4 Shear Force and Bending Moment

DMV 4343JAN ~ JUN `07

DEPARTMENT MANUFACTURING / PRODUCT DESIGN / MOULD / TOOL AND DIE

SEMESTER 4 / 6

COURSE MECHANICS OF MATERIALS DURATION 8 hrsCOURSE CODE DMV 4343 / DMV 5343 REF. NO.VTO’S NAME MISS AFZAN BINTI ROZALI

MR RIDHWAN BIN RAMELIPAGE 18

TOPICSHEAR FORCE AND BENDING MOMENT

SUB TOPIC4.1 Types of Beams and Loadings4.2 Shear Force and Bending Moment Distribution 4.3 Relation between Distributed Load, Shear Force and Bending Moment4.4 Shear Force and Bending Moment Diagrams

Chapter 4 SHEAR FORCE AND BENDING MOMENT p0

INFORMATION SHEET

REF NO. :PAGE : 18




4.0 SHEAR FORCE AND BENDING MOMENT

Introduction

In previous chapter we have discussed the behavior of slender members subjected to axial

loading and to torsional loading. Now we turn our attention to the problem of determining the

stress distribution in, and the deflection of, beam.

A beam is a structural member that is designed to support transverse loads, that is, loads

that act perpendicular to the longitudinal axis of the beam. A beam resists applied loads by a

combination of internal transverse shear force and bending moment.

4.1 Types of Beams and Loadings

Beams

There are two classifications of beams:

Beams can be classified according to their cross-sectional shapes. For example, an I

beam and a T beam have cross sections geometrically formed like the letters I and T.

Beams of steel, aluminum, and wood are manufactured in standard sizes; their dimensions

and properties are listed in engineering handbooks. Tables B.2 through B.6 (App. B) present

several common cases of steel sections. These include wide-flange shapes (W beams),

(a) (b) (c)

FIGURE 4.1 Types of beams (a) I-shape, (b) L-shape, and (c) C-shape

I shapes (also called S beams), C shapes (also referred to as channels), and L shapes, or

angle sections. Note that the web is a thin vertical part of a beam. Thin horizontal parts of

the beam are termed flanges. Interestingly, the cross section of a beam is described as

doubly symmetric, singly symmetric, or asymmetric in accordance with whether it has two,

one, or no axis of symmetry.


web


Beams can also be classified according to the way they are supported.

Lets first describe some common types of support.

Types of common support

Roller Support—prevents displacement in the transverse (i.e., y) direction, but permits

z-rotation and displacement in the axial direction; the reaction is a force in the +y or —y

direction. The support at end A in Figure 4.2a is a roller support.

Pin Support—prevents displacement in the axial direction and in the transverse

direction, but permits z-rotation; the reaction is a force with both axial and transverse

components. The support at end B in Figure 4.2a is a pin support.

Cantilever Support (or Fixed End)—prevents displacement in the axial direction and in

the transverse direction, and also prevents z-rotation; the reaction consists of a force

with both axial and transverse components, plus a couple. The support at end C in

Figure 4.2c is a cantilever support.

The types of support are summarized in following Table 4.1.



TABLE 4-1 Reactions for types of connection



Beams are normally classified by the manner in which they are supported.

(a) Simply Supported Beam—a beam with a pin support at one end and a roller support

at the other end. The beam in Figure 4.2a is a simply supported beam. Simply supported

beams are statically determinate.

(b) Fixed end (Propped) Cantilever Beam—a beam with a cantilever support (i.e., fixed

end) at one end and a roller at the other end. The beam in Figure 4.2b is a cantilever

beam. Propped cantilever beams are statically indeterminate.

(c) Cantilever Beam—a beam with a cantilever support (i.e., fixed end) at one end and

free at the other end. The beam in Figure 4.2c is a cantilever beam. Cantilever beams

are statically determinate.

(d) Fixed, simply supported beam –The beam in Figure 4.2d.

(e) Overhanging Beam—a beam that extends beyond the support at one end (or at both

ends). The beam in Figure 4.2e would be an overhanging beam if the roller support at

end A were to be moved to the right, leaving a part of the beam to the left of the roller as

an overhang.

(f) Continuous Beam—a beam with a pin support at one end, a roller support

at the other end, and one or more intermediate roller supports. The beam in Figure

4.2f is a continuous beam. Continuous beams are statically indeterminate.

