In this article Learn :cantilever beam Bending moment diagram B.M.D. and shear force diagram S.F.D. of a cantilever beam having point load at the end,several point loads,U.D.L. Over Whole Span ,U.D.L. not over the whole span,U.D.L. from support to some distance,U.D.L. Somewhere on the beam,Combination of Point Loads and U.D.L.

BENDING MOMENT AND SHEAR FORCE DIAGRAMS OF A CANTILEVER BEAM

A shear force diagram is the graphical representation of the variation of shear force along the length of the beam and is abbreviated as S.F.D.

A bending moment diagram is the graphical representation of the variation of he bending moment along the length of the beam and is abbreviated as B.M.D.

We will take different cases of beams and loading for plotting S.F. D and B.M.D.

1. Cantilever : Point Load at the End (Fig. 3.8) At section x from the end A, Fx = – W1 and is constant for any position of the section. The S.F.D. will, therefore, be rectangle of height W. Bending moment at a section x from end A is given by

Mx = + W, x              ………(straight line)

At x=0, MA=D ; x=1, MB=WL.

The B.M.D. will thus be a triangle having zero ordinate at A and WL at B.

1. Cantilever : Several Point Loads (Fig. 3.9) S.F.D. : Between A and C,

Fx= – 3 kN (constant)

Between C and D,

Fx = -3-3=-6 kN (constant)

Between D and B,

Fx = -3-3-2=-8 kN (constant)

The S.F.D. will, therefore, consist of several rectangles having different ordinates, as shown in Fig. 3.9(b).

B.M.D. : Between A and C,

Mx = 3x                                  ….. (linear)

When x=0, MA=0. When x=2m, MC=3×2=6 kN.m

Between C and D.

Mx=3x+3(x-2)                      ……(linear)

When x=2, MC=6 kN.m (as before)

When x=4, MD=(3×4)+(3×2)=18 kN.m

Between D and B,

Mx=3x+3(x-2)+2(x-4)         …… (linear)

When x=4, MD=12+6=18 kN.m (as before)

When x=5, MB=15+9+2=26 kN.m.

The B.M.D. is shown in Fig. 3.9 (c).

1. Cantilever : U.D.L. Over Whole Span (Fig. 3.10). S.F.D. At any section x from A

Fx=-wx                                   (linear)

When x=0, FA=0 ; when x = L, FB = – wL.

The S.F.D. will be triangle, as shown in Fig. 3.10(b),

B.M.D.                                  (parabolic)

When x=0, MA=0 ; When x=L, .

The B.M.D. will be a parabola, as shown in Fig. 3.10(e).

1. Cantilever : U.D.L. not over the whole span

(a)       U.D.L. from support to some distance (Fig. 3.11). Since there is no loading from A to C, the shear force diagram will be triangle having zero ordinate at C and – 4 kN ordinate at B. The B.M.D. will be parabolic having zero ordinate at C and 2×2×1)-+4 kN.m ordinate at B

(b)       U.D.L. from the end to some distance (Fig. 3.12). S.F.D. : For AC, Fx=-20x                           (linear)

At x=0, FA=0 ; At x=2, FC=-20×2=-40 kN

For CB, Fx =-(20×2)= – 40 kN                    (constant)

The S.F.D. is shown in Fig. 3.12 (b).

B.M.D. : For AC,                     (parabolic)

At x=0, MA=0 ; At x=2, MC=40 kN.m

For CB, Mx=40 (x-1)                                   (linear)

At x=2, MC=40 kN.m ; At x=3, MB=80 kN.m

The B.M.D. is shown in Fig. 3.12(c).

(c)       U.D.L. Somewhere on the beam (Fig. 3.13). The S.F.D. will be having zero ordinate from A to C, triangular from C to D and rectangular from D to B.

The B.M.D. will be having zero ordinate from A to C, parabolic from C to D and linear from D to B.

1. Cantilever : Combination of Point Loads and U.D.L. The same principle will be followed. The S.F.D. will be rectangular between point load to point load and triangular for U.D.L. Similarly, the B.M.D. will be triangular between point loads and parabolic for U.D.L. The S.F.D. and B.M.D. have been drawn indicating principal values. The students are advised to write the proper equations and work out numerical values.