Mechanics

Module VII: Analysis of Beams

Lesson 35: Shear Force and Bending Moment - II

Figure 1:

Consider a beam with a transverse loading w(x), as shown in Fig. 1. We

consider a small element of length x as shown. From force equilibrium in

the transverse direction

V (x +x) V (x) + w(x)x = 0 dVdx

= w(x) taking x 0.

Thus, the shear force distribution is obtained by integrating this differential

equation as

V (x) = V (x0)

x

x0

w(x)dx.

We observe that, if w(x) = 0, dVdx

= 0, i.e., the shear force V (x) will remain

(locally) unchanged (constant/local extremum). Thus, the slope of the SFD

reflects the load distribution on the beam.

The moment equilibrium of the small element in Fig. 1 in the limit x 0leads to

dM

dx= V (x).

Thus,

M(x) = M(x0)

x

x0

V (x)dx.

Here, we observe that, when V (x) = 0, dMdx

= 0, i.e., the bending moment

remains (locally) unchanged (constant/local extremum). The slope of the

BMD is the negative of the shear force value at a cross-section.

Combining the differential equations for the shear force and bending mo-

ment, we have

d2M

dx2= w(x).

It may be noted that the differential equations for shear force and bending

moment are all linear. In certain cases, one can use the principle of superposi-

tion to solve for V (x) and M(x) for complex loadings involving concentrated

and distributed forces on the beam.

2

Figure 2:

Problem 1

Determine the shear force and bending moment distributions and draw the

SFD and BMD of the simply supported beam loaded by the force distribution

as shown in Fig. 2.

Solution

The FBD of the beam with equivalent loading is shown in Fig. 3. The

equivalent concentrated force magnitude is easily obtained as P = qL/2,

which acts at the centroid of the distribution as shown. Using the equations

of equilibrium RA = qL/6 and RB = qL/3.

Now, for shear force distribution, we consider the FBD with the distributed

force as shown in Fig. 4. The shear force distribution is given by

dV

dx= w(x) = q

Lx V (x) = qx

2

2L+ D1

3

Figure 3:

Figure 4:

4

Now, V (0) = qL/6, implying D1 = qL/6. Hence,

V (x) =qx2

2L qL

6.

It may be noted that dVdx|x=0 = 0, and V (L/

3) = 0.

Figure 5:

The bending moment distribution is obtained as

d2M

dx2= w(x) = q

Lx

5

M(x) = qx3

6L+ C1x + C2

Now, M(0) = 0 C2 = 0, and M(L) = 0; C1 = qL/6. Hence,

M(x) = qx3

6L+

qL

6x.

The SFD and BMD are presented in Fig. 5.

Figure 6:

Problem 2

The free end of a cantilever beam is hinged to a beam with a simply supported

end and loaded, as shown in Fig. 6. Determine the shear force and bending

moment distributions, and draw the SFD and BMD.

Solution

The composite beam is an indeterminate structure. So, we consider the

FBD of the individual beams, as shown in Fig. 7. From the equations of

equilibrium, R = F = 10 kN, V0 = 42 kN, and M0 = 208 kNm. In the

6

Figure 7:

cantilever beam span (I), the shear force distribution is obtained as

dV

dx= 4 kN/m V (x) = 4x+ D1

Now, V (0) = 42 kN, implying D1 = 42 kN. Hence,

V (x) = 4x 42 kN

The bending moment distribution in (I) is obtained as

dM

dx= V (x) = 4x + 42 kN

M(x) = 2x2 + 42x+ C1

Now, M(0) = 208 kNm, implying

M(x) = 2x2 + 42x 208 kNm

In section (II), the shear force and bending moment distributions can be

obtained in a straightforward manner. The SFD and BMD are presented in

Fig. 8.

7

Figure 8:

8

Figure 9:

Problem 3

A thin and narrow 9 m long barge floating in water is loaded appropriately

to keep it horizontal, as shown in Fig. 9. Treating the barge as a beam,

determine the shear force and bending moment distributions, and draw the

SFD and BMD.

Solution

We assume that the barge is uniformly supported from below due to the

buoyancy force. The uniform support force distribution is obtained as

w =Total load

length of barge=

10

3kN/m.

The FBD of the barge is shown in Fig. 10. We divide the barge into four

sections, as shown.

Section I:

9

Figure 10:

dVIdx

= 103

kN/m VI(x) = 103

x + D1

VI(0) = 0, implies

VI(x) = 103

x kN

dMIdx

= VI(x) = 103

x kN MI(x) = 53x2 + C1 kNm

MI(0) = 0, implies

MI(x) = 53x2 kNm

Section II:

dVIIdx

= (10

3 5

)=

5

3kN/m VII(x) = 5

3x +D1

VII(5/3) = VI(5/3) = 5 kN, implying

VII(x) =5

3x 15

2kN

10

dMIIdx

= VII(x) = 53x +

15

2kN MII(x) = 5

6x2 +

15

2x + C1 kNm

MII(3/2) = MI(3/2) = 15/2 kNm, implying

MII(x) = 56x2 +

15

2x 15

8kNm

One can continue this and obtain the shear force and bending moment

distributions as shown in Fig. 11.

11

Figure 11:

12