Mechanics Of Solids|Shear And Bending Moments

Mechanics Of Solids

Solved Numericals on Shear And Bending Moments

Q-1 For the state of plane stress shown, determine the maximum shearing stress when (a) σy = 20 MPa, (b) σy = 140 MPa. (Hint: Consider both in-plane and out-of-plane shearing stresses).

(A) σ_y = 20MPa

σ_y = 20-80

σ_y =-60MPa

(B) σ_y =140MPa

σ_y =140-80

σ_y =60MPa

Q-2 For the aluminium shaft show (G = 27 GPa), determine (a) the torque T which causes an angle of twist of 5°, (b) the angle of twist caused by the same torque T in a solid cylindrical shaft of the same length and cross-sectional area.

SOL:

PART – (a)

Given :-

Length ( L ) = 1.2 m

=1200 mm

Angle of deflection (ø) = 5̊

Modulus of rigidity (g) = 27GPa = 27x10^⁹ N/㎡

= 27 x 10^³ N/m㎡

Ø in radians = (5 x 3.14) /180

= 0.0872

Polar moment of inertia (Ip) = pi/32 [40^⁴ – 24^⁴ ]

= 218644.48 mm^⁴

Now according to formula:

T/Ip =CØ/L

=> T = ( 27 x 10^³ x 0.0872 x 218644.48) / 1.2 x 10^³ .

= 428980.4698 Nmm

= 428.980 Nm

PART - (b):

Given :-

Torque (T) = 428.980 Nm

L = 1.2 m

Outer radius (Ro) = 20 mm

Inner radius (Ri) = 12mm

Area of crossection (A) = 3.14 ( 20^² – 12^² )

= 803.84 m㎡

Angle of twist = ?

For a solid cylinder of same crossectional Area

We have pi/4 x d^² where d is the diameter of the solid cylinder.

(3.14 x d^² ) / 4 = 803.84

=> d^² = 1024

=> d = 32

So we have

428980.4698/6430.72 = 27 x 10^³ x Ø /1200

= 2.9648 radians

Ans :- 1. Torque = 428.980 Nm

2. Angle of twist = 169.95̊ = 2.9648 radians

Q-3: Determine the torque T which causes a maximum shearing stress of 70 MPa in the steel cylindrical shaft shown.

SOL:

Given :-

Diameter (d) = 36 mm

Max. shaearing stress = 70 MPa = 70 x10^⁶ N/㎡

= 70 N/m㎡

Polar moment of inertia (I) = 3.14 /32 x 36^⁴

Torque (T) = 70 x 164812.32/18

=> T = 640.936 kNmm

Q-4 Draw the shear and bending-moment diagrams for the beam and loading shown, and determine the maximum absolute value (a) of the shear, (b) of the bending moment.

SOL:

Calculating reactions:

R_a+R_b=68+(30x1.2)

R_a+R_b=68+36

R_a+R_b=108KN

Moment at point B:

R_ax3.6=(68x1.2)+36(2.4+1.2/2)

R_ax3.6=(68x1.2)+(30x3.6)

R_a=189.6/3.6

R_a=52.66 KN

R_b=108-52.66 KN

R_b=51.33 KN

(A) Shear Force :

At A (L.H.S)=0

(R.H.S)=52.66 kN

At C (L.H.S)=52.66-(30x1.2)=16.66 KN

(R.H.S)=16.66 KN

D (L.H.S)=16.66 KN

(R.H.S)=16.66-68=-51.34 KN

B (L.H.S)= -51.34 KN

(R.H.S)= -51.34+51.34=0

(B) The bending moment

A=0

C=(52.66x1.2)-(30x1.2x1.2/2)-(68x1.2)=40.008 KNm

D=51.34x1.2=61.608 KNm

B=0

Q-5 A beam AB 10 meters long has supports at its ends A and B. It carries a point load of 5 KN at 3 meters from A and a point load of 7 KN from A and UDL of 1 KN/m between the point loads. Draw SF and BMD diagrams for the beam.?

SOL: Firstly , we will take the reactions that are:

Here,the loading is symmetrical reaction at each support equals half the total.Therefore,

V_a =V_b= 5+5+1x4/2

= 7KN

SHEAR FORCE(S.F) ANALYSIS:

S.F at any section in AD =+7 KN

S.F at just on RHS of AD =+7 – 5 =+2 KN

S.F at just on LHS of E =-7+5=-2 KN

S.F at any section in EB = -7 KN

S.F at centre C=0

BENDING MOMENT(B.M). ANALYSIS:

B.M at A = 0