4. I TYPES OF BEAMS -A structural member that is designed to resist forces acting laterally to its axis is called a beam. Beams differ from bars in tension and torsion primarily because of the directions of the loads that act upon them. A bar in tension is subjected to loads directed along the axis, and a bar in torsion is subjected to torques having their vectors along the axis. By contrast, the loads on a beam are directed normal to the axis, as illustrated by the force P1 act ing on beam AB of Fig. 4-1a.
In this chapter we will consider only the simplest types of beams, such as those illustrated in Fig. 4-1. These beams are planar structures because allloads act in the plane of the figure and all deftections occur in that same plane, which is called the plane of bending. The beam of Fig. 4-1a, which has a pin support at one end and a roller support at the other, is called a simply supported beam or a simple beam. The essential feature of a pin support is that it restrains the beam from translating both horizontally and vertically but does not prevent rotation. Thus, end A of the beam does not translate, but the longitudinal axis of the beam may rotate in the plane of the figure. Therefore, a pin support is capable of developing a reactive force with both horizontal and vertical components, but there will be no moment reaction. At the roller support B, translation is prevented in the vertical direction but not in the horizontal direction; hence this support can resist a vertical force but not a horizontal force. Of course, the beam axis is free to rotate at B in the same manner as at A. The vertical reactions
Fig. 4-1 Types of beams: (a) simple beam, (b) eantilever beam, and (e) beam with an overhang
4.1 Types of Beams 221
at the supports of a simple beam may act either upward or downward, as required for equilibrium.
When making a sketch of a beam, we show the types of supports by conventional symbols that indicate the manner in which the beam is restrained. These symbols also indicate the nature of the reactive forces. However, the symbols are not intended to represent the actual physical construction. For instance, the end of a beam resting on a wall and bolted down against uplift is represented in a sketch as a pin connection, but in reality there is no pin at the end of the beam.
The beam shown in Fig. 4-1 b, which is built-in or fixed at one end and free at the other, is called a cantilever beam. At the fixed (or clamped) support, the beam can neither translate nor rotate, whereas at the free end it may do both. Consequently, both force and moment reactions may exist at the fixed support. The third example in the figure is a beam with an overhang (Fig. 4-1c). This beam is simply supported at A and B, but it also projects beyond the support to point C, which is a free end.
Many other arrangements of supports for beams are possible, depending upon the particular applications. However, the examples given here and in later sections are sufficient to illustrate the basic concepts.
Loads acting on beams may be of several kinds, as illustrated in Fig. 4-1. Concentrated loads are forces such as P1 and P2 • Distributed loads act over a distance, as shown by the load q in Fig. 4-1a. Such loads are measured by their intensity, which is expressed in units of force per unit distance along the axis of the beam (for example, newtons per meter or pounds per foot). A uniformly distributed load, or uniform load, has constant intensity q per unit distance. A varying load has an intensity that changes with distance along the axis; for instance, the linearly varying load of Fig. 4-1b has an intensity that varies from ql to q2' Another kind of \oad is a couple, illustrated by the coup\e of moment M1 acting on the overhanging beam (Fig. 4-1c).
As mentioned previousiy, we assume in this discussion that the loads on the beams act in the plane of the figure. This assumption requires that all forces have their vectors in the plane of the figure and that all couples have their moment vectors perpendicular to that plane. Furthermore, the beams must be symmetric about that plane; that is, the plane of bending must be a plane of symmetry of the beam itself, which means that the cross section of each beam must have a vertical axis of symmetry. Under these conditions, the beam will deflect only in the plane of bending. However, if these conditions are not met, the beam will bend out of its plane and a more general bending analysis is required.
The beams shown in Fig. 4-1 are statically determinate and their reactions can be determined from equilibrium equations. For instance, in the case of the simple beam AB of Fig. 4-1a, we note first that the only reactions are the vertical forces Ra and Rb at the ends. (If a horizontalload were to act on the beam, a horizontal reaction would also be induced at support A.) By summing moments about point B, we can calculate the
222 CHAPTER 4 Shear Force and Bending Moment
reaction at A, and vice versa. The results are
Pl(L - a) qb 2
R - +-a-L 2L R _ Pla qb(2L - b)
b - L + 2L
An equation of equilibrium of forces in the vertical direction provides a check on these reslllts.
