1. A screw driver with a 1.5 inch diameter handle is used to install a 1/4-20 UNC screw. Determine the circumference where the e
1 answer:
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Answer:
a) 0.978
b) 0.9191
c) 1.056
d) 0.849
Explanation:
Given data :
Stiffness of each bolt = 1.0 MN/mm
Stiffness of the members = 2.6 MN/mm per bolt
Bolts are preloaded to 75% of proof strength
The bolts are M6 × 1 class 5.8 with rolled threads
Pmax =60 kN, Pmin = 20kN
<u>a) Determine the yielding factor of safety</u>
------ ( 1 )
Sp = 380 MPa, At = 20.1 mm^2, C = 0.277, Pmax = 7500 N, Fi = 5728.5 N
Input the given values into the equation above
equation 1 becomes ( np ) = = 0.978
note : values above are derived values whose solution are not basically part of the required solution hence they are not included
<u>b) Determine the overload factor of safety</u>
------- ( 2 )
Sp = 380 MPa, At = 20.1 mm^2, C = 0.277, Pmax = 7500 N, Fi = 5728.5 N
input values into equation 2 above
hence : = 0.9191
<u>C) Determine the factor of safety based on joint separation</u>
Fi = 5728.5 N, Pmax = 7500 N, C = 0.277,
input values into equation above
Hence = 1.056
<u>D) Determine the fatigue factor of safety using the Goodman criterion.</u>
nf = 0.849
attached below is the detailed solution .
The question is incomplete. We have to calculate :
a). the cutting force
b). volumetric metal removal rate, MRR
c). the horsepower required at the cut
d). if the power efficiency of the machine tool is 90%, determine the motor horsepower
Solution :
Given :
Cutting velocity (v) = 500 ft/min
= 500 x 12 in/min
= 6000 in/min
Feed , f = 0.025 in/rev
Depth of cut, d = 0.2 in
b). Volumetric material removal rate, MRR = v.f.d
= 6000 x 0.025 x 0.2
= 30
c). Horsepower required = MRR x unit horsepower
= 30 x 1.6
= 48 hp
a). Cutting force,
(1 hp = 550 ft lbf /sec)
= 3168 lbf
d). Machine HP required
= 53.33 HP