Answer:
the work done by the linear spring will be equal to 23.04 k J
Explanation:
given,
k = 32 k N /m
deflected spring = 120 cm
= 1.2 m
work done by spring can be calculate as
=
=
= 23.04 k J
hence, the work done by the linear spring will be equal to 23.04 k J
The asymptotes of the open loop transfer are:
The open loop transfer function is given as:
f(s) = 100(s + 1)/((s + 10)(s + 100))
Set the numerator of the function to 0.
So, we have:
f(s) = 0/((s + 10)(s + 100))
Evaluate
f(s) = 0
This means that, the vertical asymptote is y = 0
Set the denominator of the function to 0.
(s + 10)(s + 100) 0
Split
s + 10 = 0 and s + 100 = 0
Solve for s
s = -10 and s = -100
This means that, the horizontal asymptotes are s = -10 and s = -100
See attachment for the graph of the asymptotes
Read more about asymptotes at:
#SPJ1
Answer:
diameter is 14 mm
Explanation:
given data
power = 15 kW
rotation N = 1750 rpm
factor of safety = 3
to find out
minimum diameter
solution
we will apply here power formula to find T that is
power = 2π×N×T / 60 .................1
put here value
15 × = 2π×1750×T / 60
so
T = 81.84 Nm
and
torsion = T / Z ..........2
here Z is section modulus i.e = πd³/ 16
so from equation 2
torsion = 81.84 / πd³/ 16
so torsion = 416.75 / / d³ .................3
so from shear stress theory
torsion = σy / factor of safety
so here σy = 530 for 1020 steel
so
torsion = σy / factor of safety
416.75 / d³ = 530 × / 3
so d = 0.0133 m
so diameter is 14 mm