Figure 4.2: Types of beams: (a) Simply supported beam; (b) fixed-end beam; (c) cantilever

beam; (d) fixed, simply supported beam; (e) overhanging beam; and (f)

Continuous beam (two span).



Loads

External Loads:

The loads that are applied to beams may be classified as:

(a) distributed transverse loads,

(b) concentrated transverse forces, or

(c) couples.

(a) A simply supported beam with

distributed load

(b) A cantilever beam with a couple at A and a

concentrated force at B

(c) A two-span continuous beam (d) A propped cantilever beam

FIGURE 4.3 Examples of several types of beams with various types of loads.

Figure 4.3a illustrated simply supported beam with distributed loading: a cantilever beam

with a concentrated force at B and a concentrated couple at A is shown in Figure 4.3b.

Consider now the simply supported beam in Figure 4.3a. The downward distributed load

gives rise to upward reactions at the supports at A and B. The roller symbol at A implies that

the reaction force can have no horizontal component. If we pass an imaginary cutting plane

at C, as indicated in Figure 4.3a, and we draw separate free-body diagrams of AC and CB

(Figure 4.4), we see that a transverse shear force VC and a bending moment MC must act on

the cross section at C to maintain the force equilibrium and moment equilibrium of these two

adjoining free bodies. Newton's Law of action and reaction determines the relationship of the

directions of VC and MC on the two free-body diagrams.

The internal stress resultants that are associated with bending of beams are shown in Figure

4.5 and are defined by the following equations:

V(x) = - ∫A τxy (x,y) dA Stress resultants

M(x) = - ∫A yσx (x,y) dA



FIGURE 4.4 The transverse shear force VC and bending moment MC at cross section C in

Figure 4.3a

FIGURE 4.5 Definition of stress resultants—transverse shear force V(x) and bending

moment M(x).

(a) Positive V and M on section ‘x’

(b) Positive shear (c) Positive moment (d) Positive V and M

FIGURE 4.6 The sign convention for internal stress resultants V(x) and M(x)

The sign conventions for the internal stress resultants in beams are illustrated in Figure

4.6. The sign conventions may be stated in words as follows:

• A positive shear force, V, acts in the —y direction on a +x face.1

• A positive bending moment, M, makes the +y face of the beam concave.

Figures 4.6b and 4.6c illustrate the physical meaning of positive shear force and positive

bending moment, while Figure 4.6d summarizes the sign conventions for the internal stress



resultants in beams. It is very important to observe these sign conventions for V and M,

because equations will be developed to relate the stress distribution in beams and the

deflection of beams to these two stress resultants.



4.2 Shear Force and Bending Moment Distribution

To determine the stress distribution in a beam or to determine the deflected shape of a beam

under load, we need to consider equilibrium, material behavior, and geometry of

deformation. In this chapter we will concentrate on equilibrium of beams. Using the method

of sections, we will draw free-body diagrams and write equilibrium equations in order to

relate the shear-force and bending-moment stress resultants on beam cross sections to the

external loads acting on the beam. Several examples that illustrate the use of finite-length

free-body diagrams are given in this section.

EXAMPLE 4.1

Determine the support reactions for a combined

beam loaded as shown in Fig. a.

Figure a



EXAMPLE 4.2

The cantilever beam AD is subjected to a concentrated

force of 5 kN at C and a couple of 4 kN • m at D.

Determine the shear VB and bending moment MB at a

section 2 m to the right of the support A.



In the preceding example, the shear force and bending moment were required at a specific

cross section. Using a finite free body permits these values to be determined directly from

the corresponding equilibrium equations. The finite-free-body approach is also useful when

expressions for V(x) and M(x) are required over some portion of the beam. Example 4.2

illustrates this type of problem and also illustrates a way to handle relatively simple

distributed loads.

EXAMPLE 4.3

The simply supported beam AC is subjected to a

distributed downward loading as shown. The load varies

linearly between B and C.

(a) Determine the reactions at A and C,

(b) determine expressions for V(x) and M(x) for 0 < x ≤ 6

ft, and

(c) determine expressions for V(x) and M(x) for 6 ft ≤ x ≤

12 ft.



In both Example 4.2 and Example 4.3 it was possible for us to solve the equations and to

determine values of (or expressions for) shear and moment, since in each case the beam is

statically determinate. The next example illustrates the type of equilibrium results that are

obtained for statically indeterminate beams.

