Question 8.CA.1: Two forces P1 and P2, with a magnitude of P1 = 15 kN and P2 ......
Two forces P _1 \text { and } P _2 \text {, with a magnitude of } P_1=15 kN \text { and } P_2 = 18 kN, are applied as shown in Fig. 8.21a to the end A of bar AB, which is welded to a cylindrical member BD of radius c = 20 mm. Knowing that the distance from A to the axis of member BD is a = 50 mm and assuming that all stresses remain below the proportional limit of the material, determine (a) the normal and shearing stresses at point K of the transverse section of member BD located at a distance b = 60 mm from end B, (b) the principal axes and principal stresses at K, and (c) the maximum shearing stress at K.
Internal Forces in Given Section. Replace the forces P _1 \text { and } P _2 by an equivalent system of forces and couples applied at the center C of the section containing point K (Fig. 8.21b). This system represents the internal forces in the section and consists of the following forces and couples:
1. A centric axial force F equal to the force P _1 with the magnitude
F = P _1 = 15 kN
2. A shearing force V equal to the force P _2 with the magnitude
V = P _2 = 18 kN
3. A twisting couple T of torque T equal to the moment of P _2 about the axis of member BD:
T = P _2 a = (18 kN)(50 mm) = 900 N·m
4. A bending couple M_y of moment M_y equal to the moment of P _1 about a vertical axis through C:
M_y=P_1 a = (15 kN)(50 mm) = 750 N·m
5. A bending couple M _z of moment M _z equal to the moment of P _2 about a transverse, horizontal axis through C:
M_z=P_2 b = (18 kN)(60 mm) = 1080 N·m
The results are shown in Fig. 8.21c.
a. Normal and Shearing Stresses at Point K. Each of the forces and couples shown in Fig. 8.21c produce a normal or shear stress at point K.
Compute each of these stresses separately and then add the normal stresses and add the shearing stresses.
Geometric Properties of the Section. For the given data, we have
\begin{aligned} & A=\pi c^2=\pi(0.020 m )^2=1.257 \times 10^{-3} m ^2 \\& I_y=I_z=\frac{1}{4} \pi c^4=\frac{1}{4} \pi(0.020 m )^4=125.7 \times 10^{-9} m ^4 \\ & J_C=\frac{1}{2} \pi c^4=\frac{1}{2} \pi(0.020 m )^4=251.3 \times 10^{-9} m ^4 \end{aligned}
Also determine the first moment Q and the width t of the area of the cross section located above the z axis. Recall that \bar{y}=4 c / 3 \pi for a semicircle of radius c, giving
\begin{aligned} Q & =A^{\prime} \bar{y}=\left(\frac{1}{2} \pi c^2\right)\left(\frac{4 c}{3 \pi}\right)=\frac{2}{3} c^3=\frac{2}{3}(0.020 m )^3 \\ & =5.33 \times 10^{-6} m ^3 \end{aligned}
and
t = 2c = 2(0.020 m) = 0.040 m
Normal Stresses. Normal stresses are produced at K by the centric force F and the bending couple M _y . However, the couple M _z does not produce any stress at K, since K is located on the neutral axis corresponding to that couple.
Determining each sign from Fig. 8.21c gives
\begin{aligned} \sigma_x & =-\frac{F}{A}+\frac{M_y c}{I_y}=-11.9 MPa +\frac{(750 N \cdot m )(0.020 m )}{125.7 \times 10^{-9} m ^4} \\ & =-11.9 MPa +119.3 MPa \\ & =+107.4 MPa \end{aligned}
Shearing Stresses. The shearing stress \left(\tau_{x y}\right)_V is due to the vertical shear V, and the shearing stress \left(\tau_{x y}\right)_{\text {twist }} is caused by the torque T. Using the values for Q, t, I_z \text {, and } J_C ,
\begin{aligned} \left(\tau_{x y}\right)_V & =+\frac{V Q}{I_z t}=+\frac{\left(18 \times 10^3 N \right)\left(5.33 \times 10^{-6} m ^3\right)}{\left(125.7 \times 10^{-9} m ^4\right)(0.040 m )} \\ & =+19.1 MPa \\ \left(\tau_{x y}\right)_{ twist } & =-\frac{T c}{J_C}=-\frac{(900 N \cdot m )(0.020 m )}{251.3 \times 10^{-9} m ^4}=-71.6 MPa \end{aligned}
Adding these provides \tau_{x y} at point K.
\begin{aligned} \tau_{x y} & =\left(\tau_{x y}\right)_V+\left(\tau_{x y}\right)_{\text {twist }}=+19.1 MPa -71.6 MPa \\ & =-52.5 MPa \end{aligned}
In Fig. 8.21d, the normal stress \sigma_x and the shearing stresses \tau_{x y} are acting on a square element located at K on the surface of the cylindrical member. Note that shearing stresses acting on the longitudinal sides of the element also are included.
b. Principal Planes and Principal Stresses at Point K. Either of the two methods from Chap. 7 can be used to determine the principal planes and principal stresses at K. Selecting Mohr’s circle, plot point X with coordinates \sigma_x=+107.4 MPa \text { and }-\tau_{x y}=+52.5 MPa and point Y with coordinates \sigma_y=0 \text { and }+\tau_{x y}=-52.5 MPa and draw the circle with the diameter XY (Fig. 8.21e). Observing that
O C=C D=\frac{1}{2}(107.4)=53.7 MPa \quad D X=52.5 MPa
we determine the orientation of the principal planes:
\begin{array}{r} \tan 2 \theta_p=\frac{D X}{C D}=\frac{52.5}{53.7}=0.97765 \quad 2 \theta_p=44.4^{\circ} ⤸ \\ \theta_p=22.2^{\circ} ⤸ \end{array}
The radius of the circle is
R=\sqrt{(53.7)^2+(52.5)^2}=75.1 MPa
and the principal stresses are
\begin{aligned} & \sigma_{\max }=O C+R=53.7+75.1=128.8 MPa \\ & \sigma_{\min }=O C-R=53.7-75.1=-21.4 MPa \end{aligned}
The results are shown in Fig. 8.21f.
c. Maximum Shearing Stress at Point K. This stress corresponds to points E and F in Fig. 8.21e
\tau_{\max }=C E=R=75.1 MPa
Observing that 2 \theta_s=90^{\circ}-2 \theta_p=90^{\circ}-44.4^{\circ}=45.6^{\circ} , the planes of maximum shearing stress form an angle \theta_s = 22.8° ⤻ with the horizontal. The corresponding element is shown in Fig. 8.21g. Note that the normal stresses acting on this element are represented by OC in Fig. 8.21e and are equal to +53.7 MPa.
