Answer:
The 5/16 – 24 UNF is stronger because it has more tensile load capacity.
Tensile load capacity for M8 -1.25 = 5670 lb
Tensile load capacity for M8 -1 = 6067 lb
Explanation:
For 5/16 - 18 UNC thread:
D = 0.3125
n = 18
Therefore the tensile load capacity is = 100000 X (0.7854 X (0.3125 - 0.9743/ 18) ^2
= 5243 lb.
Similarly for 5/16 - 24 UNF , only the n value changes to 24
we get the tensile load capacity = 5806.6 lb
Hence the 5/16 – 24 UNF is stronger because it has more tensile load capacity.
For metric Bolts:
We have to consider all values in SI units
Strength = 689 MPa
We get for M8 -1.25:
Tensile load capacity as = 689 X 36.6 = 25223 N = 5670 lb
For M8 -1:
Tensile load capacity as = 689 X 39.167 = 26986 N = 6067lb
Answer: 10.29 sec.
Explanation:
Neglecting drag and friction, and at road level , the energy developed during the time the car is accelerating, is equal to the change in kinetic energy.
If the car starts from rest, this means the following:
ΔK = 1/2 m*vf ²
As Power (by definition) is equal to Energy/Time= 75000 W= 75000 N.m/seg, in order to get time in seconds, we need to convert 100 km/h to m/sec first:
100 (Km/h)*( 1000m /1 Km)*(3600 sec/1 h)= 27,78 m/sec
Now, we calculate the change in energy:
ΔK= 1/2*2000 Kg. (27,78)² m²/sec²= 771,728 J
<h2>If P= ΔK/Δt, </h2><h2>Δt= ΔK/P= 771,728 J / 75,000 J/sec= 10.29 sec.</h2>
Answer:
the saturated density should be
Explanation:
The question is incomplete. The complete question is :
The solid rod shown is fixed to a wall, and a torque T = 85N?m is applied to the end of the rod. The diameter of the rod is 46mm .
When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the strains vary linearly along radial lines. Within the proportional limit, the stress also varies linearly along radial lines. If point A is located 12 mm from the center of the rod, what is the magnitude of the shear stress at that point?
Solution :
Given data :
Diameter of the rod : 46 mm
Torque, T = 85 Nm
The polar moment of inertia of the shaft is given by :
J = 207.6 
So the shear stress at point A is :
Therefore, the magnitude of the shear stress at point A is 4913.29 MPa.
