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.
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).
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 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.
Data for this bolt:
At=0.663 in2Sp=85000 psiSy=92000 psi
The initial load in the bolt is:
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.
The additional fluctuation in the bolt load is:
When the external load is applied, the additional stress at the bolt root is:
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:
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
The fatigue strength is
The factor of safety is:
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:
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:
The appropriate bolt is a Grade 8 ¾ “ –10 UNC
Problem #M4
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:
The direct shear stress
The
total force is:
The shear stress in each bolt using the shank diameter is: