Question 1.11: A basketball player hangs on the rim after dunking the ball....
A basketball player hangs on the rim after dunking the ball. The player applies a downward force at point A with an estimated magnitude of P = 400 lb (Fig.1-56a). Later, the player dunks again and hangs on the rim with two hands: one at A and one at B (see Fig.1-56b, c).The rim and support bracket are bolted to the backboard using four bolts with washers (Fig.1-56d). Find and compare the stresses in the bolted connection at bolt location 2 for the one-handed and two-handed load cases. Assume that the backboard is a fixed support and that bolt and washer diameters are d_{b} = 3/8 in. and d_{w} = 5/8 in., respectively. The support bracket has thickness t = 3/16 in.
Use the following four-step problem-solving approach.
1. Conceptualize: Find reaction forces and moments at support point O and then distribute forces and moments to each bolt location. The rim and bracket act as a cantilever beam. From the free-body diagrams in Fig.1-57, the reactions for each load case are as follows.
Reactions—Load case 1: Load P is applied in the (-y) direction at A. Sum
forces and moments to find:
\Sigma F_{y}=0 \quad R_{y}=P=400 lb
\Sigma M_{z}=0 \quad M_{z 1}=P(5+18) \text { in. }=(400)(23) \text { lb-in. }=9200 \text { lb-in. } (a)
Reactions—Load case 2: Loads P/2 are applied at both A and B. Reactions at O are
\Sigma F_{y}=0 \quad R_{y}=P=400 lb
\Sigma M_{z}=0 \quad M_{z 2}=\frac{P}{2}(5+9) \text { in. }+\frac{P}{2}(23 \text { in. })=[200(14)+200(23)] \text { lb-in. }=7400 lb – in . (b)
\Sigma M_{x}=0 \quad T_{x}=-\frac{P}{2}(9 in .)=-200(9) lb – in .=-1800 lb – in .Forces at bolt 2—Load case 1: Use the negatives of the reactions at O to find normal and shear forces acting on the bolts. From reaction R_{y}, downward shear force P/4 = 100 lb acts at each of the four bolt locations (Fig.1-58a). Replace moment M_{z1} [Eq.(a)] with two force couples, each equal to (N_{1} )(h) (see Fig.1-58a) so normal force N_{1} acts in the (+x) direction at bolt 2 and is computed as
N_{1}=\frac{M_{z 1}}{2 h}=\frac{23 P}{2(3 in .)}=\frac{9200 lb – in .}{6 in .}=1533 lb (c)
Forces at bolt 2—Load case 2: Reaction R_{y} is the same in load cases 1 and 2, so downward shear force P/4 = 100 lb acts at bolt 2 (Fig.1-58b). Replace moment M_{z2} [Eq.(b)] with two force couples (Fig.1-58b), so the tension force on bolt 2 is
N_{2}=\frac{M_{z 2}}{2 h}=\frac{7400 lb – in .}{6 in .}=1233 lb (d)
Load case 2 also creates a torsional reaction moment T_{x} [see Eq.(b)] that can be replaced by two counterclockwise force couples each equal to (V)(d) (see Fig.1-58b) where d=\sqrt{b^{2}+h^{2}}=4.0697 in. Compute the additional in-plane shear force on bolt 2 as
V=\frac{T_{x}}{2 d}=\frac{1800 lb – in .}{2(4.0697 in .)}=221.15 lb (e)
The line of action of force V is shown in Fig.1-58(b) at angle \theta=\tan ^{-1}\left(\frac{b}{h}\right)=42.51^{\circ}. The total in-plane shear force on bolt 2 is the resultant R of forces V and P/4 computed as
R=\sqrt{(V \cos \theta)^{2}+\left(\frac{P}{4}+V \sin \theta\right)^{2}}=\sqrt{(163.02)^{2}+(100+149.43)^{2}} lb =298 lb (f)
Resultant R also can be found using the parallelogram law, as shown in Fig.1-59 with β = 14.32° and \gamma=132.51^{\circ}.
2. Categorize: Use the forces acting on bolt 2 in simple formulas to compute average stresses in the bolt and on the washer and support bracket at bolt location 2. The five connection stresses of interest are (a) normal stress in bolt (Fig.1-60a); (b) shear stress in bolt (Fig.60b); (c) bearing stress on shank of bolt (Fig.1-60c); (d) bearing stress on washer (Fig.1-60d); and (e) shear stress through bracket on periphery of washer (Fig.1-60e).
3. Analyze: The five connection stresses at bolt location 2 are listed in Table 1-2 for load cases 1 and 2. Numerical dimensions for the bolt, washer, and bracket are d_{b} = 3/8 in. , d_{w} = 5/8 in., and t = 3/16 in.
Areas needed in stress calculations are
Cross-sectional area of bolt:
A_{b}=\frac{\pi}{4} d_{b}^{2}=0.1104 in ^{2}Surface area of washer:
A_{w}=\frac{\pi}{4}\left(d_{w}^{2}-d_{b}^{2}\right)=0.1963 in ^{2}Cylindrical area through bracket on periphery of washer:
A_{p}=\pi d_{w} t=0.3682 in ^{2}4. Finalize: Shear and bearing stresses on bolt 2 (items b and c in Table 1-2) are increased three-fold for load case 2 when torsional moment T_{x} [see Eq. (b)] is applied to the bracket. The other three bolt stresses differ by about 25% for the two load cases. The average stresses illustrated in Fig. 1-60 and listed in Table 1-2 are only approximations to the true state of stress at one of the four bolt locations on the support bracket. The true maximum stresses are likely to be higher for a variety of reasons, such as localized stress concentrations, pretensioning of bolts, and impact aspects of the loading. The stress levels computed here are low. If stress values are higher, a more detailed and sophisticated analysis using computer models employing the finite element method may be required.
Table 1-2 Five connection stresses at bolt location 2
| Connection Stress | Load Case 1 | Load Case 2 |
| a. Normal stress in bolt | \frac{N_{1}}{A_{b}}=13,880 psi | \frac{N_{2}}{A_{b}}=11,164 psi |
| b. Shear stress in bolt | \frac{P}{4 A_{b}}=905 psi | \frac{R}{A_{b}}=2698 psi |
| c. Bearing stress on shank of bolt | \frac{P}{4 d_{b} t}=1422 psi | \frac{R}{d_{b} t}=4238 psi |
| d. Bearing stress on washer | \frac{N_{1}}{A_{w}}=7808 psi | \frac{N_{2}}{A_{w}}=6280 psi |
| e. Shear stress through bracket on periphery of washer |
\frac{N_{1}}{A_{p}}=4164 psi | \frac{N_{2}}{A_{p}}=3349 psi |
