Concept:

Shear force at any section is the summation of all transverse loading, either to the left or to the right of any section.

Bending moment at any section is the summation of all transverse loading and couple either to the left or to the right of any section.

Relationship between bending moment, shear force, and loading:

(1) Rate of change of shear force at any section is equal to load intensity at that section.

\(\frac{{dv}}{{dx}} = \omega \Rightarrow dV = \omega dx \Rightarrow V = \smallint \omega dx\)

i.e. Change of shear force between two sections is equal to the area under load intensity diagram between those two sections.

(2) The rate of change of bending moment at any section is equal to shear force at that section.

\(\frac{{dM}}{{dx}} = V \Rightarrow dM = Vdx \Rightarrow M = \smallint Vdx\)

i.e. Change of bending moment between two sections is equal to the area under shear force diagram between those two sections

The presence of a concentrated couple leads to sudden rise or fall of the Bending moment diagram depending upon the sign and direction of the couple.

Calculation:

Depending on the nature of the couple and sign convention adopted, change in bending moment between two sections (1) and (2) =Area under SFD between those two sections ± Concentrated couple = α ± M

Concept:

Shear force at any section is the summation of all transverse loading, either to the left or to the right of any section.

Bending moment at any section is the summation of all transverse loading and couple either to the left or to the right of any section.

Calculation:

Bending moment produced at support due to load \( = 2 \times 5 \times \frac{5}{2}\) = 25 kN-m (hogging)

Due to the symmetry of load and structure, both the support reactions will be the same. Let the reactions at support be R.

Bending moment at the center of load P:

B.M \( = 2 \times 5 \times \left( {5 + \frac{5}{2}} \right) - R \times 5 = 75 - 5R\)

∵ Both the bending moments are numerically equal

⇒ 75 – 5R = 25  ⇒ R = 10 kN

For vertical equilibrium, ∑F_y = 0

⇒ R + R = (2 × 5) + (2 × 5) + P

⇒ 10 + 10 = 10 + 10 + P

⇒ P = 0

Concept:

The kinetic energy of the moving mass is converted into the potential energy of the spring.

Kinetic energy = \(\frac{1}{2}m{v^2}\;\)

Potential energy of spring = \(\frac{1}{2}k{x^2}\)

Stiffness: The force required to produce unit deflection in the spring is called as stiffness of the spring.

Calculation: 

Deflection (x) = 150 mm = 0.15 m 

\(\begin{array}{l} \frac{1}{2}k{x^2} = \frac{1}{2}m{v^2}\\ \Rightarrow k = m{\left( {\frac{v}{x}} \right)^2} = 2000 × {\left( {\frac{2}{{15}}} \right)^2} = 355.556\ kN/m \end{array}\)

Stiffness of one spring (k) \(= \frac{{355.556}}{2} = 177.78\ kN/m\)

Maximum force = k × x \(= 177.78 × {10^3} × 0.15 = 26666.67\ N\)

Concept:

Equivalent torque:

It is the torque which while acting alone produces maximum shear stress equal to maximum shear stress due to combined action of moment and torque.

The magnitude of \({T_{eq}} = \sqrt {{M^2} + {T^2}}\)

Given:

The moment about x-axis produces torque and moment about z-axis produces a bending moment.

∴ T = 10 kN-m, M = 20 kN – m

Using formula, \({T_{eq}} = \sqrt {{M^2} + {T^2}} = \sqrt {{{20}^2} + {{10}^2}} \)

\( = 10\sqrt 5 \;\;kN - m\) = 22.36 kN – m

Important Point:

Equivalent Bending moment:

It is the bending moment which while acting alone produces maximum bending stress equal to the maximum principal stress produced due to combined action of moment and torque.

The magnitude of \({M_{eq}} = \frac{1}{2}\left[ {M + \sqrt {{M^2} + {T^2}} } \right]\)

Concept:

Flexure formula:- \(\frac{M}{I} = \frac{\sigma }{y} = \frac{E}{R}\)

Where, σ = Bending Stress, y = Distance from the bending axis, R = Radius of Curvature

This formula is theoretically valid for pure bending, but if we apply it in non-uniform bending, the error introduced is negligible. Analysis of pure bending for symmetrical conditions yields flexure formula –

\(\because \frac{M}{I} = \frac{\sigma }{y} \Rightarrow \sigma = \frac{{M y}}{I}\)

Using, Torsion formula

\(\frac{{\rm{T}}}{{{{\rm{J}}_{\rm{}}}}} = \frac{{\rm{τ }}}{{\rm{R}}} = \frac{{{\rm{G\theta }}}}{{\rm{L}}}\)

Where, T = Applied Torsion, J = Polar moment of inertia, R = Radius of the shaft, G = Shear modulus, θ = permissible angle of twist, L = Length of the shaft, τ = permissible shear stress in the shaft

