Question 1.12: A steel bar serving as a vertical hanger to support heavy ma...
A steel bar serving as a vertical hanger to support heavy machinery in a factory is attached to a support by the bolted connection shown in Fig.1-61. Two clip angles (thickness t_{c} = 9.5 mm) are fastened to an upper support by bolts 1 and 2 each with a diameter of 12 mm; each bolt has a washer with a diameter of d_{w} = 28 mm. The main part of the hanger is attached to the clip angles by a single bolt (bolt 3 in Fig.1-61a) with a diameter of d = 25 mm. The hanger has a rectangular cross section with a width of b_{1} = 38 mm and thickness of t =13 mm, but at the bolted connection, the hanger is enlarged to a width of b_{2} = 75 mm. Determine the allowable value of the tensile load P in the hanger based upon the following considerations.
(a) The allowable tensile stress in the main part of the hanger is 110 MPa.
(b) The allowable tensile stress in the hanger at its cross section through the bolt 3 hole is 75 MPa. (The permissible stress at this section is lower because of the stress concentrations around the hole.)
(c) The allowable bearing stress between the hanger and the shank of bolt 3 is 180 MPa.
(d) The allowable shear stress in bolt 3 is 45 MPa.
(e) The allowable normal stress in bolts 1 and 2 is 160 MPa.
(f) The allowable bearing stress between the washer and the clip angle at either bolt 1 or 2 is 65 MPa.
(g) The allowable shear stress through the clip angle at bolts 1 and 2 is 35 MPa.
Use a four-step problem-solving approach to find the allowable value of the tensile load P in the hanger based upon a variety of different allowable stresses in the different connection components.
1. Conceptualize: Start by sketching a series of free-body diagrams to find the forces acting on each connection component. Express the force on each component in terms of an allowable stress times the associated area upon which it acts. This force is the allowable value of applied load P for that stress condition. Each of the seven stress states [(a)–(g) in the problem statement] and the associated applied load are illustrated in this example’s figures; each is adjacent to the corresponding calculations in Step (3).
2. Categorize: Compute seven different values of the allowable load P, each based on an allowable stress and a corresponding area. The minimum value of load P will control.
Numerical data for the hanger connection design shown in Fig.1-61 are as follows.
(a) Connection component dimensions:
t_{c}=9.5 mm \quad t=13 mm \quad b_{1}=38 mm \quad b_{2}=75 mm
d_{1}=12 mm \quad d=25 mm \quad d_{w}=28 mm(b) Allowable stresses:
\sigma_{a}=110 MPa \quad \sigma_{a 3}=75 MPa \quad \sigma_{b a 3}=180 MPa \quad \tau_{a 3}=45 MPa
\tau_{a 1}=35 MPa \quad \sigma_{a 1}=160 MPa \quad \sigma_{b a 1}=65 MPa3. Analyze:
Part (a): Find the allowable load based upon the stress in the main part of the hanger (Fig.1-61c). This is equal to the allowable stress in tension (110 MPa) times the cross-sectional area of the hanger (Eq.1-30):
P_{\text {allow }}=\sigma_{\text {allow }} A (1-30)
P_{a}=\sigma_{a} b_{1} t=(110 MPa )(38 mm \times 13 mm )=54.3 kNA load greater than this value will overstress the main part of the hanger (that is, the actual stress will exceed the allowable stress), thereby reducing the factor of safety.
Part (b): Find the allowable load based upon the allowable tensile stress (75 MPa) in the hanger at its cross section through the bolt 3 hole.
At the cross section of the hanger through the bolt hole (Fig. 1-61d), make a similar calculation but with a different allowable stress and a different area. The net cross-sectional area (that is, the area that remains after the hole is drilled through the bar) is equal to the net width times the thickness. The net width is equal to the gross width b_{2} minus the diameter d of the hole. Thus, the equation for the allowable load P_{b} at this section is
P_{b}=\sigma_{a 3}\left(b_{2}-d\right) t=(75 MPa )(75 mm -25 mm )(13 mm )=48.8 kNPart (c): Now find the allowable load based upon the allowable bearing stress (180 MPa) between the hanger and the shank of bolt 3.
The allowable load based upon bearing between the hanger and the bolt ( Fig.1-61e) is equal to the allowable bearing stress times the bearing area. The bearing area is the projection of the actual contact area, which is equal to the bolt diameter times the thickness of the hanger. Therefore, the allowable load (Eq.1-32) is
P_{\text {allow }}=\sigma_{b} A_{b} (1-32)
P_{c}=\sigma_{b a 3} d t=58.5 kN =(180 MPa )(25 mm )(13 mm )=58.5 kNPart (d): Determine the allowable load based upon the allowable shear stress (45 MPa) in bolt 3.
The allowable load P_{d} based upon shear in the bolt (Fig.1-61f) is equal to the allowable shear stress times the shear area (Eq.1-31). The shear area is twice the area of the bolt because the bolt is in double shear; thus,
P_{\text {allow }}=\tau_{\text {allow }} A (1-31)
P_{d}=2 \tau_{a 3}\left(\frac{\pi}{4} d^{2}\right)=2(45 MPa )\left[\frac{\pi}{4}(25 mm )^{2}\right]=44.2 kNPart (e): Find the allowable load based upon the allowable normal stress (160 MPa) in bolts 1 and 2.
The allowable normal stress in bolts 1 and 2 is 160 MPa. Each bolt carries one half of the applied load P (see Fig.1-61g). The allowable total load P_{e} is the product of the allowable normal stress in the bolt and the sum of the crosssectional areas of bolts 1 and 2:
P_{e}=\sigma_{a 1}(2)\left(\frac{\pi}{4} d_{1}^{2}\right)=(160 MPa )(2)\left[\frac{\pi}{4}(12 mm )^{2}\right]=36.2 kNPart (f): Now find the allowable load based upon the allowable bearing stress (65 MPa) between the washer and the clip angle at either bolt 1 or 2.
The allowable bearing stress between the washer and the clip angle at either bolt 1 or 2 is 65 MPa. Each bolt (1 or 2) carries one half of the applied load P (see Fig.1-61h). The bearing area here is the ring-shaped circular area of the washer (the washer is assumed to fit snugly against the bolt). The allowable total load P_{f} is the allowable bearing stress on the washer times twice the area of the washer:
\begin{aligned} P_{f}=\sigma_{b a 1}(2)\left[\frac{\pi}{4}\left(d_{w}^{2}-d_{1}^{2}\right)\right] &=(65 MPa )(2)\left\{\frac{\pi}{4}\left[(28 mm )^{2}-(12 mm )^{2}\right]\right\} \\ &=65.3 kN\end{aligned}Part (g): Finally, determine the allowable load based upon the allowable shear stress (35 MPa) through the clip angle at bolts 1 and 2.
The allowable shear stress through the clip angle at bolts 1 and 2 is 35 MPa. Each bolt (1 or 2) carries one half of the applied load P (see Fig.1-61i). The shear area at each bolt is equal to the circumference of the hole (\pi \times d_{w} ) times the thickness of the clip angle (t_{c} ).
The allowable total load P_{g} is the the allowable shear stress times twice the shear area:
4. Finalize: All seven conditions were used to find the allowable tensile loads in the hanger. Comparing the seven preceding results shows that the smallest value of the load is P_{allow} = 36.2 kN. This load is based upon normal stress in bolts 1 and 2 [see part (e)] and is the allowable tensile load for the hanger.
A more refined analysis that includes the weight of the entire hanger assembly can be carried out as shown in Example 1-6. As in Example 1-11, note that these computed stresses are average values only and do not include localized effects such as stress concentrations around bolt holes.
