Bending moment and shear force diagram of a cantilever beam

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.


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.

Similar Posts


  1. It’s really important that one is able to determine the BMD of any beam at any given moment. But it is important also to realize that most of these situations have been listed in beam diagrams that will help you determine the moment and shear force in more complicated situations.

    Most of these situations are superpositions of simpler load cases listed in the beam diagrams.

  2. Shear force refers to the force acting on a surface. Basically, the force does not tilt / angle the surface it is acting on. Within a beam, the shear force at any section is basically the algebraic sum of the lateral forces acting on either side of the section.

    On the other hand, the bending moment refers to the internal rotational moments that cause a section to bend. In the case of a beam, it can be calculated as the algebraic sum of the moments on the section of all the forces acting on each side of the section, where a falling moment will make the beam concave (positive) upwards in that section, and vice versa for a hoarding moment.

  3. All that said and done, I am not yet convinced as to why the B.M for a cantilever is positive yet it makes the beam to bend downwards (hog). Someone to help me please

    1. “Positive” refer to convention only.

      As long as you interpret the reaction correctly (as you do) and thus apply the correct calculations and placing of the’ll be fine.

Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.