The cantilever beam of Fig. 4-1b is subjected to a linearly varying distributed load; hence the diagram of load intensity is trapezoidal. This load is equilibrated by a vertical force Ra and a couple Ma acting at the fixed support. From the equilibrium of forces in the vertical direction, we get
in which we have utilized the fact that the resultant of the distributed load is equal to the area of the load-intensity diagram. The moment reaction Ma is found from equilibrium of moments. It is convenient in this example to sum moments about point A in order to eliminate Ra from the moment equation. For the purpose of obtaining the moment about A of the distributed load, we divide the trapezoidal diagram into two triangles, as shown by the dashed line in Fig. 4-1 b. Then the moment about A of the lower triangular part of the load is
~(qlb)(a+~) in which q l b/2 is the resultant force (equal to the area of the lower triangular load diagram) and a + b/3 is the moment arm of the resultant. A similar procedure may be used to obtain the moment of the upper triangular portion of the load, and the final result is
This reactive moment acts counterclockwise, as shown in the figure. The beam with an overhang (Fig. 4-1c) is subjected to a vertical force
P2 and a couple of moment Ml . Taking moments about points B and A gives the following equations of static equilibrium (counterclockwise moments are positive):
Again, summation of forces in the vertical direction provides a check on the results.
r 1'" " I----x---I
b.~1 IV
I----x----i
(b)
V
.\1 (' + 1r--------1~ (e)
Fig. 4-2 Stress resultants V and M
4.2 Shear Force and Bending Moment 223
The preceding examples illustrate how the reactions (forces and couples) of statically determinate beams are ca\culated from equilibrium equations. Of course, the reactions of statically indeterminate beams cannot be found from equilibrium alone; their ca\culation requires consideration of the deflections caused by bending. These subjects are discussed in Chapters 7 and 8.
The support conditions shown in Fig. 4-1 are idealizations of actual conditions encountered in practice. Due to lack of perfect rigidity in the supporting structures or foundations, there may be a small amount of translation at a pin support or a small rotation at a fixed support. AIso, it is rare to find complete lack of restraint against horizontal translation (as assumed for a roller support); instead, a small force may deve\op due to friction and other effects. Under most conditions, especially for statically determinate beams, these minor deviations from idealized conditions ha ve little effect on the action of the beam and can be disregarded.
4.2 SHEAR FORCE ANO BENDING MOMENT -When a beam is loaded by forces or couples, internal stresses and strains are created. To determine these stresses and strains, we first must find the internal forces and internal couples that act on cross sections of the beam. As an illustration, consider a cantilever beam acted upon by a vertical force P at its free end (Fig. 4-2a). Now imagine that we cut through the beam at a cross section mn located at distance x from the free end and isolate the left-hand part of the beam as a free body (Fig. 4-2b). The free body is held in equilibrium by the force P and by the stresses that act over the cut cross section mn. These stresses represent the action of the righthand part of the beam on the left-hand part. Of course, at this stage of our discussion, we do not know the distribution of the stresses acting over the cross section; ali we know is that the resultant of these stresses must be such as to maintain equilibrium of the free body we selected.
It is convenient to reduce the resultant to a shear force V acting paralle\ to the cross section and a bending couple of moment M. Because the load P is transverse to the axis of the beam, no axial force exists at the cross section. Both the shear force and the bending couple act in the plane of the beam, which means that the moment vector for the couple is perpendicular to the plane of the figure. The moment of the bending couple is called the bending moment M. Because shear forces and bending moments, like axial forces in bars and twisting couples in shafts, are the resultants of stresses distributed over the cross section, they are known collectively as stress resultants.
The stress resultants in statically determinate beams can be calculated from equations of static equilibrium. As an example, consider again the cantilever beam of Fig. 4-2a. From the free-body diagram of
224 CHAPTER 4 Shear Force and Bending Moment
Fig. 4-3 Sign conventions for shear force V and bending moment M
(a)
(b)
Fig. 4-4 Deformations (highly exaggerated) of an element caused by: (a) shear forces, and (b) bending moments
Fig. 4-2b, we obtain
V=p M=Px
where X is the distance from the free end to section mn. Thus, through the use of a free-body dia gram and equations of equilibrium, we are able to calculate the stress resultants without difficulty. In the next chapter, we will see how to determine the internal stresses associated with V and M.
