Question 8.6: A sign with dimensions of 2.0 m × 1.2 m is supported by a ho...
A sign with dimensions of 2.0 m × 1.2 m is supported by a hollow circularpole having an outer diameter of 220 mm and an inner diameter of 180 mm (Fig. 8-27). The sign is offset 0.5 m from the centerline of the pole, and its lower edge is 6.0 m above the ground.
(a) Determine the principal stresses and maximum shear stresses at points A and B at the base of the pole due to a wind pressure of 2.0 kPa against the sign.
(b) Compare the circular pole stresses at the base and twist at the top to stresses and twist of a square tube with the same height, same wall thickness, and same cross-sectional area.
(a) Circular pole: Stress resultants. The wind pressure against the sign pro-duces a resultant force W that acts at the midpoint of the sign (Fig. 8-28a) and is equal to the pressure p times the area A over which it acts:
W = pA = (2.0 kPa)(2.0 m × 1.2 m) = 4.8 kN
The line of action of this force is at height h = 6.6 m above the ground and at distance b = 1.5 m from the centerline of the pole.
The wind force acting on the sign is statically equivalent to a lateral force W and a torque T acting on the pole (Fig. 8-28b). The torque is equal to the force W times the distance b:
T = Wb = (4.8 kN)(1.5 m) = 7.2 kN . m
The stress resultants at the base of the pole (Fig. 8-28c) consist of a bending moment M, a torque T, and a shear force V. Their magnitudes are
M = Wh = (4.8 kN)(6.6 m) = 31.68 kN . m
T = 7.2 kN . m V = W = 4.8 kN
Examination of these stress resultants shows that maximum bending stresses occur at point A and maximum shear stresses at point B.
Therefore, A and B are critical points where the stresses should be deter-mined. (Another critical point is diametrically opposite point A, as explained in the Note at the end of part (a) of this example.
Stresses at points A and B. The bending moment M produces a ten-sile stress \sigma_{A} at point A (Fig. 8-28d) but no stress at point B (which is located on the neutral axis). The stress \sigma_{A} is obtained from the flexure formula:
\sigma_{A}=\frac{M\left(d_{2} / 2\right)}{I}in which d_{2} is the outer diameter (220 mm) and I is the moment of iner-tia of the cross section. The moment of inertia is
I=\frac{\pi}{64}\left(d_{2}^{4}-d_{1}^{4}\right)=\frac{\pi}{64}\left[(220 \mathrm{~mm})^{4}-(180 \mathrm{~mm})^{4}\right]=63.46 \times 10^{-6} \mathrm{~m}^{4}in which d_{1} is the inner diameter. Therefore, the stress \sigma_{A} is
\sigma_{A}=\frac{M d_{2}}{2 l}=\frac{(31.68 \mathrm{~kN} \cdot \mathrm{~m})(220 \mathrm{~mm})}{2\left(63.46 \times 10^{6} \mathrm{~m}^{4}\right)}=54.91 \mathrm{~MPa}The torque T produces shear stresses \tau_{1} at points A and B (Fig. 8-28d). We can calculate these stresses from the torsion formula:
\tau_{1}=\frac{T\left(d_{2} / 2\right)}{I_{P}}in which I_{P } is the polar moment of inertia:
I_{P}=\frac{\pi}{32}\left(d_{2}^{4}-d_{1}^{4}\right)=2 I=126.92 \times 10^{-6} \mathrm{~m}^{4}Thus,
\tau_{1}=\frac{T d_{2}}{2 I_{P}}=\frac{(7.2 \mathrm{~kN} \cdot \mathrm{~m})(220 \mathrm{~mm})}{2\left(126.92 \times 10^{-6} \mathrm{~m}^{4}\right)}=6.24 \mathrm{~MPa}Finally, we calculate the shear stresses at points A and B due to the shear force V. The shear stress at point A is zero, and the shear stress at point B (denoted \tau_{2} in Fig. 8-28d) is obtained from the shear formula for a circular tube [Eq. (5-48) \tau_{\max }=\frac{V Q}{I b}=\frac{4 V}{3 A}\left(\frac{r_{2}^{2}+r_{2} r_{1}+r_{1}^{2}}{r_{2}^{2}+r_{1}^{2}}\right) of Section 5.9]
\tau_{2}=\frac{4 V}{3 A}\left(\frac{r_{2}^{2}+r_{2} r_{1}+r_{1}^{2}}{r_{2}^{2}+r_{1}^{2}}\right) (a)
in which r_{2} and r_{1} are the outer and inner radii, respectively, and A is the cross-sectional area:
\begin{aligned}&r_{2}=\frac{d_{2}}{2}=110 \mathrm{~mm} \quad r_{1}=\frac{d_{1}}{2}=90 \mathrm{~mm} \\&A=\pi\left(r_{2}^{2}-r_{1}^{2}\right)=12,570 \mathrm{~mm}^{2}\end{aligned}Substituting numerical values into Eq. (a), we obtain
\tau_{2}=0.76 \mathrm{~MPa}The stresses acting on the cross section at points A and B now have been calculated.
