Question 1.13: Continuous cable ADB runs over a small frictionless pulley a...
Continuous cable ADB runs over a small frictionless pulley at D to support beam OABC, which is part of an entrance canopy for a building (see Fig.1-62). Load P = 1000 lb is applied at the end of the canopy at C. Assume that the canopy segment has weight W = 1700 lb.
(a) Find cable force T and pin support reactions at O and D.
(b) Find the required cross-sectional area of cable ADB if the allowable normal stress is 18 ksi.
(c) Determine the required diameter of the pins at O, A, B, and D if the allowable stress in shear is 12 ksi.
(Note: The pins at O, A, B, and D are in double shear. Also, consider only load P and the weight W of the canopy; disregard the weight of cable ADB.)
Use the following four-step problem-solving approach.
1. Conceptualize: Begin with a free-body diagram of beam OABC (Fig.1-63a). Also sketch free-body diagrams of the entire structure (Fig.1-63b) and of joint D alone (Fig.1-63c). Show cable force T and all applied and reaction force components.
2. Categorize: First, use the free-body diagram of beam OABC (Fig.1-63a) to find cable force T and reaction force components at O. Then use Fig.1-63b or Fig.1-63c to find reaction forces at D. Use cable force T and the allowable normal stress to find the required cross-sectional area of the cable. Also use force T and the allowable shear stress to find the required diameter of the pins at A and B. Use the resultant reaction forces at O and D to find pin diameters at these locations.
3. Analyze:
Cable force T: First, find required distances and angles in Fig.1-63a:
A D=\sqrt{5^{2}+8^{2}-2(5)(8) \cos 70^{\circ}} ft =7.851 ft \quad B D=\sqrt{10^{2}+8^{2}-2(10)(8) \cos 70^{\circ}} ft =10.454 ft\theta_{A x}=\sin ^{-1}\left(\frac{8 ft }{A D} \sin 70^{\circ}\right)-20^{\circ}=53.241^{\circ} \quad \theta_{B x}=\sin ^{-1}\left(\frac{8 ft }{B D} \sin 70^{\circ}\right)-20^{\circ}=25.983^{\circ}
Now sum moments about O in Fig.1-63a to find tension T in continuous cable ADB:
T=\frac{W(7.5 ft )\left(\cos 20^{\circ}\right)+P(15 ft )\left(\cos 20^{\circ}\right)}{d_{1}+d_{2}}=2177 lb (a)
where
d_{1}=\cos \theta_{A x}\left(8 ft -A D \sin \theta_{A x}\right)+\sin \theta_{A x}\left(5 ft \cos 20^{\circ}\right)=4.788 ftd_{2}=\cos \theta_{B x}\left(8 ft -B D \sin \theta_{B x}\right)+\sin \theta_{B x}\left(10 ft \cos 20^{\circ}\right)=7.191 ft
Reaction force at O: Sum forces in Fig.1-63a to find reaction force components at O:
\Sigma F_{x}=0 \quad R_{O x}=T\left(\cos \theta_{A x}+\cos \theta_{B x}\right)=3260 lb\Sigma F_{y}=0 \quad R_{O y}=-T\left(\sin \theta_{A x}+\sin \theta_{B x}\right)+W+P=2 lb
The resultant reaction force at O is R_{\text {Ores }}=\sqrt{R_{O x}^{2}+R_{O y}^{2}}=3260 lb (b)
Reaction force at D: Sum forces in Fig.1-63c to find reaction force components at D:
\Sigma F_{x}=0 \quad D_{x}=-T\left(\cos \theta_{A x}+\cos \theta_{B x}\right)=-3260 lb\Sigma F_{y}=0 \quad D_{y}=T\left(\sin \theta_{A x}+\sin \theta_{B x}\right)=2698 lb
The resultant reaction force at D is
D_{ res }=\sqrt{D_{x}^{2}+D_{y}^{2}}=4231 lb (c)
Cross-sectional area of cable ADB: Use the allowable normal stress of 18 ksi and cable force T = 2177 lb [Eq.(a)] to find the required cross-sectional area of the cable:
A_{\text {cable }}=\frac{2177 lb }{18 ksi }=0.121 in ^{2}Required diameter of the pins at O, A, B, and D: All pins are in double shear. The allowable shear stress is \tau_{allow} = 12 ksi. Required diameters of each pin are computed as
Pins A , B : A_{\text {reqd }}=\frac{T}{2 \tau_{\text {allow }}}=\frac{2177 lb }{2(12 ksi )}=0.091 in ^{2} \text { so } d=\sqrt{\frac{4}{\pi}\left(0.091 in ^{2}\right)}=0.340 in
Pin O : A_{\text {reqd }}=\frac{R_{\text {Ores }}}{2 \tau_{\text {allow }}}=\frac{3260 lb }{2(12 ksi )}=0.136 in ^{2} \text { so } d=\sqrt{\frac{4}{\pi}\left(0.136 in ^{2}\right)}=0.416 in
Pin D : A_{\text {reqd }}=\frac{D_{\text {res }}}{2 \tau_{\text {allow }}}=\frac{4231 lb }{2(12 ksi )}=0.176 in ^{2} \text { so } d=\sqrt{\frac{4}{\pi}\left(0.176 in ^{2}\right)}=0.474 in
4. Finalize: In practice, other loads besides the weight of the canopy would have to be considered before making a final decision about the sizes of the cables and pins. Loads that could be important include wind loads, earthquake loads, and the weights of objects that might have to be supported temporarily by the structure. In addition, if cables AD and BD are separate cables (instead of one continuous cable ADB), the forces in the two cables are not equal in magnitude. The structure is now statically indeterminate, and the cable forces and the reactions at O and D cannot be determined using the equations of static equilibrium alone. Problems of this type are discussed in Chapter 2, Section 2.4 (see Example 2-7).
