Question 9.3: An axial compressive load of 800 kips is applied eccentrical......
An axial compressive load of 800 kips is applied eccentrically to a short rectangular compression member, as shown in Fig. 1. (The effect of eccentric compressive loading on longer members is treated in Section 10.4.) Determine the distribution of normal stress on a cross section, say ABCD, that is far enough from the point of load application that stress concentration effects may be neglected. Sketch the stress distribution and identify the location of the neutral axis in cross section ABCD.
Plan the Solution The eccentric load P produces axial deformation plus bending about the y and z axes. Therefore, this problem involves a superposition of the stresses due to F, M_y, and M_z (Eq. 9.10).
σ_x = \frac{F}{A}= \frac{M_yz}{I_y}- \frac{M_zy}{I_z} (9.10)
Stress Resultants: Figure 2 shows the stress resultants on cross section ABCD. The sign conventions for stresses and stress resultants is the same as the sign conventions adopted previously in Chapters 2 and 6.
Applying equilibrium to a free-body diagram of the member above section ABCD we get the following expressions for the stress resultants:
F = -P
M_y = -Pe_z (1)
M_z = Pe_yIndividual Normal Stresses: Combining Eq. 2.4, for the normal stress due to the axial force F, with Eq. 6.30, for the normal stress due to the bending-moment components, we have Eq. 9.10:
σ(x) = \frac{F(x)}{A(x)} Axial Stress Equation (2.4)
σ_x = \frac{M_yz}{I_y} – \frac{M_zy}{I_z} Flexure Formula (6.30)
σ_x = \frac{F}{A}= \frac{M_yz}{I_y}- \frac{M_zy}{I_z} (2)
Taking each stress contribution separately, and combining Eqs. (1) and (2), we obtain the following:
(σ_x)_F = \frac{-P}{A}= \frac{-800 kips}{(40 in.)(20 in.)} = -1000 psi
(σ_x)_{My} = \frac{(-Pe_z)z}{I_y}= \frac{(-800 kips)(5 in.)z}{[\frac{1}{12}(40 in.)(20 in.)^3 ]} = -150z psi (3)
(σ_x)_{Mz} = \frac{-(Pe_y)y}{I_z}= \frac{(-800 kips)(10 in.)y}{[\frac{1}{12}(20 in.)(40 in.)^3 ]} = -75y psi
These stress contributions are sketched in Fig. 3.The maximum values of the bending stresses are
Max|(σ_x)M_y| = 150z|_{z=10 in} = 150(10 in.) = 1500 psi
Max|(σ_x)M_z| = 75y|_{y=20 m} = 75(20 in.) = 1500 psi
Superposition of Stresses: Using Figs. 3a through 3c, we can combine, algebraically, the individual stress contributions at four corners to get
(σ_x)_A = -1000 + 1500 – 1500 = -1000 psi
(σ_x)_B = -1000 – 1500 – 1500 = -4000 psi
(σ_x)_C = -1000 – 1500 + 1500 = -1000 psi (4)
(σ_x)_D = -1000 + 1500 + 1500 = 2000 psi
With the aid of these corner stresses, we can sketch the combined stress distribution. Since Eq. (2) is linear in y and z, the stress surface will be a plane (which has been “folded” to show tension and compression as they are shown in Fig. 3). In Fig. 4 the member has been rotated about the x axis to provide a better perspective for viewing the stress surface and the neutral axis, which is the intersection, RS, of the stress surface with the ABCD plane.
The equation of the neutral axis line is given by setting σ_x(y*, z*) = 0, where (y*, z*) are coordinates of points on the neutral axis.
Combining Eqs. (2) and (3) we get
σ_x(y*, z*) = -1000 – 75y* – 150z* = 0 (5)
The intersection, R, of the neutral axis with edge CD is given by setting y* = -20 in. This gives (y*, z*)_R = (-20 in., 3.33 in.). Similarly, (y*, z*)_S = (6.67 in., 10 in.).
Review the Solution It is obvious from the location of the compressive force P in Fig. 1 that corner B will have the highest compressive stress of any point in the cross section. This is confirmed by Eq. (4b) and by Fig. 4. The load P is far enough from the axis of the member that it actually causes tension at D. This means that between B and D the stress changes from compression to tension. Hence, there is a neutral axis (σ_x = 0) that passes between A and D. Therefore, Fig. 4 appears to represent accurately the stress distribution due to the eccentric load P.
