Question 18.CS.2: Design Analysis of Arm CD The arm CD of a winch crane is rep...
Design Analysis of Arm CD
The arm CD of a winch crane is represented schematically in Figure 18.2. Determine the maximum stress and the factor of safety against yielding. What is the deflection under the load using the method of superposition?
Given: The geometry and loading are known from Case Study 18.1. The frame is made of ASTM-A36 structural steel tubing. From Table B.1,
S_{y} = 250 MPa E = 200 GPa
Assumptions: The loading is static. The displacements of welded joint C are negligibly small; hence, part CD of the frame is considered a cantilever beam.
See Figures 18.2 and 18.3; and Table B.1, Section 3.7.
We observe from Figure 18.2 that the maximum bending moment occurs at points B and C and M_{B} = M_{C} = M. Since two vertical beams resist moment at B, the critical
section is at C of cantilever CD carrying its own weight per unit length w and concentrated load P at the free end (Figure 18.3).
The bending moment M and shear force V at the cross section through the point C, from static equilibrium, have the following values:
\begin{aligned} M &=P L_1+\frac{1}{2} w L_1^2 \\ &=3000(1.5)+\frac{1}{2}(130)(1.5)^2=4646 N \cdot m \\ V &=3 kN \end{aligned}
The cross-sectional area properties of the tubular beam are
\begin{aligned} A &=b h-(b-2 t)(h-2 t) \\ &=50 \times 100-38 \times 88=1.66\left(10^{-3}\right) m ^2 \end{aligned}
\begin{aligned} I &=\frac{1}{12} b h^3-\frac{1}{12}(b-2 t)(h-2 t)^3 \\ &=\frac{1}{12}\left[\left(50 \times 100^3\right)-(38)(88)^3\right]=2.01\left(10^{-6}\right) m ^4 \end{aligned}
where I represents the moment of inertia about the neutral axis.
Therefore, the maximum bending stress at the top of outer fiber of section through C equals
\sigma_{\max }=\frac{M c}{I}=\frac{4646(0.05)}{2.01\left(10^{-6}\right)}=115.6 MPa
where the shear strain is zero. The highest value of the shear stress occurs at the neutral axis. Referring to Figure 18.3 and Equation 3.21, the first moment of the area about the NA is
Q=\int_{A^*} y d A=A^* \bar{y} (3.21)
\begin{aligned} Q_{\max } &=b\left\lgroup \frac{h}{2} \right\rgroup \left\lgroup \frac{h}{4} \right\rgroup -(b-2 t)\left\lgroup \frac{h}{2}-t \right\rgroup \left\lgroup \frac{h / 2-t}{2} \right\rgroup \\ &=50(50)(25)-(38)(44)(22)=25.716\left(10^{-6}\right) m ^3 \end{aligned}
Hence,
\begin{aligned} \tau_{\max } &=\frac{V Q_{\max }}{I(2 t)} \\ &=\frac{3000(25.716)}{2.01(2 \times 0.006)}=3.199 MPa \end{aligned}
The factor of safety against yielding is then equal to
n=\frac{S_y}{\sigma_{\max }}=\frac{250}{115.6}=2.16
This is satisfactory because the frame is made of average material operated in ordinary environment and subjected to known loads.
Comments: At joint C, as well as at B, a thin (about 6 mm) steel gusset should be added at each side (not shown in the figure). These enlarge the weld area of the joints and help reduce stress in the welds. Case Study 18.9 illustrates the design analysis of the welded joint at C.
When the load P and the weight w of the cantilever depicted in the figure act alone, displacements at D (from cases 1 and 3 of Table A.9) are P L_1^3 / 3 E I \text { and } w L_1^4 / 8 E I, , respectively. It follows that the deflection \upsilon _{D} at the free end owing to the combined loading is
\upsilon _{D}=-\frac{P L_1^3}{3 E I}-\frac{w L_1^4}{8 E I}
Substituting the given numerical values into the preceding expression, we have
\begin{aligned} \upsilon _{D} &=-\frac{1}{200\left(10^3\right)(2.01)}\left[\frac{3000(1.5)^3}{3}+\frac{130(1.5)^4}{8}\right] \\ &=-8.6 mm \end{aligned}
Here, the minus sign means a downward displacement.
Comment: Since \upsilon _{D} \ll h / 2 , the magnitude of the deflection obtained is well within the acceptable range (see Section 3.7).
