2 I}=\frac{(3307.5 \mathrm{~N} \cdot \mathrm{~m})(150 \mathrm{~mm})}{2\left(22.49 \times 10^{-6} \mathrm{~m}^{4}\right)}=11.03 \mathrm{~MPa}(3) The shear force P_2 (Fig. 8-31b) 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. (5-55) \tau_{\text {aver }}=\frac{V}{t h_{1}} in Section 5.10]. Thus, the shear stress produced at point B by the force {P_{2}} is
\tau_{P_{2}}=\frac{P_{2}}{A_{\text {web }}}=\frac{P_{2}}{2 t(b-2 t)}=\frac{3.6 \mathrm{~kN}}{2(13 \mathrm{~mm})(150 \mathrm{~mm}-26 \mathrm{~mm})}=1.12 \mathrm{~MPa}The stress \tau_{P_{2}} acts at point B in the direction shown in Fig. 8-31c.
If desired, we can calculate the shear stress \tau_{P_{2}} from the more accurate formula of Eq. (5-53a) \tau_{\max }=\frac{V}{8 I t}\left(b h^{2}-b h_{1}^{2}+t h_{1}^{2}\right) in Section 5.10. The result of that calculation is \tau_{P_{2}} = 1.13 MPa , which shows that the shear stress obtained from the approximate formula is satisfactory.
(4) The bending moment {M_{2}} (Fig. 8-31b) 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}{2 I}=\frac{(4.68 \mathrm{~kN} \cdot \mathrm{~m})(150 \mathrm{~mm})}{2\left(22.49 \times 10^{-6} \mathrm{~m}^{4}\right)}=15.61 \mathrm{~MPa}This stress is also shown in Fig. 8-31c.
Stress elements. The next step is to show the stresses acting on stress ele-ments at points A and B (Figs. 8-31d and e). Each element is oriented so that the y axis is vertical (that is, parallel to the longitudinal axis of the post)the x axis is horizontal. At point A the only stress is a compressive stress \sigma_{A} in the y direction (Fig. 8-31d):
Thus, this element is in uniaxial stress.
At point B, the compressive stress in the y direction (Fig. 8-31e) is
and the shear stress is
\tau_{P_{2}}=1.12 \mathrm{~MPa}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. 8-32), we write the stresses for element A (Fig. 8-31d) as follows:
Since the element is in uniaxial stress, the principal stresses are
\sigma_{1}=0 \quad \sigma_{2}=-28.7 \mathrm{~MPa}and the maximum in-plane shear stress [Eq. (7-26)] is
\tau_{\max }=\frac{\sigma_{1}-\sigma_{2}}{2}=\frac{-28.7 \mathrm{~MPa}}{2}=14.4 \mathrm{~MPa}The maximum out-of-plane shear stress [Eq. (7-28a) \sigma_{x_{1}}=\tau_{x y} \sin 2 \theta] has the same magnitude.
Principal stresses and maximum shear stresses at point B. Again using the standard notation for plane stress (Fig. 8-32), we see that the stresses at point B (Fig. 8-31e) are
\sigma_{x}=0 \quad \sigma_{y}=-\sigma_{B}=-13.1 \mathrm{~MPa} \quad \tau_{x y}=-\tau_{P_{2}}=-1.12 \mathrm{~MPa}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_{x y}^{2}} (a)
Substituting for \sigma_{x}, \sigma_{y}, and \tau_{xy}, we get
\sigma_{1,2}=-6.55 \mathrm{~MPa} \pm 6.65 \mathrm{~MPa}or
\sigma_{1}=0.1 \mathrm{~MPa} \quad \sigma_{2}=-13.2 \mathrm{~MPa}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_{x y}^{2}} (b)
This term was evaluated previously, so we see immediately that
\tau_{\max }=6.65 \mathrm{~MPa}Because 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 6.65 MPa.
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 diag-onally opposite point A (Fig. 8-31c), 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}}=-2.06 \mathrm{~MPa}+11.03 \mathrm{~MPa}+15.61 \mathrm{~MPa}=24.58 \mathrm{~MPa}The stresses acting on a stress element at that point (see Fig. 8-32) are
\sigma_{x}=0 \quad \sigma_{y}=24.58 \mathrm{~MPa} \quad \tau_{x y}=0and therefore the principal stresses and maximum shear stress are
\sigma_{1}=24.58 \mathrm{~MPa} \quad \sigma_{2}=0 \quad \tau_{\max }=12.3 \mathrm{~MPa}Thus, the largest tensile stress anywhere at the base of the post is 24.58 MPa, the largest compressive stress is 28.7 MPa, and the largest shear stress is 14.4 MPa. (Keep in mind that only the effects of the loads P_{1} and P_{2} are con-sidered in this analysis. Other loads, such as the weight of the structure, also produce stresses at the base of the post.)
