Answer:
screw is the answer of the question
Answer:
maximum isolator stiffness k =1764 kN-m
Explanation:
mean speed of rotation 


=65.44 rad/sec


= 0.1*(65.44)^2
F_T =428.36 N
Transmission ratio 
also
transmission ratio ![= \frac{1}{[\frac{w}{w_n}]^{2} -1}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B1%7D%7B%5B%5Cfrac%7Bw%7D%7Bw_n%7D%5D%5E%7B2%7D%20-1%7D)
![0.7 =\frac{1}{[\frac{65.44}{w_n}]^2 -1}](https://tex.z-dn.net/?f=0.7%20%3D%5Cfrac%7B1%7D%7B%5B%5Cfrac%7B65.44%7D%7Bw_n%7D%5D%5E2%20-1%7D)
SOLVING FOR Wn
Wn = 42 rad/sec

k = m*W^2_n
k = 1000*42^2 = 1764 kN-m
k =1764 kN-m
Answer:
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This question is incomplete, the complete question is;
Determine the design moment strength (ϕMn) for a W21x73 steel beam with a simple span of 18 ft when lateral bracing for the compression flange is provided at the ends only (i.e., Lb = 18 ft). Report the result in kip-ft.
Use Fy=50 ksi and assume Cb=1.0 (if needed).
Answer: the design moment strength for the W21x73 steel beam is 566.25 f-ft
Explanation:
Given that;
section W 21 x 73 steel beam;
now from the steel table table for this section;
Zx = Sx = 151 in³
also given that; fy = 50 ksi and Cb = 1.0
QMn = 0.9 × Fy × Zx
so we substitute
QMn = 0.9 × 50 × 151
QMn = 6795 k-inch
we know that;
12inch equals 1 foot
so
QMn = 6795 k-inch / 12
QMn = 566.25 f-ft
Therefore the design moment strength for the W21x73 steel beam is 566.25 f-ft