Answer:
a) 53 MPa, 14.87 degree
b) 60.5 MPa
Average shear = -7.5 MPa
Explanation:
Given
A = 45
B = -60
C = 30
a) stress P1 = (A+B)/2 + Sqrt ({(A-B)/2}^2 + C)
Substituting the given values, we get -
P1 = (45-60)/2 + Sqrt ({(45-(-60))/2}^2 + 30)
P1 = 53 MPa
Likewise P2 = (A+B)/2 - Sqrt ({(A-B)/2}^2 + C)
Substituting the given values, we get -
P1 = (45-60)/2 - Sqrt ({(45-(-60))/2}^2 + 30)
P1 = -68 MPa
Tan 2a = C/{(A-B)/2}
Tan 2a = 30/(45+60)/2
a = 14.87 degree
Principal stress
p1 = (45+60)/2 + (45-60)/2 cos 2a + 30 sin2a = 53 MPa
b) Shear stress in plane
Sqrt ({(45-(-60))/2}^2 + 30) = 60.5 MPa
Average = (45-(-60))/2 = -7.5 MPa
Answer:
the elevation at point X is 2152.72 ft
Explanation:
given data
elev = 2156.77 ft
BS = 2.67 ft
FS = 6.72 ft
solution
first we get here height of instrument that is
H.I = elev + BS ..............1
put here value
H.I = 2156.77 ft + 2.67 ft
H.I = 2159.44 ft
and
Elevation at point (x) will be
point (x) = H.I - FS .............2
point (x) = 2159.44 ft - 6.72 ft
point (x) = 2152.72 ft
Answer: I would help you but I don’t know the answer, so sorry
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