Answer:
No it is not a problem
Explanation:
It is not a problem because the stress intensity factor K would approach infinity as you get close to a crack tip and the intensity factor would approach Zero as you get too far away from the crack tip and this is simply because a crack is a notch with zero tip radius .
and The application of stress intensity factor k in respect to present fatigue crack tip is termed " linear elastic fracture mechanics "
Answer:
22.90 × 10⁸ kg
Explanation:
Given:
Diameter, d = 0.02 m
ωₙ = 0.95 rad/sec
Time period, T = 0.35 sec
Now, we know
T= 
where, L is the length of the steel cable
g is the acceleration due to gravity
0.35= 
or
L = 0.0304 m
Now,
The stiffness, K is given as:
K = 
Where, A is the area
E is the elastic modulus of the steel = 2 × 10¹¹ N/m²
or
K = 
or
K = 20.66 × 10⁸ N
Also,
Natural frequency, ωₙ = 
or
mass, m = 
or
mass, m = 
mass, m = 22.90 × 10⁸ kg
Answer:
In the clevis shown in Fig. 1-11b, find the minimum bolt diameter and the minimum thickness of each yoke that will support a load P = 14 kips without exceeding a shearing stress of 12 ksi and a bearing stress of 20 ksi.
127-clevis-double-shear-bolt.gif
Solution 127
Hide Click here to show or hide the solution
127-fbd-clevis-double-shear-bolt.gifFor shearing of rivets (double shear)
P=τA
14=12[2(14πd2)]
d=0.8618in → diameter of bolt answer
For bearing of yoke:
P=σbAb
14=20[2(0.8618t)]
t=0.4061in → thickness of yoke answer
