Question 2.13: The stress distribution on the rectangular cross section sho......
The stress distribution on the rectangular cross section shown in Fig. 1 is given by
σ_x = (800y – 400z + 1200) psi
τ_{xy} = 0, τ _{xz} = 300(9 – z²) psi
Determine the net internal force system (i.e., the resultant forces and moments) on this cross section. Let b = 6 in. and h = 8 in.
Plan the Solution At the cross section there can, in general, be three components of force—F, V_y, and V_z—and three moments—T, M_y, and M_z. Using the given distribution of stresses and using Eqs. 2.33 and 2.34, we can calculate these stress resultants.
F(x) = \int_A{σ_x dA}V_y(x)=\int_A{τ _{xy} dA} Force Resultants (2.33)
V_z(x)= \int_A{τ _{xz} dA}T(x)=\int_A{yτ _{xz} dA} – \int_A{zτ _{xy dA}}
M_y(x)=\int_A{zσ _{x} dA} Moment Resultants (2.34)
M_z(x)=-\int_A{yσ _{x} dA}F = \int_{A}{\sigma _x dA= 800}\int_{A}{y dA – 400} \int_{A}{z dA + 1200} \int_{A}{dA}
= 800 \bar{y}A – 400 \bar{z}A + 1200A
= (1200 psi)(8 in.)(6 in.) = 57,600 lb
(Since the origin of y and z is at the centroid of the cross section, \bar{y} = \bar{z} = 0.) Continuing,
V_y = \int_{A}{ τ_{xy} dA } = 0V_z = \int_{A}{ τ_{xz} dA } = 2700 \int_{A}{dA – 300 }\int_{A}{z^2 dA}
From Appendix C.2, for the rectangular cross section of “base” b and “height” h, \int_{A}{z^2 dA} = \frac{1}{12} hb^3 and \int_{A}{y^2 dA} = \frac{1}{12} bh^3. Therefore,
V_z = 2700bh – \frac{300} {12} (hb³)
= (2700 psi)(8 in.)(6 in.) – (25 lb/in^4)(8 in.)(6 in.)³
= 86,400 lb
T = \int_{A}{y τ_{xz} dA } – \int_{A}{ zτ_{xy} dA }= 2700 \int_{A}{y dA} – 300 \int_{A}{y z^2 dA}
= 2700\bar{y}A – 300 \int^3_{-3}{z^2}\int^4_{-4}{y dy dz} = 0
M_y = \int_A{zσ_x dA}= 800\int_A yz dA – 400 \int _A {z^2 dA + 1200}\int_A {z dA}
= -\frac{400}{12}(hb^3) = -\left(\frac{400}{12} lb/in^3\right) (8 in.)(6 in.)³
= -57,600 lb · in.
+400 \int_A {yz dA} – 1200 \int_A{y dA}
-\frac{800}{12}(hb^3) = -\left(\frac{800}{12} lb/in^3\right) (6 in.)(8 in.)³
= -204,800 lb · in.
In summary,
F = 57.6 kips, V_y = 0, V_z = 86.4 kips
T = 0, M_y = -57.6 kip · in., M_z = -205 kip · in
Review the Solution The only way to check the above answers is to go back over the calculations, using information from Appendix C to check all of the integrals. We can also spot-check some of the magnitudes.
For example, the 1200-psi term in σ_x represents constant normal stress on the cross section, which would produce an axial force F = σ_xA = (1200 psi)(8 in.)(6 in.) = 57.6 kips. We can also see that 0 ≤ τ _{xz} ≤ 2700 psi everywhere on the cross section. Hence, V_z must be less than (2700 psi)(8 in.)(6 in.) = 129.6 kips. So the value of V_z = 86.4 kips seems reasonable.
Stress distributions like the one in this example will arise in Chapter 6 in the study of bending of beams.