EXAMPLE 4.4

The propped cantilever beam AC has a couple applied at its

center B. Determine expressions for the reactions (i.e., the

shear force and bending moment) at C in terms of the

applied couple M0 and the reaction at A.



4.3 Relation between Distributed Load, Shear Force and Bending Moment

In the previous section we used finite free-body diagrams to determine values of shear force

and bending moment at specific cross sections, and to determine expressions for V(x) and

M(x) over specified ranges of x. Here we use infinitesimal free-body diagrams to obtain

equations that relate the external loading to the internal shear force and bending moment.

These expressions will be especially helpful in Section 4.4, where we discuss shear and

moment diagrams. In addition to the sign conventions for shear force and bending moment,

given in Figure 4.6, we need to adopt a sign convention for external loads (Figure 4.7).

• Positive distributed loads and positive concentrated loads act in the +y direction (e.g.,

loads p(x) and P0 in Figure 4.7).

• A positive external couple acts in a right-hand-rule sense with respect to the z axis, that

is, counterclockwise as viewed in the xy plane (e.g., the external couple M0 in Figure

4.7b).

(a) Distributed load (b) Concentrated force and couple

FIGURE 4.7 The sign convention for external loads on a beam

First, let us consider a portion of the beam where there are no concentrated external loads,

and let us establish equilibrium equations for an infinitesimal free-body diagram. Take the

segment of beam from x to (x + Δx) in Figure 4.7a, as redrawn in Figure 4.8a.

(a) General FBD (b) Shear-jump FBD (c) Moment-jump FBD

FIGURE 4.8 Infinitesimal free-body diagrams



For equilibrium of the free body in Figure 4.8a

+↑∑Fy = 0 ; V(x) + p(x)Δx + O (Δx2) – V (x + Δx) = 0

where O(…) means "of the order of," and Δp ~ Ax. Collecting terms and dividing by Δx we

get

V (x + Δx) -

V(x) = p(x) + O (Δx)

Δx

By taking the limit as Δx → 0, we get

dV= p(x) (4.2)

dx

since the limit of the O(.) term is zero. To satisfy moment equilibrium for the free-body in

Figure 4.8a, we can take moments about point C at (x + Δx).

+ (∑MC ) = 0 ;M(x) - M(x+Δx) +

p(x)

(Δx2)+ O (Δp.Δx2) + V(x)Δx = 0

2

Dividing through by Δx and taking the limit as Δx → 0, we obtain

dM= V(x) (4.3)

dx

Wherever there is an external concentrated force, such as P0 in Figure 4.7b, or a

concentrated couple, such as M0 in Figure 4.7b, there will be a step change in shear or

moment, respectively. From the partial free-body diagram in Figure 4.8b (moments have

been omitted for clarity),

+↑∑Fy = 0 ; VA- -(VA- + ΔVA) + P0 = 0

ΔVA = P0

where VA- is the (internal) shear force just to the left of the point XA where P0 is applied. That

is, a concentrated force P0 at coordinate XA will cause a step change ΔVA in shear having the

same sign as P0.

An external couple M0 at coordinate XB causes a step change in the moment at xB. From

Figure 4.8c (shear forces have been omitted for clarity),

+ (∑M)B = 0 ; MB- -(MB- + ΔMB) - M0 = 0

or



ΔMB = - M0

Equations 4.2 and 4.3 are differential equations relating the distributed load p(x) to the shear

force V(x), and the shear force V(x) to the bending moment M(x). Let x1 ≤ x ≤ x2 be a portion

of the beam that is free of concentrated forces or couples (Figure 4.9).

FIGURE 4.9 A free-body diagram of a finite portion of a beam

We can integrate Eq. 4.2 over this portion of the beam to get

∫x2 dV

dx = V1 – V2 = ∫x2

p(x) dxx1 dx x1

V1 – V2 = ∫x2

p(x) dx (4.6)x1

Similarly, from Eq. 4.3,

M1 – M2 = ∫x2

V(x) dx (4.7)x1

Equations 4.6 and 4.7 can be slated in words as follows:

• The change in shear from Section 1 to Section 2 is equal to the area under the load

curve from 1 to 2. (The "area" that results from negative p(x) is negative.)

• The change in moment from Section 1 to Section 2 is equal to the area under the

shear curve from 1 to 2. (The "area" that results from negative V(x) is negative.)

