## 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.

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Fig. 21.1

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

Shear Force (S.F) at X = Sx = + P

Bending Moment (B.M) at X = Mx = - 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 = Sx= 100 kN

B.M = Mx = - 100x

At x = 0, Mx = 0

At x = 1, Mx = -100 kNm

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

S.F = Sx = 100 + 50 = + 150 kN

B.M = Mx = - 100x – 50(x-1)

= -150x + 50

At x = 1, Mx = -100 KNm

At x = 2, Mx = -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 = Sx = + px

B.M at X = Mx = - px . ${x \over 2}$

Therefore, Mx = - ${{p{x^2}} \over 2}$

• 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, Sx = 0 and also Mx = 0

At x = l, Sx = + wl and Mx = - ${{p{l^2}} \over 2}$

(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, Sx = + px

B.M, Mx = - ${{p{x^2}}\over 2}$

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, Sx = 0 and Mx = 0

At x = a, Sx = +pa and Mx = -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, Mx = -pa $\left({a -{a\over 2}}\right)$ = ${{w{a^2}}\over 2}$

At x = l, Mx = - 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.

Sx = wx + P

And Mx = - $\left( {{{w{x^2}} \over 2} + Px} \right)$

At x = 0, Sx = + P and Mx = 0

At x = l, Sx = + (wl + P)

Mx = - $\left( {{{w{x^2}} \over 2} + Pl} \right)$

(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 wx per unit run.

Therefore wx  = $x \over l}w$  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

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

Sx $+{{w{x^2}} \over {2l}}$

Mx

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

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

Mx = $-{{w{x^3}} \over {6l}}$

At x = 0, Sx = 0 and Mx = 0

x = l, Sx = $+ {{wl} \over 2}$ and Mx = $- {{w{l^2}} \over 6}$

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 Ma be the reacting moment of fixing moment A.

Therefore Ma = moment of the total load amount A

= ${{wl}\over 2}.{{2l} \over 3}={{w{l^2}} \over 3}$

Va = vertical reaction at A

= Total load on the cantilever

Va = ${{wl}\over 2}$

Consider any section X distant x from the fixed end A

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

Sx = ${{wl}\over 2} - {x \over 2}.{{wx} \over l}$

Sx = ${{wl}\over 2} - {{w{x^2}} \over {2l}}$

B.M at X = algebraic sum of forces and reactions on AX above X

Mx = ${{wl} \over 2}x - {{w{x^2}} \over {2l}}.{x \over 3} - {M_a}$

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

At x = 0 i.e at A,           Sx= + ${{wl} \over 2}$  and  Mx = - ${{w{l^2}} \over 3}$

At x = l i.e at B,             Sx = ${{wl} \over 2}$${{w{l^2}} \over 2l}$ = 0    and   Mx=  ${{wl} \over 2}l - {{w{l^3}} \over {6l}} - {{w{l^3}} \over 3}=0$

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 = Sx = +500 N

B.M = Mx = -500x

At x = 0, Mx = 0

x = 0.5m, Mx = -250 Nm

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

S.F = Sx = +500 + 500 = + 1000 N

B.M = Mx = -500x – 500(x - 0.5)

= -1000x + 250

At x = 0.5, Mx = -250 Nm

At x = 1m , Mx = -750 Nm

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

S.F = Sx = +500 + 500 + 400 = + 1400 N

B.M = Mx = -500x – 500(x - 0.5) – 400(x – 1)

= -1400x + 650

At x = 1m, Mx = -750 Nm

At x = 1.5m , Mx = -1450 Nm

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

S.F = Sx = +500 + 500 + 400 + 400 = + 1800 N

B.M = Mx = -500x – 500(x - 0.5) – 400(x – 1) – 400 (x – 1.5)

= -1800x + 1250

At x = 1.5m, Mx = -1450 Nm

At x = 2m , Mx = -2350 Nm