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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 21.

**21.1 Shear Force and Bending Moment diagrams for Cantilever subjected to different types of loading**

(i) Cantilever subjected to concentrated load ‘P’ at the free end

Fig.21.1 shows a cantilever *AB* fixed at *A* and free at *B *and carrying a load *W* at the free end *B*.

\

**Fig. 21.1**

Consider a section *X* at a distance *x* from the free end

Shear Force (S.F) at *X* = *S _{x}*

_{ }= +

*P*

Bending Moment (B.M) at *X* = *M _{x}* = -

*Px*

So, the S.F is constant at all the sections of the member between *A *and *B*.

But the B.M at any section is proportional to the section from the free end.

At *x* = 0 (at *B*), B.M = 0

At *x* = *l* (at *A*), B.M = -*Pl*

(ii) Cantilever with more than one concentrated load

**Fig. 21.2**

Suppose a cantilever *AC* is *2 m* long and is subjected to two concentrated loads.

Consider the section between *B* and *C*, distance *x* from *C*.

S.F = *S _{x}*= 100 kN

B.M = *M _{x}* = - 100

*x*

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

_{ }= 0

At *x* = 1, *M _{x}* = -100

*kNm*

Now consider section *A* and *B*, distance *x* from *C*.

S.F = *S _{x}* = 100 + 50 = + 150

*kN*

B.M = *M _{x}* = - 100

*x*– 50(

*x*-1)

= -150*x* + 50

At *x* = 1, *M _{x}* = -100

*KNm*

At *x* = 2, *M _{x}* = -250

*KNm*

(iii) Cantilever subjected to uniformly distributed load of p per unit run over the whole length

* *

Fig.21.3 shows a cantilever *AB* fixed at *A* and free at *B* carrying a uniformly distributed load of *w* per unit run over the whole length.

Consider any section *X* distant *x* from the end *B*.

S.F at *X* = *S _{x}* = +

*px*

B.M at *X* = *M _{x}* = -

*px*. \[{x \over 2}\]

Therefore,

*M*

_{x}_{ }= - \[{{p{x^2}} \over 2}\]

- The variation of S.F is according to a linear law, while the variation of the bending moment is according to a parabolic law.

At *x* = 0, *S _{x}* = 0 and also

*M*= 0

_{x}At *x* = *l*, *S _{x}* = +

*wl*and

*M*

_{x}_{ }= - \[{{p{l^2}} \over 2}\]

(iv) Cantilever subjected to a uniformly distributed load of p per unit run for a distance a from the free end

**Fig.21.4**

Fig.21.4 shows a cantilever *AB* fixed at *A* and free at *B *and carrying *udl* over a distance *a.*

Consider any section between *C* and *B *distant *x* from the free end *B*.

S.F, *S _{x}*

_{ }= +

*px*

B.M, *M _{x}* = - \[{{p{x^2}}\over 2}\]

The above relations are good for all the values of

*x*between

*x*= 0 and

*x*=

*a*

The variation of S.F will be linear and the variation of B.M for this section will be parabolic.

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

_{ }= 0 and

*M*= 0

_{x}At *x *= *a*, *S _{x}* = +

*pa*and

*M*= -

_{x}*pa*\[\left({x-{a\over 2}}\right)\]

For section

*A*and

*C*, S.F is constant at +

*pa*but the B.M varies according to linear law.

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

_{ }= -

*pa \[\left({a -{a\over 2}}\right)\]*= \[{{w{a^2}}\over 2}\]

At

*x*=

*l*,

*M*

_{x}_{ }= -

*pa*

*\[\left( {L - {a \over 2}} \right)\]*(v) Cantilever subjected to a uniformly distributed load w per unit run over the whole length and a concentrated load P at the free end*.*

**Fig. 21.5 and 21.6**

Fig.21.5 shows a cantilever *AB *fixed at *A* and free at *B* and carrying the load system mentioned above. Consider any section *X* distant *x* from the end *B*. The S.F and the B.M at the section *X*, are respectively given by.

*S _{x}* =

*wx*+

*P*

And *M _{x}* = - \[\left( {{{w{x^2}} \over 2} + Px} \right)\]

At

*x*= 0,

*S*= +

_{x}*P*and

*M*

_{x}_{ }= 0

At *x *= *l*, *S _{x}*

_{ }= +

*(wl + P)*

*M _{x}*

_{ }= - \[\left( {{{w{x^2}} \over 2} + Pl} \right)\]

(vi) Cantilever subjected to a load whose intensity varies uniformly from zero at the free end to w per unit run at the fixed end

Fig.21.6 shows a cantilever *AB *of length *l* fixed at *A* and free at *B* carrying the load as mentioned above.