Equations 4.2 through 4.7 will he very useful to us in next Sections 4.4 where we draw shear

and moment diagrams. And we can employ modification of Eqs. 4.6 and 4.7 to determine

expressions for V(x) and M(x). Thus,

V(x) = V1 + ∫x

p( ξ ) dξ (4.8)x1

and

M(x) = M1 + ∫x

V( ξ ) dξ (4.9)x1

Equations 4.2 through 4.9 will he used extensively in Section 4.4 in constructing shear and

moment diagrams (Examples 4.7 through 4.9). Also, whenever shear and moment



expressions must be obtained for a beam with a distributed load other than a simple uniform

load or a triangular distributed load, it is much easier to use Eqs. 4.2 through 4.9 than to use

a finite free-body diagram, as is illustrated in Example 4.6.



4.4 Shear Force and Bending Moment Diagrams

In this section, shear and moment diagrams will be used to graphically represent V(x) and

M(x). In Example 4.2 we obtained expressions for V(x) and M(x) for a simply supported

beam with distributed loading. However, to design a beam (i.e., to select a beam of

appropriate material and cross section) we need to ask questions like "What is the maximum

value of the shear force, and where does it occur?" and "What is the maximum value of the

bending moment, and where does it occur?" These questions are much more readily

answered if we have a plot of V(x) and a plot of M(x). These plots are called the shear

diagram and the moment diagram, respectively. As you study the example problems in this

section, observe that maximum positive and negative bending moments can occur at any of

the following sections of a beam:

(1) a cross section where the shear force is zero (Examples and 4.4 and 4.9),

(2) a cross section where a concentrated couple is applied (Examples 4.5, 4.6 and 4.8),

(3) a cross section where a concentrated load is applied and where the shear force

changes sign (Example 4.7), and

(4) a point of support where there is a reaction force and where the shear force changes

sign (Example 4.9).

In this section two methods for constructing shear and moment diagrams are described:

• Method 1—Equilibrium Method: Use finite free-body diagrams or Eqs. 4.8 and 4.9 to

obtain shear and moment functions. V(x) and .M(x): then plot these expressions.

• Method 2—Graphical Method: Make use of Eqs. 4.2 through 4.7 to sketch V(x) and

M(x) diagrams.

The following examples illustrate both procedures. Examples 4.4 through 4.6 illustrate the

first method; Examples 4.7 through 4.9 illustrate the second approach.

Shear-Force and Bending-Moment Diagrams: Equilibrium Method.

Examples 4.4 illustrates the Equilibrium Method for constructing shear and moment

diagrams.



EXAMPLE 4.5

Figure 1 shows the simply supported

beam of Example 4.2, including the

reactions.

The expressions for V(x) and M(x)

obtained in Example 4.2 are Figure 1

V1 = (220 – 40x)lb, 0 < x ≤ 6 ft

V2 = [ - 140 + (10/3)(12 – X)2] lb, 6 ≤ x < 12 ft

M1 = (220x – 20x2) Ib • ft, 0 < x ≤ 6 ft

M2 = [140(12 - x) – (10/9)(12 –x)3] lb • ft, 6 ≤ x < 12 ft

a) Using the above expressions for V(x) and M(x), plot shear and moment diagrams for

this simply supported beam,

b) Determine the location of the section of maximum bending moment, and calculate

the value of the maximum moment.

Plan the Solution

It is straightforward to plot the shear and moment diagrams from the given expressions

(e.g., using a computer). From the shear diagram, the location of the section where V(x) =

0 can be determined. Then the appropriate moment equation can be used to determine the

value of the moment at this critical section. Since there is more load over the left half of the

beam than over the right half, the maximum moment should occur to the left of x = 6 ft.



Review the Solution

The downward load on this simply support beam bends it downward, so it is concave upward

everywhere. This is consistent with the fact that the bending moment is positive for the entire

length of the beam. As expected, since there is more load over the left half of the beam than

over the right half, the maximum moment does occur to the left of x = 6 ft. This is an

example of a maximum moment that occurs where V = dM/dx = 0.

EXAMPLE 4.6

Determine the reactions and sketch the shear and

moment diagrams for the beam shown. (This beam is

said to have an overhang BC.) Show all significant

values (that is, maxima, minima, positions of maxima

and minima, etc.) on the diagrams.



Common features of shear and moment diagram is summarized in Table 4.2 below.

TABLE 4.2 Shear Moment and Diagram Features


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4 Shear Force and Bending Moment