Question 18.CS.1: Entire Frame Load Analysis Consider the crane winch depicted...
Entire Frame Load Analysis
Consider the crane winch depicted in Figure 18.1. The entire frame of this machine is illustrated in Figure 18.2. Determine
a. The design load on the front and rear wheels
b. The factor of safety n_{t} for the crane tipping forward from the loading
Given: The geometry of each element is known. The cable and hook are rated at 15 kN, which gives a safety factor of 5. The 85 mm diameter drum is about 20 times the cable diameter. The crane frame carries the load P, counterweight W_{C}, weights of parts W_{i} (i = 1, 2, 3, 4, 5), and the push force F, as shown in Figure 18.2. The frame is made of b = 50 mm and h = 100 mm structural steel tubing of t = 6 mm thickness with weight w Newton per meter (Table A.4).
Data:
\begin{array}{rlrlrl} P & =3 kN & F & =100 N & W_C & =2.7 kN \\ w & \approx 130 N / m & a & =0.8 m & H & =1 m \\ L_1 & =1.5 m & L_2 & =2 m & L_3 & =1 m \\ L_4 & =0.5 m & L_5 & =0.65 m & & \end{array}
and
\begin{array}{l} W_1=w L_1=130(1.5)=195 N \quad W_2 \approx w L_2=260 N \\ W_3=130 N \quad W_4=65 N \quad W_5=84.5 N \end{array}
For dimensions and properties of a selected range of frequently used crane members, refer to manufacturers’ catalogs
Assumptions:
1. A line speed of 0.12 m/s is used, as suggested by several catalogs for lifting. The efficiency of the speed reduction unit or gearbox is 95%. The electric motor has 0.5 hp capacity to lift 3 kN load for the preceding line speed and efficiency and includes an internal brake to hold the load when it is inoperative. The gear ratios (see Case Study 18.4) satisfy the drive system requirements.
2. Only the weights of concrete counterbalance and main frame parts are considered. All frame parts are weld connected to one another.
3. Compression forces caused by the cable running along the members are ignored. All forces are static; F is x directed (horizontal) and remaining forces are parallel to the xy plane. Note that the horizontal component of the reaction at B equals F/2, not indicated in Figure 18.2.
See Figure 18.2; Section 1.9.
a. Reactional forces R_{A} and R_{B} acting on the wheels are determined by applying conditions of equilibrium, \sum M_z=0 and \sum F_y=0 , to the free-body diagram shown in the figure with F = 0. Therefore,
\begin{array}{l} R_A=\frac{1}{2}\left\lgroup P \frac{L_1}{L_3}+\frac{1}{2} W_1 \frac{L_1}{L_3}+\frac{1}{4} W_C+W_3+\frac{1}{2} W_5 \right\rgroup \\ R_B=-R_A+\frac{1}{2} P+\frac{1}{2} W_1+W_2+W_3+\frac{1}{2} W_4+\frac{1}{2} W_C+\frac{1}{2} W_5 \end{array} (18.1)
Substitution of the given data into the foregoing results in
R_A=\frac{1}{2}\left[3000(1.5)+\frac{1}{2}(195)(1.5)+\frac{1}{4}(2700)+130+\frac{1}{2}(84.5)\right]=2747 N
R_B=-2747+\frac{1}{2}(3000)+\frac{1}{2}(195)+260+130+\frac{1}{2}(65)+\frac{1}{2}(2700)+\frac{1}{2}(84.5)=665 N
Note that, when the crane is unloaded (P = 0), Equations 18.1 give
R_A=497 N \quad R_B=1415 NComment: Design loads on front and rear wheels are 2747 N and 1415 N, respectively.
b. The factor of safety n_{t} is applied to tipping loads. The \sum M_z=0 condition at point A is
\begin{aligned} n_t\left[P\left(L_1-L_3\right)+F H\right]=& W_1\left(L_3-\frac{1}{2} L_1\right)+2 W_2 L_3+W_4 L_3+\frac{3}{4} W_C L_3 \\ &+\frac{1}{2}\left(2 W_3+W_5\right) L_3 \end{aligned} (18.2)
Introducing the given numerical values,
\begin{aligned} n_t[3000(0.5)+100(1)]=& 195(0.25)+2(260)(1)+65(1)+\frac{3}{4}(2700)(1) \\ &+\frac{1}{2}[2(130)+84.5](1) \end{aligned}
from which n_{t} = 1.77.
Comments: For the preceding forward tipping analysis, the rear wheels are assumed to be locked and the friction is taken to be sufficiently high to prevent sliding. Side-toside tipping may be checked similarly.