Let the intensity of loading at *X*, at a distance *x* from the free end *B* be *w _{x} *per unit run.

Therefore *w _{x }* = \[x \over l}w\] since the intensity of load increases uniformly from zero at the free end to w at the fixed end.

The total downward load on the free body, equal to the area of the triangular loading diagram (Fig.21.6), is

\[+{1\over 2}\left({{{{w_x}}\over l}}\right)\left( x \right)=+{{w{x^2}} \over {2l}}\]

*S _{x \[+{{w{x^2}} \over {2l}}\]}M_{x}* = Moment of the load acting on

*XB*about

*X*

= area of the load diagram between *X* and *B* × distance of the centroid of this diagram from *X*

\[=-{{w{x^2}} \over {2l}}.{x \over 3}\]

*M _{x}*

_{ }= \[-{{w{x^3}} \over {6l}}\]

At

*x*= 0,

*S*= 0 and

_{x}*M*= 0

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

*M*= \[- {{w{l^2}} \over 6}\]

_{x}The S.F varies following a parabolic law while the B.M follows cubic law.

(vii) Cantilever subjected to a load whose intensity varies unifrormly from zero at the fixed end to w per unit run at the free end

* *

* ***Fig. 21.7**

Fig.21.7 shows a cantilever *AB* of length *l* and fixed at *A* and free at *B* and carrying the loading as mentioned above

Let *M _{a}* be the reacting moment of fixing moment

*A*.

Therefore *M _{a}* = moment of the total load amount

*A*

= \[{{wl}\over 2}.{{2l} \over 3}={{w{l^2}} \over 3}\]*V _{a}* = vertical reaction at

*A*

= Total load on the cantilever

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

Consider any section

*X*distant

*x*from the fixed end

*A*

S.F at *X* = algebraic sum of forces on *AX*

*S _{x}* = \[{{wl}\over 2} - {x \over 2}.{{wx} \over l}\]

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

_{x}B.M at

*X*= algebraic sum of forces and reactions on

*AX*above

*X*

*M _{x}* = \[{{wl} \over 2}x - {{w{x^2}} \over {2l}}.{x \over 3} - {M_a}\]

= \[{{wl} \over 2}x - {{w{x^3}} \over {6l}} - {{w{l^3}} \over 3}\]

At

*x*= 0 i.e at

*A*,

*S*= + \[{{wl} \over 2}\] and

_{x}*M*= - \[{{w{l^2}} \over 3}\]

_{x}At

*x*=

*l*i.e at

*B*,

*S*= \[{{wl} \over 2}\] - \[{{w{l^2}} \over 2l}\] = 0 and

_{x}*M*= \[{{wl} \over 2}l - {{w{l^3}} \over {6l}} - {{w{l^3}} \over 3}=0\]

_{x}**Example:** Fig.21.8 shows a cantilever subjected to a system of loads. Draw S.F and B.M diagrams.

Solution: At any section between D and E, distance x from E

S.F = S_{x} = +500 N

B.M = M_{x} = -500x

At x = 0, M_{x} = 0

x = 0.5m, M_{x} = -250 Nm

At any section between C and D, distance x from E

S.F = S_{x} = +500 + 500 = + 1000 N

B.M = M_{x} = -500x – 500(x - 0.5)

= -1000x + 250

At x = 0.5, M_{x} = -250 Nm

At x = 1m , M_{x} = -750 Nm

At any section between B and C, distance x from E

S.F = S_{x} = +500 + 500 + 400 = + 1400 N

B.M = M_{x} = -500x – 500(x - 0.5) – 400(x – 1)

= -1400x + 650

At x = 1m, M_{x} = -750 Nm

At x = 1.5m , M_{x} = -1450 Nm

At any section between A and B, distance x from E

S.F = S_{x} = +500 + 500 + 400 + 400 = + 1800 N

B.M = M_{x} = -500x – 500(x - 0.5) – 400(x – 1) – 400 (x – 1.5)

= -1800x + 1250

At x = 1.5m, M_{x} = -1450 Nm

At x = 2m , M_{x} = -2350 Nm