Solution of Machine Design-I
Problem #M1a : Given: Two plates are bolted with initial clamping force of
2250 lbs. The bolt stiffness is twice the clamping material stiffness. Find:
External separating load that would reduce the clamping force to 225 lbs.
Find the maximum bolt force at minimum clamping force.
Fc  Fi 
kc
Fe
kb  k c
225  2250 
1
Fe
1 2
Fe  6075 lbs
Fb ,max  Fi 
kb
2
Fe  2250  ( )(6075)  6300 lbs
kb  k c
3
Problem #M1b :Select a bolt that would withstand 6300 lbs nominal load in
direct tension. Apply a factor of safety of 2.5. Use a bolt with SAE strength
grade of 2 (which has a proof strength of 55 ksi).
Fb , all  6300(2.5)  15750 lbs
Fb , all  At S p
15750  At (55000)  At  0.286 in 2
A ¾” 10-UNC bolt has a tensile area of 0.336 square inches.
Problem #M1c: A 1”-12 UNF steel bolt of SAE grade 5 is under direct shear
loading. The coefficient of friction between mating surfaces is 0.4. The bolt
is tightened to its full proof strength. Tensile area is 0.663 in 2 Proof strength
is 85 kpsi, and yield strength is 92 kpsi
a) What shear force would the friction carry?
b) What shear load can the bolt withstand w/o yielding if the friction
between clamped members is completely lost? Base the calculation
on the thread root area.
Data for this bolt:
At=0.663 in2 Sp=85000 psi
The initial load in the bolt is:
Sy=92000 psi
Fi=AtSp=.663(85000)=56355 lbs
The friction force
Ff=Fi = 0.4(56355)=22542 lbs
The direct shear stress force w/o yielding is:
Fs = At(Sys)=At(0.58Sy)=0.663(0.58)(92000)=35370 lbs
If the bolt is subject to shear in the shank area, then use the larger shank area.
Problem #M2 : Solve the same problem as in the example above but for bolts ¾ “ UNF
Grade 2 bolts (Sy=57 ksi, and Su=74 ksi). Create a scaled drawing of the Goodman
graph and lable points 1,2,3, and 4 along with all the numerical values for stress labels in
the graph. Answer: Factor of Safety is about 2.8
The tensile area and proof strength are At = .373 in2 and Sp=55 ksi. The initial tension is:
Fi = .9*(0.373)(55) = 15.1 kips
Under this initial load, the root of the thread yields – Point-1 in the figure.
Sy
Sn
4
Sf
a
3
2
1
Sy
Su
The additional fluctuation in the bolt load is:
Fb 
Kb
1
Fe  (6150)  1230 lbs
Kb  Kc
5
When the external load is applied, the additional stress at the bolt root is:
 (
Fb
1230
)K f  (
)3  9890 psi
At
.373
The root stress, however, remains at yield point as this additional stress is applied. The
additional load is simply picked up by the cross-section that is stressed nominally (force
divided by area). But when the load is removed, the stress backs off from the yield point
and goes back into elastic region by exactly the same amount of stress that is removed.
In this case, the stress goes back to (57000-9890)=47110 psi. This point is shown as
Point 2 in the graph.
On subsequent application and removal of the load, the stress point moves between Point
2 and Point 3 on a 45-degree line. The reason is that half of the stress adds to the mean
stress and the other half becomes the amplitude of the alternating stress as shown below:
57 ksi
As the alternating load is increased, Line2-3 moves left until it reaches Point-4 which is
the eventual failure point. The margin or factor of safety is determined by the distance
between a and Sf
Endurance limit
Sn  0.5SutCLCGCS  0.5(74)(1)(0.90)(1)  33.3 ksi
The fatigue strength is
Sf 
Su  S y
Su  S n
Sn 
74  57
(33)  13.7 ksi
74  33
The factor of safety is:
FS 
Sf
a

13700
 2.8
4945
Problem #M3: Consider the bracket shown above. Assume the bracket is rigid and the
shear loads are carried by friction. The bracket is bolted by four bolts. The following is
known: F=5400 lbs, L=40 inches, D=12 inches, d=4 inches. Find appropriate UNC bolt
specifications for bolts of 120 Ksi proof strength using a factor of safety of 4.
Assuming a rigid bracket, the top bolts elongate or strain 3 times more than the lower
bolts. Taking moments about the pivot point we get:
F * L  FT ( D)  FB (d )
5400(40)  FT (12) 
FB
(4)
3
FT  16200 lbs
The load per bolt is half of this amount or 8100 lbs. Incorporating a factor of safety of 4,
the design load per bolt is 4(8100)=32400 lbs. The required tensile area for each bolt is:
F
32400

 0.27 in 2
S p 120000
The appropriate bolt is a Grade 8 ¾ “ –10 UNC
At 
Problem #M4
L
The bolts are ½”-13UNC. The distance between bolts is 1.25”. The load is
2700 lbs and L=8”. Find the shear stress on each bolt.
The torsional shear force (ballancing the force couple) is:
F1 
PL 27000(8)

 8640 lbs
2d
2(1.25)
The direct shear stress
F2 
F
 900 lbs
3
The total force is:
F  F12  F22  8686 lbs
The shear stress in each bolt using the shank diameter is:

F
8686

 44237
A  (.5) 2
4
psi