In order to develop this problem it is necessary to take into account the concepts related to fatigue and compression effort and Goodman equation, i.e, an equation that can be used to quantify the interaction of mean and alternating stresses on the fatigue life of a materia.
With the given data we can proceed to calculate the compression stress:
Through Goodman's equations the combined effort by fatigue and compression is expressed as:
Where,
Fatigue limit for comined alternating and mean stress
Fatigue Limit
Mean stress (due to static load)
Ultimate tensile stress
Security Factor
We can replace the values and assume a security factor of 1, then
Re-arrenge for 
We know that the stress is representing as,
Then,
Where 
=Max Moment
I= Intertia
The inertia for this object is
Then replacing and re-arrenge for
Thereforethe moment that can be applied to this shaft so that fatigue does not occur is 3.2kNm
Answer:
modulus of elasticity for the nonporous material is 340.74 GPa
Explanation:
given data
porosity = 303 GPa
modulus of elasticity = 6.0
solution
we get here modulus of elasticity for the nonporous material Eo that is
E = Eo (1 - 1.9P + 0.9P²) ...............1
put here value and we get Eo
303 = Eo ( 1 - 1.9(0.06) + 0.9(0.06)² )
solve it we get
Eo = 340.74 GPa
Answer:
your answer is correct
Explanation:
You have the correct mapping from inputs to outputs. The only thing your teacher may disagree with is the ordering of your inputs. They might be written more conventionally as ...
A B Y
0 0 1
0 1 0
1 0 0
1 1 1
That is, your teacher may be looking for the pattern 1001 in the last column without paying attention to what you have written in column B.
Answer:
water rise = 22 mm
Explanation:
weight of packet IN AIR = 40 *9.81 =392.4 N
weight of packet IN WATER= 18 *9.81 =176.58 N
by Archimedi's principle
difference in weight = weight of displaced water
w_a - w_w = \rho_w v_d g
392.4 - 176.58 = 1000* v_d* 9.81
v_d = 0.022 m^3
v_d = A*H_rise
0.022 =1*H_rise
H_rise = 0.022 m = 22 mm
water rise = 22 mm
Answer:
true
Explanation:
it depends on how fast the car goes because the gas burns faster or slower depending on the speed you go or drive. if your drive 50 MPH, the gas will burn slowly. if your drive 45 MPH, the gas will burn faster if you go slow too much. if you press the glass button and hold it for 5 minutes, the more you hold the gas for too long, the more the gas will burn inside the gas engine. go look at your car or your parents car and see how it goes and that will help.
