Answer:
The reaction at support B
Rb= 235440N
The reaction at support C
RC= 29430N
Explanation : check attachment
Team members and the work from ther
Answer:
It will not experience fracture when it is exposed to a stress of 1030 MPa.
Explanation:
Given
Klc = 54.8 MPa √m
a = 0.5 mm = 0.5*10⁻³m
Y = 1.0
This problem asks us to determine whether or not the 4340 steel alloy specimen will fracture when exposed to a stress of 1030 MPa, given the values of <em>KIc</em>, <em>Y</em>, and the largest value of <em>a</em> in the material. This requires that we solve for <em>σc</em> from the following equation:
<em>σc = KIc / (Y*√(π*a))</em>
Thus
σc = 54.8 MPa √m / (1.0*√(π*0.5*10⁻³m))
⇒ σc = 1382.67 MPa > 1030 MPa
Therefore, the fracture will not occur because this specimen can handle a stress of 1382.67 MPa before experience fracture.
Answer:
0 m/s , 3 m/s , 2 m/s^2
Explanation:
Given : s(t) = ( t^2 - 6t + 5)
v(t) = ds / dt = 2t - 6
s(0) = 5 m
s(6) = (6)^2 - 6*6 + 5 = 5 m
Vavg = ( s(6) - s(0) ) / 2 = 0 m\s
Find the turning point of particle:
ds/dt = 0 = 2t - 6
t = 3 sec
s(3) = 3^2 -6*3 + 5 = - 4
Total distance = 5 - (-4) + (5 - (-4)) = 18 m
Total time = 6s
Average speed = Total distance / Total time = 18 / 6 = 3 m/s
Taking derivative of v(t) to obtain a(t)
a (t) = dv(t) / dt = 2 m/s^2