If the bolt head and the supporting bracket are made of the same material having a failure shear stress of 'Tra;i = 120 MPa, det
ermine the maximum allowable force P that can be applied to the bolt so that it does not pull through the plate. Apply a factor of safety of F.S. = 2.5 against shear failure.
1 answer:
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
You might be interested in
Answer:
Explanation:
In an equilateral trinagle the center of mass is at 1/3 of the height and horizontally centered.
We can consider that the weigth applies a torque of T = W*b/2 on the right corner, being W the weight and b the base of the triangle.
The weigth depends on the size and specific gravity.
W = 1/2 * b * h * L * SG
Then
Teq = 1/2 * b * h * L * SG * b / 2
Teq = 1/4 * b^2 * h * L * SG
The water would apply a torque of elements of pressure integrated over the area and multiplied by the height at which they are apllied:
The term sin(30) is because of the slope of the wall
The pressure of water is:
p(y) = SGw * (h - y)
Then:
T1 = SGw * sin(30) * L * (h^2*y - h*y^2 + 1/3*y^3)(evaluated between 0 and h)
T1 = SGw * sin(30) * L * (h^2*h - h*h^2 + 1/3*h^3)
T1 = SGw * sin(30) * L * (h^3 - h^3 + 1/3*h^3)
T1 = 1/3 * SGw * sin(30) * L * h^3
To remain stable the equilibrant torque (Teq) must be of larger magnitude than the water pressure torque (T1)
1/4 * b^2 * h * L * SG > 1/3 * SGw * sin(30) * L * h^3
In an equilateral triangle h = b * cos(30)
1/4 * b^3 * cos(30) * L * SG > 1/3 * SGw * sin(30) * L * b^3 * (cos(30))^3
SG > SGw * 4/3* sin(30) * (cos(30))^2
SG > 1/2 * SGw
For the dam to hold, it should have a specific gravity of at leas half the specific gravity of water.
This is avergae specific gravity, including holes.
Answer:
X_cp = c/2
Explanation:
We are given;
Chord = c
Angle of attack = α
p u (s) = c 1
p1(s)=c2,
and c2 > c1
First of all, we need to find the resultant normal force on the plate and the total moment about leading edge.
I've attached the solution
Answer:
Hello your question is incomplete attached below is the missing part
a) p1 = 454.83 kPa, p2 = 283.359 Kpa , p3 = 536.423 kpa , p4 = 860.959kPa
b) W12 = 3.4 kJ, W23 = -3.5875 KJ, W34 = -4.0735 KJ, W41 = 3.5875 KJ
c) 5
Explanation:
Given data:
mass of air ( m ) = 1/10 kg
adiabatic index ( k ) = 1.4
temperature for isothermal expansion = 250K
rate of heat transfer ( Q12 ) = 3.4 KJ
temperature for Isothermal compression ( T4 ) = 300k
final volume ( V4 ) = 0.01m ^3
a) Calculate the pressure, in Kpa, at each of the four principal states
from an ideal gas equation
P4V4 = mRT4 ( input values above )
hence P4 = 860.959kPa
attached below is the detailed solution
b) Calculate work done for each processes
attached below is the detailed solution
C) Calculate the coefficient of performance
attached below is detailed solution
