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MODULE 1. BASIC CONCEPTS

MODULE 2. SYSTEM OF FORCES

MODULE 3.

MODULE 4. FRICTION AND FRICTIONAL FORCES

MODULE 5.

MODULE 6.

MODULE 7.

MODULE 8.

## LESSON 22.

**22.1 Shear Force and Bending Moment diagrams for Simply Supported Beams subjected to different types of loading**

(i) Simply Supported Beam subjected to concentrated load at mid span

**Fig. 22.1 and 22.2**

As shown in Fig.22.1, a beam *AB* simply supported at the ends *A* and *B*. The length of the span is *l*. *P* is the concentrated load applied at the centre.

The load is placed symmetrical on the span,

*V _{a}* and

*V*are the vertical reactions at

_{b}*A*and

*B*.

The reaction at each support will be \[{P \over 2}\]

Therefore, *V _{a}* =

*V*= \[{P \over 2}\]

_{b}For any section between *A* and *C*,

S.F = *S _{x} *= + \[{P \over 2}\]

Similarly for section *C* and *B*,

S.F = *S _{x}* = - \[{P \over 2}\]

At the point *C*, the S.F changes from + \[{P \over 2}\] to - \[{P \over 2}\]

For B.M, at any section between *A* and *C* distant *x* from the end *A*,

*M _{x}* = + \[{P \over 2}\] x (sagging moment)

At *x* = 0, *M _{x}* = 0

At *x* = , *M _{x}* = \[{{Pl} \over 4}\]

Hence B.M increases uniformly from zero at

*A*to \[{{Pl} \over 4}\] at

*C*.

Similarly, B.M decreases uniformly from \[{{Pl} \over 4}\] at *C* to zero at *B*.

Note: It should be noted that maximum bending moment occurs at mid span (at C), where the shear force changes its sign.

(ii) Simply supported beam subjected to a concentrated load placed eccentrically on the span

Fig.22.2 shows a simply supported beam *AB* of span *l* carrying a concentrated load *P* at *C* eccentrically. Where *AC* = *a* and *CB* = *b*

For the equilibrium of the beam,

Taking moments of the forces on the beam about *A*,

We have *V _{b}*

*. l*=

*Pa*

*V _{b}* = \[{{Pa} \over l}\]

*V*=

_{a}*P*- \[{{Pa} \over l} = {{P\left( {l - a} \right)} \over l}\]

*V*= \[{{Pb} \over l}\]

_{a}We know that *a + b = l*

At any section between *A* and *C*,

S.F, *S _{x}* =

*V*= + \[{{Pb} \over l}\]

_{a}for section *C* and *B*,

S.F, *S _{x}* = -

*V*= - \[{{Pb} \over l}\]

_{b}Similarly, Bending Moment for any section between *A* and *C*

*M _{x}*

_{ }= + \[{{Pb} \over l}x\] (sagging)

At *x *= 0, *M _{x}* = 0

At *x* = *a*, *M _{x}*

_{ }= \[{{Pab} \over l}\]

Hence the B.M increases uniformly from zero at the left end *A* to \[{{Pab} \over l}\] at *C*. Similarly the B.M will decrease uniformly from \[{{Pab} \over l}\] at *C* to zero at the right end *B*.

(iii) Simply supported beam subjected to more than one concentrated load* *

**Fig. 22.3**

Fig.22.3 shows a simply supported beam *AB* of span 8m subjected to concentrated loads of 4*kN*, 8k*N* and 6k*N* at distances of 2m, 4m and 6m respectively from support *A*.

To find the vertical reactions *Va* and *Vb* at the supports *A* and *B* respectively.

For the equilibrium of beam, Taking moment of the forces on the beam about the left support,

We have,

*V _{b} *× 8 = 6 × 6 + 8 × 4 + 4 × 2

*V _{b}* = 9.5

*kN*

*V _{a}*

_{ }= total load on the beam –

*V*

_{b} = 18 - 9.5 = 8.5 *kN*

We will start from the support *A*,

S.F between *A* and *C* = 8.5 *kN*

S.F between *C* and *D *= 8.5 – 4 = 4.5 *kN*

S.F between *D* and *E *= 8.5 – 4 – 8 = -3.5 *kN*

S.F between *E *and *B *= 8.5 – 4 – 8 – 6 = -9.5 *kN*

B.M at *A* = 0

B.M at *C* = 8.5 × 2 = 17 *kNm* (sagging)

B.M at *D* = 8.5 × 4 – 4 × 2 = 26*kNm* (sagging)

B.M at *E* = 8.5 × 6 – 4 × 4 – 8 × 2 = 19*kNm* (sagging)

Note: it can be observed from the S.F and B.M diagrams that the maximum B.M occurs at *D* where the S.F changes its sign.

(iv)Simply supported beam subjected to uniformly distributed load of w per unit run over the whole span

* ***Fig. 22.4 and 22.5**

Fig.22.4 shows a simply supported beam *AB* of span *l* subjected to uniformly distributed load *w* per unit run over the whole span.