Calculation:

M_x = 50 N – m ---- Torsional

M_y = 0

M_z = 100 N – m -----Bending moment

Shear stress

\(\tau = \frac{T}{J} \times r\)

\(= \frac{{50}}{{2 \times \pi \times {r^3}t}} \times r\)             [∴ J = 2π r³t  for circular fillet weld]

\(\tau = \frac{{50}}{{2\pi t{r^2}}}\)            --------1)

Bending Stress

\({\sigma _b} = \frac{M}{I} \times Y\)

\({\sigma _b} = \frac{{100}}{{\pi {r^3}t}} \times r\)           ---------2) [∴ I = π r³t for circular fillet weld]

From 1) & 2)

\(\begin{array}{l} \frac{{{\sigma _b}}}{\tau } = \frac{{\left( {\frac{{100}}{{\pi {r^3}t}} \times r} \right)}}{{\left( {\frac{{50}}{{2\pi t{r^2}}}} \right)}}\\ \frac{{{\sigma _b}}}{\tau } = 4 \end{array}\)

Concept: 

Point of Contraflexure ⇒ The point where the bending moment changes its sign is termed as a point of contraflexure.

Calculation:

Bending moment between B & C is given by

\({M_x} = 2 \times \left( {2 + x} \right)-10x + 2 \times x \times \frac{x}{2}\)

= 4 – 8x – x²

At point of contra flexure,

M_x = 0

⇒ 4 – 8x – x2 = 0 ⇒ x = 0.54 m

∴ Point of contra flexure is (2 + 0.54) = 2.54 m from point D.

Important Point:

The points where the deflected shape changes its curvature is known as the point of inflection.

Concept:

Using, Torsion formula

\(\frac{{\rm{T}}}{{{{\rm{J}}_{\rm{}}}}} = \frac{{\rm{τ }}}{{\rm{R}}} = \frac{{{\rm{G\theta }}}}{{\rm{L}}}\)

Where, T = Applied Torsion, J = Polar moment of inertia, R = Radius of the shaft, G = Shear modulus, θ = permissible angle of twist, L = Length of the shaft, τ = permissible shear stress in the shaft

Calculation:

Equation of equilibrium:

The load or torque T produces reactions T_AB and T_BC at the fixed end of the shaft

∴ T_AB + T_BC = T                       ------------------------------ (1)

Because there are two unknowns in this equation (and no other useful equation of equilibrium), the composite shaft is statically indeterminate.

Equation of compatibility:

            We now separate the bar from its support at end C and obtain a bar that is fixed at end A and free at end C.

Let f₁ = an angle of twist at end C when the load T acts alone

f₂  = an angle of twist at end C when the reactive torque T_BC acts alone

The angle of twist at end C in the original bar, equal to the sum of f₁and f₂ is zero

∴ the equation of compatibility is

∅₁ + ∅₂ = 0             ------------------------------- (2)

\({\emptyset _1} = \frac{{Tl}}{{2GJ}}\;\& \;{\emptyset _2} = \frac{{ - {T_{BC}}l}}{{2GJ}} - \frac{{{T_{BC}}l}}{{2GJ}}\)

\(\therefore \left( 2 \right) \Rightarrow \frac{{Tl}}{{2GJ}} - \frac{{{T_{BC}}l}}{{2GJ}} - \frac{{{T_{BC}}l}}{{2GJ}} = 0\)

⇒ T = 2T_BC           -------------------------------- (3)

Substitute equation (3) in equation (1)

(1)  ⇒ T_AB + T_BC = 2 T_BC

⇒ T_AB = T_BC

\(\Rightarrow \frac{{{T_{AB}}}}{{{T_{BC}}}} = 1:1\)

Let force in spring 1 = P₁ and elongation in spring 1 = Δ₁

Let force in spring 1 = P₂ and elongation in spring 1 = Δ₂

for bar to remain horizontal

L₁ + Δ₁ = L₂ + h + Δ₂

Δ₁ = Δ₂ + 30                              ---- (1)

FBD of spring system is

ΣF_V = 0

P₁ + P₂ = W

K₁Δ₁ + K₂Δ₂ = 25

0.3Δ₁ + 0.4 (Δ₁ – 30) = 25

Δ₁ = 52.85 mm

So, Δ₂ = Δ₁ – 30 = 22.85 mm

Also, ΣM_B = 0

\(\begin{array}{l} {P_1}\left( {L - b} \right)\; = \;W \times \frac{L}{2}\\ \Rightarrow {K_1}{{\rm{\Delta }}_1}\left( {L - b} \right)\; = \;W \times \frac{L}{2}\\ \Rightarrow 0.3 \times 52.85\left( {350 - b} \right)\; = \;\frac{{25 \times 350}}{2} \end{array}\)