Stress elements. The next step is to show these stresses on stress ele-ments (Figs. 8-28e and f). For both elements, the y axis is parallel to the longitudinal axis of the pole and the x axis is horizontal. At point A, the stresses acting on the element are
At point B, the stresses are
\sigma_{x}=\sigma_{y}=0 \quad \tau_{x y}=\tau_{1}+\tau_{2}=6.24 \mathrm{~MPa}+0.76 \mathrm{~MPa}=7.00 \mathrm{~MPa}Since there are no normal stresses acting on the element, point B is in pure shear.
Now that all stresses acting on the stress elements (Figs. 8-28e and f) are known, we can use the equations given in Section 7.3 to determine the principal stresses and maximum shear stresses. Principal stresses and maximum shear stresses at point A. The prin-cipal stresses are obtained from 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}} (b)
Substituting \sigma_{x}=0, \sigma_{y}=54.91 \mathrm{~MPa}, \text { and } \tau_{x y}=6.24 \mathrm{~MPa} , we get
\sigma_{1,2}=27.5 \mathrm{~MPa} \pm 28.2 \mathrm{~MPa}or
\sigma_{1}=55.7 \mathrm{~MPa} \quad \sigma_{2}=-0.7 \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}} (c)
This term was evaluated previously, so we see immediately that
\tau_{\max }=28.2 \mathrm{~MPa}Because the principal stresses σ_1 \text{ and } σ_2 have opposite signs, the maxi-mum in-plane shear stresses are larger than the maximum out-of-plane shear stresses [see Eqs. (7-28a, b, and c) \begin{aligned}&\left(\tau_{\max }\right)_{x_{1}}=\pm \frac{\sigma_{2}}{2} \quad\left(\tau_{\max }\right)_{y_{1}}=\pm \frac{\sigma_{1}}{2} \\&\left(\tau_{\max }\right)_{z_{1}}=\pm \frac{\sigma_{1}-\sigma_{2}}{2}\end{aligned}] and the accompanying discus-sion. Therefore, the maximum shear stress at point A is 28.2 MPa.
Principal stresses and maximum shear stresses at point B. The stresses at this point are \sigma_{x}=0, \sigma_{y}=0, \text { and } \tau_{x y}=7.0 \mathrm{~MPa}. Since the element is in pure shear, the principal stresses are
\sigma_{1} = 7.0 \mathrm{~MPa} \quad \quad \sigma_{2} = -7.0 \mathrm{~MPa}and the maximum in-plane shear stress is
\tau_{\max }=7.0 \mathrm{~MPa}The maximum out-of-plane shear stresses are half this value.
Note: If the largest stresses anywhere in the pole are needed, then we must also determine the stresses at the critical point diametrically opposite point A, because at that point the compressive stress due to bending has its largest value. The principal stresses at that point are
and the maximum shear stress is 28.2 MPa. Therefore, the largest tensile stress in the pole is 55.7 MPa, the largest compressive stress is – 55.7 MPa and the largest shear stress is 28.2 MPa. (Keep in mind that only the effects of the wind pressure are considered in this analysis. Other loads, such as the weight of the structure, also produce stresses at the base of the pole.
