Question 7.6: A tubular post of square cross section supports a horizontal...
A tubular post of square cross section supports a horizontal platform (Fig. 7-21). The tube has outer dimension b = 150 mm and wall thickness t = 13 mm. The platform has dimensions 175 mm × 600 mm and supports a uniformly distributed load of 140 kPa acting over its upper surface. The resultant of this distributed load is a vertical force P_1:
P_1 = (140 kPa)(175 mm × 600 mm) = 14.7 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 and P_2.
Stress resultants. The force P_1 acting on the platform (Fig. 7-21) is statically equivalent to a force P_1 and a moment M_1 = P_1d acting at the centroid of the cross section of the post (Fig. 7-22a). 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. 7-22b. 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_1d = (14.7 kN)(225 mm) = 3307.5 N.m
3. A shear force P_2 = 800 lb
4. A bending moment M_2 produced by the force P_2:
M_2=P_2h = (3.6 kN)(1.3 m) = 4.68 kN.m
Examination of these stress resultants (Fig. 7-22b) 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. 7-22b) produces uniform compressive stresses throughout the post. These stresses are
\sigma _{P_1}=\frac{P_1}{A}
in which A is the cross-sectional area of the post:
A = b² – (b – 2t)² = 4t(b – t)
= 4(13 mm)(150 mm – 13 mm) = 7124 mm²
Therefore, the axial compressive stress is
\sigma _{P_1}=\frac{P_1}{A} =\frac{14.7 kN}{7124 mm^2} =2.06 MPa
The stress \sigma_{P_1} is shown acting at points A and B in Fig. 7-22c.
(2) The bending moment M_1 (Fig. 7-22b) produces compressive stresses \sigma_{M_1} at points A and B (Fig. 7-22c). These stresses are obtained from the flexure formula:
\sigma _{M_1}=\frac{M_1(b/2)}{I} =\frac{M_1b}{2I}
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} [(150 mm)^4-(124 mm)^4]= 22.49 \times 10^{-6 } m^4
Thus, the stress \sigma_{M_1} is
\sigma _{M_1}=\frac{M_1b}{2I}=\frac{(3307.5 N.m)(150 mm)}{2(22.49 \times 10^{-6} m^4)} = 11.03 MPa
(3) The shear force P_2 (Fig. 7-22b) 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 5.9), we know that an approximate value of the shear stress can be obtained by dividing the shear force by the web area (see Eq. 5-47 in Section 5.9). Thus, the shear stress produced at point B by the force P_2 is
\tau _{aver} = \frac{V}{th_1} (5-47)
\tau _{P_2}=\frac{P_2}{A_{web}} =\frac{P_2}{2t(b-2t)} =\frac{3.6 kN}{2(13 mm)(150 mm – 26 mm)} = 1.12 MPa
The stress \tau_{P_2} acts at point B in the direction shown in Fig. 7-22c.
If desired, we can calculate the shear stress \tau_{P_2} from the more accurate formula of Eq. (5-45a) in Section 5.9. 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.
\tau _{max}=\frac{V}{8It} (bh^2 – bh_1^2 + th_1^2) (5-45 a)
(4) The bending moment M_2 (Fig. 7-22b) 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_2b}{2I}=\frac{(4.68 kN.m)(150 mm)}{2(22.49 \times 10^{-6} m^4)} = 15.61 MPa
This stress is also shown in Fig. 7-22c.
Stress elements. The next step is to show the stresses acting on stress elements at points A and B (Figs. 7-22d 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. 7-22d):
\sigma_A = \sigma_{P_1} + \sigma_{M_1} + \sigma_{M_2}
= 2.06 MPa + 11.03 MPa + 15.61 MPa = 28.7 MPa (compression)
Thus, this element is in uniaxial stress.
At point B the compressive stress in the y direction (Fig. 7-22e) is
\sigma_B = \sigma_{P_1} + \sigma_{M_1} = 2.06 MPa + 11.03 MPa = 13.1 MPa (compression)
and the shear stress is
\tau_{P_2} = 1.12 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. 7-23), we write the stresses for element A (Fig. 7-22d) as follows:
\sigma_x= 0 \sigma_y = – \sigma_A = – 28.7 MPa \tau_{xy} = 0
Since the element is in uniaxial stress, the principal stresses are
\sigma_1 = 0 \sigma_2 =- 28.7 MPa
and the maximum in-plane shear stress (Eq. 7-26) is
\tau _{max}=\frac{\sigma _1-\sigma _2}{2} =\frac{-28.7 MPa}{2} = 14.4 MPa
The maximum out-of-plane shear stress (Eq. 6-28a) has the same magnitude.
(\tau_{max})_{x_1} =\pm \frac{\sigma_2}{2} (6-28 a)
Principal stresses and maximum shear stresses at point B. Again using the standard notation for plane stress (Fig. 7-23), we see that the stresses at point B (Fig. 7-22e) are
\sigma_x = 0 \sigma_y = – \sigma_B = -13.1 MPa \tau_{xy} = -\tau_{P_2} = -1.12 MPa
To obtain the principal stresses, we use Eq. (6-17), which is repeated here:
\sigma _{1,2}=\frac{\sigma _x+\sigma _y}{2} \pm \sqrt{(\frac{\sigma _x-\sigma _y}{2} )^2+\tau _{xy}^2} (6-17) (m)
Substituting for \sigma_x, \sigma_y , and \tau_{xy} , we get
\sigma _{1,2} = – 6.55 MPa ± 6.65 MPa
or
\sigma _1 = 0.1 MPa \sigma _2 = – 13.2 MPa
The maximum in-plane shear stresses may be obtained from Eq. (6-25):
\tau _{max}=\sqrt{(\frac{\sigma _x-\sigma _y}{2} )^2+\tau _{xy}^2} (6-25) (n)
This term was evaluated previously, so we see immediately that
\tau_{max} = 6.65 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. 6-28a, b, and c and the accompanying discussion). Therefore, the maximum shear stress at point B is 6.65 MPa.
(\tau_{max})_{x_1} =\pm \frac{\sigma_2}{2} (\tau_{max})_{y_1} =\pm \frac{\sigma_1}{2} (\tau_{max})_{z_1} =\pm \frac{\sigma_1-\sigma_2}{2} (6-28 a, b , 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. 7-22c), 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 MPa + 11.03 MPa + 15.61 MPa = 24.58 MPa
The stresses acting on a stress element at that point (see Fig. 7-23) are
\sigma _x =0 \sigma _y = 24.58 MPa \tau _{xy} = 0
and therefore the principal stresses and maximum shear stress are
\sigma _1 = 24.58 MPa \sigma _2 = 0 \tau_{max} = 12.3 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 considered in this analysis. Other loads, such as the weight of the structure, also produce stresses at the base of the post.)