⇒ b = 74 mm

Concept:

Maximum bending across shaft occurs at either top or bottom fiber and it is given by the bending formula:

\({{\rm{\sigma }}_{{\rm{max}}}} = \frac{{\rm{M}}}{{\rm{Z}}} = \frac{{2{\rm{M}}}}{{\frac{{\rm{\pi }}}{{64}}{{\rm{d}}^4}}} \times \frac{{\rm{d}}}{2} = \frac{{64{\rm{M}}}}{{{\rm{\pi }}{{\rm{d}}^3}}}\)

Maximum shear stress is given by the twisting formula:

\({{\rm{\tau }}_{{\rm{max}}}} = \frac{{{\rm{Tr}}}}{{\rm{I}}} = \frac{{\left( {\frac{1}{2}} \right) \times \left( {\frac{{\rm{d}}}{2}} \right)}}{{\frac{{\rm{\pi }}}{{32}}{{\rm{d}}^4}}} = \frac{{8{\rm{T}}}}{{{\rm{\pi }}{{\rm{d}}^3}}}\)

\(\frac{{{{\rm{\sigma }}_{{\rm{\;max}}}}}}{{{{\rm{\tau }}_{{\rm{max}}}}}} = \frac{{\frac{{64{\rm{M}}}}{{{\rm{\pi }}{{\rm{d}}^3}}}}}{{\frac{{8{\rm{T}}}}{{{\rm{\pi }}{{\rm{d}}^3}}}}} = \frac{{8{\rm{M}}}}{{\rm{T}}}\)

Concept:

Shear force at any section is the summation of all transverse loading, either to the left or to the right of any section.

Bending moment at any section is the summation of all transverse loading and couple either to the left or to the right of any section.

Relationship between bending moment, shear force, and loading:

(1) The rate of change of shear force at any section is equal to load intensity at that section.

\(\frac{{dV}}{{dx}} = w \Rightarrow dV = w dx \Rightarrow V = \smallint w dx\)

i.e. The change of shear force between two sections is equal to the area under load intensity diagram between those two sections.

(2) The rate of change of bending moment at any section is equal to shear force at that section.

\(\frac{{dM}}{{dx}} = V \Rightarrow dM = Vdx = M = \smallint Vdx\)

i.e.The change of bending moment between two sections is equal to the area under the shear force diagram between those two sections.

The loading diagram obtained from SFD is generally found to satisfy the force equilibrium but may fail to satisfy moment equilibrium.

The presence of any concentrated couple will not affect the SFD.

Hence from the SFD, the loading diagram is obtained first, and then the bending moment diagram is obtained after verification for force and moment equilibrium.

Calculation:

Arguments for load diagram from SFD:

At A: SFD jumps up by 20 kN

⇒ There is an upward point load of 20 kN at A.

A to B: The slope of SFD is constant and negative.

⇒ A uniformly distributed downward load acts from A to B. Let the magnitude of this UDL be w.

\(\because \frac{{dV}}{{dx}} = w \Rightarrow \frac{{10 - 20}}{5} = w \Rightarrow w = - 2\;kN/m\;\)

 Hence a UDL of 2 kN/m acts from A to B in the downward direction.

At B: There is a sudden jump of SFD from +10 kN to -10 kN.

⇒ A downward point load of (- 10 - 10) = -20 kN acting at point B.

From A to C: The slope of SFD is constant and negative

⇒ A uniformly distributed load of acts from B to C.

\(\because \frac{{dv}}{{dx}} = w \Rightarrow \left( {\frac{{ - 15 - \left( { - 10} \right)}}{5}} \right) = w \Rightarrow w = - 1\;kN/m\)

Hence a UDL of a 1 kN/m acts from B to C.

At C: SFD jumps up by 15 kN.

∴ Upward point load of 15 kN acts at C.

Check for equilibrium of the beam:

∑F_y = 20 – 2 × 5 – 20 – 1 × 5 + 15 = 0

Since ∑F_y = 0 ⇒ Force equilibrium is satisfied.

\(\sum {M_A}\left( {in\;clockwise\;direction} \right)\; = \left( {2 \times 5 \times \frac{5}{2}} \right) + \left( {20 \times 5} \right) + \left( {1 \times 5 \times 7.5} \right) - 15 \times 10\)

∑M_A   = 12.5 kN - m   

Hence to counteract this moment a anticlockwise moment of 12.5 kN-m must act to satisfy moment equilibrium. This moment can act at any point in the beam and any specific point cannot be referred to using the SFD only.

Hence the correct load diagram can be:

Note that the moment 12.5 kN-m can be applied anywhere in the beam.