Answer:
=> The total numbers of cylinders on the engine.
=> Total number of lobes in the cam ring.
=> The direction at which the cam ring rotates.
Explanation:
A cam is a kind of ring and a device that is being used in engines. One or the main purpose of using cams in engines us because it helps in the change or transformation of the rotational movement of the engine to a translational one. Instead of using a cam ring, a cam roller can be used in place.
There are three things that can be used to to determine the speed of cam ring speed and they are given below as;
=> The total numbers of cylinders on the engine.
=> The Total number of lobes in the cam ring.
=> The direction at which the cam ring rotates.
The three suspender bars AB, CD, and EF are made of A-36 steel and have equal cross-sectional areas of 500 mm2. Determine the average normal stress in each bar if the rigid beam is subjected to a force of P
Answer:
hello a diagram attached to your question is missing attached below is the missing diagram
The three suspender bars AB, CD, and EF are made of A-36 steel and have equal cross-sectional areas of 500 mm2. Determine the average normal stress in each bar if the rigid beam is subjected to a force of P = 70kN , let d = 2.4 m , L = 4m
Answer :
Stress = force / area
for Bar 1 = (40.03 * 1000) / 500 = 80.06 MPa
Bar 2 = ( 23.33 * 1000 ) / 500 = 46.66 MPa
for Bar 3 = ( 5.33 * 1000 ) / 500 = 10.66 MPa
Explanation:
Given data:
Type of steel = A-36
cross-sectional area = 500 mm^2
Calculate the average normal stress in each bar
we have to make some assumptions
assume forces in AB, CD, EF to be p1,p2,p3 respectively
∑ Fy = 0 ; p1 + p2 + p3 = 70kN ---------- ( 1 )
∑ Mc = 0 ; P1 * d - p * d/2 - p3 * d = 0
where d = 2.4 hence ; p1 - p3 = 35 -------- ( 2 )
Take ; Tan∅
Tan∅ = MN / 2d = OP/d
i.e. s1 - 2s2 - s3 = 0
[tex]\frac{P1L}{AE} - \frac{2P2}{AE} + \frac{P3L}{AE} = 0[/tex]
L , E and A are the same hence
P1 - 2p2 + p3 = 0 ----- ( 3 )
Next resolve the following equations
p1 = 40.03 kN, p2 = 23.33 kN, p3 = 5.33 kN
Stress = force / area
for Bar 1 = (40.03 * 1000) / 500 = 80.06 MPa
Bar 2 = ( 23.33 * 1000 ) / 500 = 46.66 MPa
for Bar 3 = ( 5.33 * 1000 ) / 500 = 10.66 MPa