The choices are:
a. bringing it to a mechanic for a tune-up
b. checking to make sure it is in working order
c. making sure you are entering your vehicle
d. giving the tires a good kick to check traction
Answer:
The answer is: letter b, checking to make sure it is in working order.
Explanation:
A "vehicle check" is essential for everyone involved. It makes the driver and the passengers safe, and also extends that safety to the people in the environment. Before driving, you need to check your car for the following: <em>fluids, wipers, spare wheels, leaks and the like</em>. This will ensure that your vehicle is in proper working order and <u>will lessen the possibility of car accident in the future.</u>
Thus, this explains the answer, letter b.
Answer:
a)σ₁ = 265.2 MPa
b)σ₂ = -172.8 MPa
c)
d)Range = 438 MPa
Explanation:
Given that
Mean stress ,σm= 46.2 MPa
Stress amplitude ,σa= 219 MPa
Lets take
Maximum stress level = σ₁
Minimum stress level =σ₂
The mean stress given as
2 x 46.2 = σ₁ + σ₂
σ₁ + σ₂ = 92.4 MPa --------1
The amplitude stress given as
2 x 219 = σ₁ - σ₂
σ₁ - σ₂ = 438 MPa --------2
By adding the above equation
2 σ₁ = 530.4
σ₁ = 265.2 MPa
-σ₂ = 438 -265.2 MPa
σ₂ = -172.8 MPa
Stress ratio
Range = 265.2 MPa - ( -172.8 MPa)
Range = 438 MPa
Answer:
3.03 INCHES
Explanation:
According to ASTM D198 ;
Modulus of rupture = ( M / I ) * y ----- ( 1 )
M ( bending moment ) = R * length of span / 2
= (120 * 10^3 ) * 48 / 2 = 288 * 10^4 Ib-in
I ( moment of inertia ) = bd^3 / 12
= ( 2 )*( d )^3 / 12 = 2d^3 / 12
b = 2 in , d = ?
length of span = 4 * 12 = 48 inches
R = P / 2 = 240 * 10^3 / 2 = 120 * 10^3 Ib
y ( centroid distance ) = d / 2 inches
back to equation ( 1 )
( M / I ) * y
940.3 ksi = ( 288 * 10^4 / 2d^3 / 12 ) * d / 2
= ( 288 * 10^4 * 12 ) / 2d^3 ) * d / 2
940300 = 34560000* d / 4d^3
4d^3 ( 940300 ) = 34560000 d ( divide both sides with d )
4d^2 = 34560000 / 940300
d^2 = 9.188 ∴ Value of d ≈ 3.03 in