Since the loading the symmetrical on the span, each vertical reaction equal half the total load on the span.

Therefore, *V _{a}*

_{ }=

*V*= \[{{wl} \over 2}\]

_{b}Consider any section

*X*distant

*x*from the left end

*A*.

S.F and B.M at the section *X* are given by,

*S _{x}* = +

*V*

_{a}_{ }–

*wx*= + \[{{wl} \over 2}\] –

*wx*

*M _{x}*

_{ }= +

*V*- \[{{w{x^2}} \over 2}\] = \[{{wl} \over 2}\]

_{a}x*x*- \[{{w{x^2}} \over 2}\]

Therefore, *M _{x}*

_{ }= + \[{{w} \over 2}\] x (l - x)

At *x* = 0, *S _{x}* = \[{{wl} \over 2}\] and

*M*= 0

_{x}At *x* = *l*, *S _{x}* = \[{{wl} \over 2}\] - wl = - \[{{wl} \over 2}\] and

*M*=0

_{x}At *x* = \[{1 \over 2}\] , *S _{x}* = \[{{wl} \over 2}\] - \[{{wl} \over 2}\] = 0

And *M _{x}*

_{ }= + \[{w \over 2}\] . \[{l \over 2}\] \[\left( {l - {l \over 2}} \right)\] = + \[{{w{l^2}} \over 8}\]

The S.F diagram is straight line. The S.F uniformly changes from + \[{{wl} \over 2}\] at

*A*to - \[{{wl} \over 2}\] at

*B*and obviously the S.F at mid span is zero.

The B.M diagram is parabola. It increases from zero at *A* to + \[{{w{l^2}} \over 8}\] at the mid span *C* and from this value it decreases to zero at *B*.

(v)Simply supported beam subjected to a uniformly distributed load over some part of the span

Fig.22.5 shows a simply supported beam as mentioned above.

Taking moments about A,

V_{b} × 10 = 10 × 4 × \[{4 \over 2}\]

Therefore V_{b} = 8 kN

V_{a} = 10 × 4 – 8 = 32 kN

Consider any section between A and C distant x from A

S.F at the section is given by, S_{x} = +32 – 10x

At x = 0, S_{x} = + 32 kN

At x = 4m, S_{x} = 32 – 40 = - 8 kN

Let the S.F be zero x meters from A. Equating the S.F to zero, we get,

32 – 10x = 0

x = 3.2 m from A

at any section in AC distant x from A the B.M is given by,

M_{x} = + 32x – 10 \[{{{x^2}} \over 2}\] = 32x - 5x^{2}

At x = 0, M_{x} = 0

At x = 4m, M_{x} = 32 × 4 - 5× 4^{2}

= 48 kNm

At x = 3.2 m, M_{x} = 32 × 3.2 – 5 × 3.2^{2}

= 51.2 kNm

B.M decreases from 48 kNm at C to zero at B according to the linear law.

Maximum B.M occurs at D where S.F is zero.

(vi)simply supported beam subjected to two couples.

**Example 1:** Draw the S.F and B.M diagrams for the beam (Fig.-------------).

Sol: By taking moments about A.

V_{b} × 10 + 100 + 80 = 20 × 10 × 5

V_{b} = 80 kN

V_{a} = (20 × 10) – 80 = 120 kN

S.F Diagram: At any section distant x from A, the shear force is given by,

S* _{x}* = 120 – 20

*x*= 0

At *x* = 0, S.F = 120 kN

At *x* = 10, S.F = -80 kN

For finding the section of zero shear, equating the general expression for shear force to zero.

120 – 20*x* = 0

*x* = 6 m

B.M Diagram: At any section distant x from the end A, the bending moment is given by,

M = 120 – 20. *x*. - 100

M = 120 – 10*x*^{2} – 100

at *x* = 0, B.M = -100 kNm

at *x* = 10, B.M = 100 kNm

at *x* = 6m, B.M = 260 kNm

For finding the point of contraflexure, equating the expression for B.M to zero.

120*x* – 10*x*^{2} – 100 = 0

*x*^{2} – 12*x* + 10 = 0

we get, *x* = 0.9m

**Example:** Fig…………shows the simply supported beam. Draw the S.F and B.M diagrams.

Solution: Let V_{a} and V_{b} be the vertical reactions at the supports A and B respectively.

By taking moments about A.

V_{b} × 8 = 5 × 1+ 10 × 3.5 + 6 × 6

V_{b} = 9.5 kN

V_{a} = 5 + 10 + 6 – 9.5 = 11.5 kN

S.F between A and C = 11.5 kN

S.F between C and D = 11.5 – 5 = 6.5 kN

S.F between D and E = 11.5 – 5 – 10 = - 4.5 kN

B.M at A = 0

B.M at C = 11.5 × 1 = 11.5 kNm

B.M at D = 11.5 × 3.5 – 5 × 2.5 = 27.75 kNm

B.M at E = 9.5 × 2 = 19 kNm * *