The shear force and bending moment are assumed to be positive when they act on the left-hand part of the beam in the directions shown in Fig. 4-2b. If we consider the right-hand part of the beam (Fig. 4-2c), then the directions of these same stress resultants are reversed. Therefore, we must recognize that the algebraic sign of a stress resultant does not depend upon its direction in space, such as upward or downward, or c\ockwise or counterc\ockwise, but rather the sign depends upon the direction of the stress resultant with respect to the material against which it acts. To make this point c\ear, the sign conventions for shear forces and bending moments are repeated in Fig. 4-3, where V and Mare shown acting on an element of the beam cut out between two cross sections that are a small distance apart.
The deformations of an element caused by both positive and negative shear forces and bending moments are sketched in Fig. 4-4. We see that a positive shear force tends to deform the element by causing the right-hand face to move downward with respect to the left-hand face, and a positive bending moment elongates the lower part of the beam and compresses the upper part. Because the signs for V and Mare related to the deformations of the material, these sign conventions are called deformation sign conventions. We previously used a deformation sign convention for axial forces (tension is positive, compression is negative). A different kind of sign convention, called a static sign convention, is used in equations of static equilibrium. When using a static sign convention, forces are taken as positive when they act in the positive direction of a coordinate axis.
To illustrate the two types of sign conventions, let us write equations of equilibrium for the two parts of the beam shown in Fig. 4-2. Note that V and Mare positive according to the deformation sign convention for stress resultants. However, if the y axis is positive upward, then the shear force V in Fig. 4-2b is given a negative sign in the equilibrium equation, which is written according to a static sign convention:
"LFy = O or P - V = O
Of course, V would be given a positive sign in an equation of equilibrium for the right-hand part of the beam. Thus, a positive shear force may appear in a force equilibrium equation with either a positive or a negative sign, depending upon the free-body dia gram that is considered. An analogous situation exists for bending moments when moment equilibrium equations are used. Difficulties with signs can be avoided by keeping in mind that two types of sign conventions are used in mechanics:
P
.4 ~ .\10
A I f)
R .. ţ L~ -L 4 .- L. 2
(a)
p
~ ~)\I A
R. t ,.
(b)
P
~ ~~)\I 4
R. t l'
(e)
Fig. 4-5 Example 1
8
~ tR,
4.2 Shear Force and Bending Moment 225
deformation sign conventions are used for stress resultants and static sign conventions are used in equations of equilibrium. The former are ba sed upon how the material is deformed, and the latter are based upon directions in space.
Example I
A simple beam AB supports two loads, a force P and a couple M o, acting as shown in Fig. 4-5a. Find the shear force V and bending moment M in the beam at cross sections located as follows: (a) a small distance to the left of the middle of the beam and (b) a small distance to the right of the middle of the beam.
The tirst step in the analysis of this beam is to tind the reactions Ra and Rb. Taking moments about ends A and B gives two equations of equilibrium, from which we tind
"3P M o R=--~
a 4 L P M o
R =-+b 4 L
Next, the beam is cut at a cross sectionjust to the left of the middle, and a free-body diagram is drawn of either half of the beam. In this example, we choose the lefthand half of the beam, and the corresponding diagram is shown in Fig. 4-5b. The force P and the reaction Ra appear in this diagram, along with the unknown shear force V and bending moment M, both of which are shown in their positive directions. The couple Mo does not appear in the tigure because the beam is cut to the left of the point of application of M o. A summation of forces in the vertical direction gives
Ra - P - V = O or P M o
V=----4 L
This result shows that when P and Mo act in the directions shown in Fig. 4-5a, the shear force is negative and acts in the direction opposite to that assumed in Fig. 4-5b. Taking moments about an axis through the cross section where the beam is cut (Fig. 4-5b) gives
PL M o M=---
8 2
The bending moment M may be either positive or negative, depending upon the relative magnitudes of the terms in this equation.
To obtain the stress resultants at a cross section just to the right of the middle, we cut the beam at that section and again draw a free-body diagram (Fig. 4-5c). The difference between this diagram and the former one is that the couple Mo now acts on the part of the beam to the left of the cut section. Again summing forces in the vertical direction, and also taking moments about an axis through the cut section, we obtain
P M o V= ---~ 4 L
PL M o M=-+~
8 2
We see from these results that when the cut section is shifted from left to right of the couple M o , the shear force does not change but the bending moment increases algebraically by an amount equal to M o.
