Answer:
t=2.025 inches
Explanation:
Given that
P = 400 Psi
Yield stress ,σ = 80 ksi
Diameter ,d= 45 ft
We know that
1 ft = 12 inches
d= 540 inches
Factor of safety ,K= 3
The required thickness given as

t=thickness


t=2.025 inches
Therefore thickness will be 2.025 inches.
Answer:
P=361.91 KN
Explanation:
given data:
brackets and head of the screw are made of material with T_fail=120 Mpa
safety factor is F.S=2.5
maximum value of force P=??
<em>solution:</em>
to find the shear stress
T_allow=T_fail/F.S
=120 Mpa/2.5
=48 Mpa
we know that,
V=P
<u>Area for shear head:</u>
A(head)=π×d×t
=π×0.04×0.075
=0.003×πm^2
<u>Area for plate:</u>
A(plate)=π×d×t
=π×0.08×0.03
=0.0024×πm^2
now we have to find shear stress for both head and plate
<u>For head:</u>
T_allow=V/A(head)
48 Mpa=P/0.003×π ..(V=P)
P =48 Mpa×0.003×π
=452.16 KN
<u>For plate:</u>
T_allow=V/A(plate)
48 Mpa=P/0.0024×π ..(V=P)
P =48 Mpa×0.0024×π
=361.91 KN
the boundary load is obtained as the minimum value of force P for all three cases. so the solution is
P=361.91 KN
note:
find the attached pic
Answer:
0.00650 Ib s /ft^2
Explanation:
diameter ( D ) = 0.71 inches = 0.0591 ft
velocity = 0.90 ft/s ( V )
fluid specific gravity = 0.96 (62.4 ) ( x )
change in pressure ( P ) = 0 because pressure was constant
viscosity = (change in p - X sin∅ )
/ 32 V
= ( 0 - 0.96( 62.4) sin -90 ) * 0.0591 ^2 / 32 * 0.90
= - 59.904 sin (-90) * 0.0035 / 28.8
= 0.1874 / 28.8
viscosity = 0.00650 Ib s /ft^2
Answer:

Explanation:
the half life of the given circuit is given by

where [/tex]\tau = RC[/tex]

Given 
resistance in the circuit is 40 ohm and to extend the half cycle we added new resister of 48 ohm. the net resitance is 40+48 = 88 ohms
now the new half life is

Divide equation 2 by 1


putting all value we get new half life