(b) Square tube. The square tube has the same height ( h = 6.6 m to the center of pressure on the sign), same wall thickness ( t = 20 mm ) , and same cross-sectional area ( A = 12.570 \mathrm{~mm}^{2}) as that of the circular pole. We thus can compute tube dimension b (along the median line of the tube, see Fig. 8-29a) as
(b+t)^{2}-(b-t)^{2}=12,570 \mathrm{~mm}^{2} \quad \text { so } \quad b=157.125 \mathrm{~mm}The torsion constant J of the tube [see Eq. (3-94) J=\frac{2 b^{2} h^{2} t_{1} t_{2}}{b t_{1}+h t_{2}} ] and the area A_m enclosed by the median line of the tube are
J=b^{3} t=7.758 \times 10^{-5} \mathrm{~m}^{4} \quad A_{m}=b^{2}=2.469 \times 10^{4} \mathrm{~mm}^{2}(We assume that the formulas for thin-walled tubes in Section 3.11 apply here and we will disregard the effects of stress concentrations at the cor-ners of the tube)
For normal and transverse shear stress calculations, we will also need the moment of inertia I_{\text {tube }} with respect to the neutral axis of the cross section for use in the flexure formula, from Eq. (5-14) \sigma_{x}=-\frac{M y}{I}, and the first moment Q_{\text {tube }} of the area with respect to the neutral axis for use in the shear formula from Eq. (5-41) \tau=\frac{V Q}{I b} . These properties are computed a
Stresses at A and B on tube. The normal tensile stress at A (see Fig. 8-29b) is computed using the flexure formula with M = 31.68 kN.m , giving
\sigma_{A}=\frac{M(b+t)}{2 I_{\text {tube }}}=53.38 \mathrm{~MPa}The normal stress at B is zero, because it is located on the neutral axis. The transverse shear stress at A is zero, and the shear stress at B is obtained from the shear formula as
\tau_{2}=\frac{V Q_{\text {tube }}}{I_{\text {tube }}(2 t)}=0.85 \mathrm{~MPa}The torque T = 7.2 kN.m produces shear stress at both A and B. Using Eq. (3-81), shear stress \tau_{1} is
\tau_{1}=\frac{T}{2 t A_{m}}=7.29 \mathrm{~MPa}The resulting stress states at locations A and B on the square tube in Fig. 8-29b are the same as that shown in Figs. 8-28e and f, where
Principal stresses and maximum in-plane shear stress at point A. Repeating the calculations for the square tube using Eq. (b), we have
Principal stresses and maximum in-plane shear stress at point B. The stress element at point B is in pure shear, so the principal stresses and maximum in-plane shear stress are
\begin{aligned}\sigma_{1} &=\tau_{x y}=8.1 \mathrm{~MPa} \\\sigma_{2} &=-\tau_{x y}=-8.1 \mathrm{~MPa} \\\tau_{\max } &=\tau_{x y}=8.1 \mathrm{~MPa}\end{aligned}These stresses at A and B on the square tube are comparable to those for the circular pole. As a final comparison, we will look at the twist dis-placement on each pole at the level of the sign center of pressure (h = 6.6 m) . The twist rotation for the circular pole is computed using Eq. (3-17) (assuming that G = 80 Gpa for steel)
\phi_{c}=\frac{T h}{G I_{p}}=4.68 \times 10^{-3} \text { radians }and that for the square tube is computed using Eq. (3-73) \phi=\frac{T L}{\left(k_{2} b t^{3}\right) G}=\frac{T L}{G J_{r}},
\phi_{t}=\frac{T h}{G J}=7.656 \times 10^{-3} \text { radians }The twist rotation for the circular pole is 39% lower than that of the square tube. (See Example 3-16 for more discussion of square and circu-lar tube stresses and twist rotations.) Both circular and square tube sign posts also displace in the direction of the wind force, but calculation of bending displacements must be delayed until beam deflections are dis-cussed in Chapter 9.
