ME 437 Notes on Machine Design
Important Topics
Power screws
Bolt design
Bolt clamping loads
Bolt shear loads
Bolt groups
Bolts in fatigue loading
Gears
Kinematics
Forces
Stresses
Belts
Welds
Welds subjected to shear
Welds subjected to torsion
Welds subjected to bending
Clutches/Brakes
Disk type / Drum type
Flywheels
Spring design
Helical
Roller/ball bearing design
Journal bearings
Shaft design
Against fatigue
Whirling
Lateral vibration
Machine Design
Recommended Text:
Fundamentals of Machine Component Design by Juvinall and Marshek – 3rd
Edition
Bolt Selections and Design
You need
 Dimensions of standard threads (UNF/UNC)
 Strength specifications (grades) of bolts.
Clamping forces
Fe
The bolt force is
kb
Fb  Fi 
Fe
kb  kc
Where Kb and KC are the bolt and the clamping material stiffness and Fi is
the initial bolt tensioning.
The clamping force is
kc
Fc  Fi 
Fe
kb  k c
Fb
Fi
Fc
Fe
Recommended initial tension : Fi = 0.90 SpAt where Sp is the proof strength
and At is the tensile area of the bolt.
Recommended tightening torque: T = 0.20 Fid Where d is the nominal bolt
size.
Design of bolts in tension
Fb = At Sp
Where At is the tensile area.
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.
Answers: 6075 lbs and 6300 lbs
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). Answer: ¾ “ 10-UNC
Bolts under shear loading
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 in2 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.
Approximate Answers: a) 22500 lbs, b)35300 lbs
Bolt design under fatigue loading
 Analysis method:
 Root stress – Goodman Criteria
Sy
Sn
4
Sf
a
3
2
1
Sy
Su
Example: Consider a ½ “ UNF Grade 7 bolt with an initial tension of
90% of its proof load. The clamping material is 4 times stiffer than the bolt.
If a cyclic load of 0 to 6150 lbs is applied, what would be the factor of safety
guarding against eventual fatigue failure?
The tensile area and proof strength are At = 1599 in2 and Sp=105 ksi. The
initial tension is: Fi = .9*(0.1599)(105) = 15.1 kips
Under this initial load, the root of the thread yields – Point-1 in the figure.
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  23000 psi
At
.1599
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 (115-23)=92 ksi. 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:
92 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(133)(1)(0.90)(1)  60 ksi
The fatigue strength is
Sf 
Su  S y
Su  S n
Sn 
133  115
(60)  14.8 ksi
133  60
The factor of safety is:
FS 
Sf
a

14.8
 1.3
11.5
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
Design of Bolt Groups in Bending




Assume bolted frame is rigid.
Use geometry to determine bolt elongations.
Assume load distribution proportional to elongations.
Assume shear loads carried by friction.
L
F
D
d
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.
Answer: ¾”- 10UNC
Design of Bolt Groups in Torsion






Assume bolted frame is rigid.
Use geometry to determine bolt distortion.
Assume torque distribution proportional to distortions.
Assume bracket rotates around the bolt group C.G.
Exclude direct shear if its magnitude is small.
Use the bolt shank area for analysis of stresses.
Example #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.
Answer: 44250 psi