The shear stress at the centre of a circular shaft under torsion is
The simple torsion equation is written as
This states that the shearing stress varies directly as the distance ‘r' from the axis of the shaft and the following is the stress distribution
Hence the maximum shear stress occurs on the outer surface of the shaft where r = R and at the centre the shear stress is zero.
Match List-I with List-II and select the correct answer using the code given below the lists.
List-I (Loaded Beam)
List-II (Maximum Bending moment)
A-3, B-1, C-4
Beam A is cantilever so maximum bending moment occurs at the fixed support
Beam B is simply supported so maximum bending moment occurs at midspan
Beam C is cantilever so maximum bending moment occurs at fixed support
A solid circular shaft of diameter d is subjected to a combined effect of bending moment M and torque T. The relation is used for designing the shaft using which one of this material property
Ssy = Torsional yield strength
A solid steel shaft is surrounded by a copper shaft, such that, Isteel = 1/2 Icopper. If Gcopper = 1/2 Gsteel, what is the ratio of Ts/Tc if the composite shaft is subjected to a twisting moment?
Both shaft will rotate through same degree.
Couple M is applied at C on a simply supported beam AB. The maximum shear force on AC will be
R1 + R2 = 0
∑MA = 0
⇒ R2 × 3 = -M
So maximum shear force = M/3
A hollow steel shaft of external diameter 100 mm and internal diameter 50 mm is to be replaced by a solid allow shaft. Assuming the same value of polar modulus for both, the diameter of the solid allows shaft will be-
Let the diameter of solid shaft is ‘d’
For a loaded cantilever beam of uniform cross-section, the bending moment (in N-mm) along the length is M(x) = 5x2 + 10x, where x is the distance (in mm) measured from the free end of the beam. The magnitude of shear force (in N) in the cross-section at x = 10 mm is ________.
M = 5x2 + 10x
Shear Force = dM/dx = 10x + 10
Shear force at X = 10 mm
= 10 x 10 + 10 = 110 N
S.F.D for the beam is:
The equation for the deflected shape of a beam carrying a U.D.L and simply supported at ends is given below The load carrying capacity of the beam is
Using the equation of deflection:
W = 40 kN/m
For the given figure, calculate the bending moment (in kN-m) at 2 m from the left end?
∑FV = 0
RA + RB = 3 + 2 × 2
RA + RB = 7 kN
∑FH = 0
HA = 0
∑Mz = 0
RA × 4 – 3 × 3 - 2 × 2 × 1= 0
RA = 3.25 kN
RB = 3.75 kN
BMD = RA × 2 - 3 × 1
BMD = 3.25 × 2 - 3 = 3.5 kN