Answer:
correct option is (A) 0.5
Explanation:
given data
axial column load = 250 kN per meter
footing placed = 0.5 m
cohesion = 25 kPa
internal friction angle = 5°
solution
we know angle of internal friction is 5° that is near to 0°
so it means the soil is almost cohesive soil.
and for a pure cohesive soil
= 0
and we know formula for
is
= (Nq - 1 ) × tan(Ф) ..................1
so here Ф is very less
should be nearest to zero
and its value can be 0.5
so correct option is (A) 0.5
Answer:
The correct answer is A : Orientation dependence of normal and shear stresses at a point in mechanical members
Explanation:
Since we know that in a general element of any loaded object the normal and shearing stresses vary in the whole body which can be mathematically represented as

And 
Mohr's circle is the graphical representation of the variation represented by the above 2 formulae in the general oriented element of a body that is under stresses.
The Mohr circle is graphically displayed in the attached figure.
Answer:
E=52000Hp.h
E=38724920Wh
E=1.028x10^11 ftlb
Explanation:
To solve this problem you must multiply the engine power by the time factor expressed in h / year, to find this value you must perform the conventional unit conversion procedure.
Finally, when you have the result Hp h / year you convert it to Ftlb and Wh

E=52000Hp.h

E=38724920Wh

E=1.028x10^11 ftlb
Answer:
C₁₀ = 6.3 KN
Explanation:
The catalog rating of a bearing can be found by using the following formula:
C₁₀ = F [Ln/L₀n₀]^1/3
where,
C₁₀ = Catalog Rating = ?
F = Design Load = 2.75 KN
L = Design Life = 1800 rev/min
n = No. of Hours Desired = 10000 h
L₀ = Rating Life = 500 rev/min
n₀ = No. of Hours Rated = 3000 h
Therefore,
C₁₀ = [2.75 KN][(1800 rev/min)(10000 h)/(500 rev/min)(3000 h)]^1/3
C₁₀ = (2.75 KN)(2.289)
<u>C₁₀ = 6.3 KN</u>