Question 9.5: Wind blowing on a sign produces a pressure whose resultant, ......
Wind blowing on a sign produces a pressure whose resultant, P, acts in the -y direction at point C, as shown in Fig. 1. The weight of the sign, W_s, acts vertically through point C, and the thin-wall pipe that supports the sign has a weight W_p.
Following the procedure outlined at the beginning of Section 9.4, determine the principal stresses at points A and B, where the pipe column is attached to its base. Use the following numerical data.
Pipe OD = 3.50 in., A = 2.23 in², I_y = I_z = 3.02 in^4, I_p = 6.03 in^4,
W_s = 125 lb, W_p = 160 lb, P = 75 lb, b = 40 in., L = 220 in.
Plan the Solution It will be a good idea to tabulate the stress resultants, stress formulas, and so forth, so that no stress contribution will be missed. The weight W_s contributes to the axial force, and it also produces a moment about the y axis. The wind force P produces a transverse shear force in the y direction, and it also causes a torque about the x axis and a moment about the z axis. A correct free-body diagram is essential.
Stress Resultants: All six stress resultants on the cross section at the base of the pipe are shown in Fig. 1. The upper portion of Fig. 1 can serve as a free-body diagram for determining these six stress resultants. The sign convention is the one introduced in Fig. 2.40. Let us tabulate the equilibrium equations and indicate what stress is produced by each stress resultant and label each individual stress.
Individual Stresses: Using the formulas from Table 9.1, we can compute the numerical value of each of the nonzero stresses listed in Table 1.
σ_{A1} = σ_{B1} = \frac{F}{A}=\frac{-(125 lb) – (160 lb)}{2.23 in^2} = -128 psi (7)
The shear stress τ_{B2} is due to the transverse shear force V_y. The basic shear stress formula is
τ_{B2} = \frac{V_yQ}{I_zt} (8)
where Q has to be calculated for the shaded area in Fig. 2. In Example Problem 6.16, it was shown that the shear stress in this case (stress at the neutral axis of a thin-wall pipe) is given by
τ = \frac{2V}{A} (9a)
Therefore,
τ_{B2} = \frac{2(75 lb)}{2.23 in^2} = 67 psi (9b)
τ_{A4} = τ_{B4} = \frac{Tr_o}{I_p}=\frac{(Pb)r_o}{I_p} (10a)
so
τ_{A4} = τ_{B4} =\frac{(75 lb)(40 in.)(1.75 in.)}{6.03 in^4} = 871 psi (10b)
The flexural stresses due to M_y and M_z are given by Eq. 6.30.
σ_x = \frac{M_yz}{I_y}-\frac{M_zy}{I_z} Flexure Formula (6.30)
σ_{B5} = \frac{M_yr_o}{I_y}= \frac{(-W_sb)r_o}{I_y} (11a)
σ_{B5} = \frac{-(125 lb)(40 in.)(1.75 in.)}{3.02 in^4} = -2897 psi (11b)
σ_{A6} = \frac{-M_zr_o}{I_z}= \frac{-(-PL)r_o}{I_z} (12a)
σ_{A6} = \frac{(75 lb)(220 in.)(1.75 in.)}{3.02 in^4} = 9561 psi (12b)
Superposition of Stresses: Using the above values, and taking proper note of the physical significance of the sign of each term by referring to Fig. 1, we get the stresses shown in Fig. 3.
Using the stresses shown in Fig. 3, we can construct a Mohr’s circle for the states of plane stress at points A and B on the pipe surface. The radial normal stress is σ_r = 0 at both places. From Fig. 4a.
R_A = \sqrt{(9433/2)^2 + (871)^2} = 4796 psi (13)
(σ_1)_A = (9433/2) + 4796 = 9513 psi
(σ_3)_A = (9433/2) – 4796 = -80 psi (14)
and, from Fig. 4b,
R_B = \sqrt{(-3025/2)^2 + (938)^2} = 1780 psi (15)
(σ_1)_B = (-3025/2) + 1780 = 267 psi
(σ_3)_B = (-3025/2) – 1780 = -3292 psi (16)
In summary, the principal stresses at points A and B, rounded to three significant figures, are:
(σ_1)_A = 9510 psi, (σ_2)_A = 0, (σ_3)_A = -80 psi
(σ_1)_B = 267 psi, (σ_2)_B = 0, (σ_3)_B = -3290 psi
Review the Solution By showing all six possible internal resultants at the cross section where stresses are to be calculated, by writing down and solving all six possible equilibrium equations, and by carefully considering what stress(es) is (are) produced by each stress resultant, we have accounted for the effects of all loads on the structure. As noted earlier, we have been careful to make sure that each stress component acts in the direction that “makes sense.” For example, the force P bends the pipe in the direction that produces tension at point A, and so forth.
The maximum flexural stress at the base occurs at neither A nor B.
Equation 6.30 could be used to combine the flexural stresses due to M_y and M_z, and we would also have to consider the effect of shear stress.
(See Homework Problem 9.4-26.)
TABLE 9 . 1 Formulas for Stresses
| Stress Resultant | Symbol | Formula | References |
| Normal force | F | σ= \frac{F}{A} | Sections 2.2, 3.2 |
| Torsional moment | T | τ =\frac{Tρ}{I_p} | Section 4.3 |
| Bending moment | M | σ = \frac{-My}{I} | Section 6.3 |
| Transverse shear force | V | τ =\frac{VQ}{It} | Section 6.8 |
TABLE 1 A Table of Stress Resultants and the Stresses Produced
| Eq. No. | Equilibrium Equation | Stress
at A |
Stress at B |
| (1) | ∑F_x = 0 F=-W_s – W_p | σ_{A1} | σ_{B1} |
| (2) | ∑F_y = 0 V_y=-P | — | τ_{B2} |
| (3) | ∑F_z = 0 V_z=0 | — | — |
| (4) | ∑M_x = 0 T = Pb | τ_{A4} | τ_{B4} |
| (5) | ∑M_y = 0 M_y=-W_sb | — | σ_{B5} |
| (6) | ∑M_z = 0 M_z = -PL | σ_{A6} | — |
