Answer:
a)Wt =25.68 lbf
b)Wt = 150 lbf
F= 899.59 N
Explanation:
Given that
m= 150 lbm
a)
Weight on the spring scale(Wt) = m g
We know that
Wt = 150 x 5.48/32 lbf
Wt =25.68 lbf
b)
On the beam scale
This is scale which does not affects by gravitational acceleration.So the wight on the beam scale will be 150 lbf.
Wt = 150 lbf
If the plane is moving upward with acceleration 6 g's then the for F
F = m a
We know that
a=6 g's
So
F = 90 x 9.99 N
F= 899.59 N
Answer:
yield strength before cold work = 370 MPa
yield strength after cold work = 437.87 MPa
ultimate strength before cold work = 440 MPa
ultimate strength after cold work = 550 MPa
Explanation:
given data
AISI 1018 steel
cold work factor W = 20% = 0.20
to find out
yield strength and ultimate strength before and after the cold-work operation
solution
we know the properties of AISI 1018 steel is
yield strength σy = 370 MPa
ultimate tensile strength σu = 440 MPa
strength coefficient K = 600 MPa
strain hardness n = 0.21
so true strain is here ∈ = = 0.223
so
yield strength after cold is
yield strength =
yield strength =
yield strength after cold work = 437.87 MPa
and
ultimate strength after cold work is
ultimate strength =
ultimate strength =
ultimate strength after cold work = 550 MPa
Answer:
the maximum length of the specimen before deformation is 0.4366 m
Explanation:
Given the data in the question;
Elastic modulus E = 124 GPa = 124 × 10⁹ Nm⁻²
cross-sectional diameter D = 4.2 mm = 4.2 × 10⁻³ m
tensile load F = 1810 N
maximum allowable elongation Δl = 0.46 mm = 0.46 × 10⁻³ m
Now to calculate the maximum length for the deformation, we use the following relation;
= [ Δl × E × π × D² ] / 4F
so we substitute our values into the formula
= [ (0.46 × 10⁻³) × (124 × 10⁹) × π × (4.2 × 10⁻³)² ] / ( 4 × 1810 )
= 3161.025289 / 7240
= 0.4366 m
Therefore, the maximum length of the specimen before deformation is 0.4366 m