A cantilever beam that is free at end A and fixed at end B is subjected to a distributed load of linearly varying intensity q (Fig. 4-6a). The maximum intensity of the load occurs at the fixed support and is denoted by qo. Find the shear force V and bending moment M at distance x from the free end.
We begin by cutting through the beam at distance x from the left-hand end and isolating part of the beam as a free body (Fig. 4-6b). As in the preced ing example, the unknown shear force V and bending moment Mare assumed to be positive. The intensity of the distributed load is q = qox/ L; therefore, the total downward load on the free body of Fig. 4-6b is equal to qox 2/2L. Equilibrium in the vertical direction gives
(a)
From this equation, we note that at the free end A (x = O) the shear force is V = O and at the fixed end B (x = L) the shear force is V = - qoL/2.
To find the bending moment in the beam, we write an equation of moment equilibrium about an axis through the cut section and sol ve for M:
(b)
Again considering the two ends of the beam, we see that the bending moment is equal to zero when x = O and equal to -qoL 2/6 when x = L. Equations (a) and (b) can be used to obtain V and M at any point in the beam, and we see from these equations that both the shear force and the bending moment reach their numerically largest values at the fixed end of the beam.
----_._~-- -----
Example 3
A beam ABC with an overhang supports a uniform load of intensity q = 6 kN/m and a concentrated load P = 28 kN (Fig. 4-7a). Calculate the shear force Vand bending moment M at a cross section D located 5 m from the left-hand support.
We begin by calculating the reactions from equations of equilibrium for the entire beam. Thus, taking moments about the support at B, we get
-Ra(8 m) + (28 kN)(5 m) + (6 kN/m)(lO m)(3 m) = O
from which Ra = 40 kN. In a similar manner, equilibrium of moments about support A yields Rb = 48 kN. (We also observe that equilibrium of forces in the vertical direction is satisfied.)
Next, we make a cut at section D and construct a free-body diagram of the left-hand part of the beam (Fig. 4-7b). When drawing this diagram, we assume that the unknown stress resuItants V and Mare positive. The equations of equilibrium for the free body are as follows:
40 kN - 28 kN - (6 kN/m)(5 m) - V = O
-(40 kN)(5 m) + (28 kN)(2 m) + (6 kN/m)(5 m)(2.5 m) + M = O
Fig. 4-7 Example 3
4.3 Relationships Between Load, Shear Force, and Bending Moment 227
(a)
28 kN
from which
V=-18kN M = 69kN·m
The minus sign in the result for V means that the shear force acts in the negative direction (opposite to the direction shown in Fig. 4-7b).
An alternative method of solution is to obtain V and M from a free-body diagram of the right-hand part of the beam (Fig. 4-7c). When drawing this diagram, we again assume that the unknown shear force and unknown bending moment are positive. Then the two equations of equilibrium are
V - (6 kN/m)(S m) + 48 kN = O
- M - (6 kN/m)(S m)(2.S m) + (48 kN)(3 m) = O
from which
V=-18kN M=69kN·m
as before.
4.3 - RELATIONSHIPS BETWEEN LOAO, SHEAR FORCE, ANO BENOING MOMENT
We will now obtain some important relationships between the loads on a beam, the shear force V, and the bending moment M. These relationships are quite useful when investigating the shear force and bending moment throughout the entire length of a beam, and they are especially helpful
228 CHAPTER 4 Shear Force and Bending Moment
'1
v
'(1 v + C/V
(a)
(b)
(e)
Fig. 4-8 Element of a beam used in deriving relationships between loads, shear forees, and bending moments
when constructing shear-force and bending-moment diagrams (see Section 4-4). As a means of obtaining the relationships, let us consider an element of a beam cut out between two cross sections that are distance dx apart (Fig. 4-8a). On the left-hand face of the element are shown the shear force V and bending moment M, acting in their positive directions. In general, V and Mare functions of the distance x measured along the axis of the beam, which means that the shear force and bending moment have values on the right-hand face of the element that are slightly different from their values on the left-hand face. If we denote the increments in V and M by dV and dM, respectively, then the corresponding stress resultants on the right-hand face are V + dV and M + dM. Again, they are shown acting in their positive directions.
The load acting on the top surface of the element may be a distributed load, a concentrated load, or a couple. Let us assume first that the load is distributed and of intensity q, as shown in Fig. 4-8a. Then, from equilibrium of forces in the vertical direction, we get
or
V - (V + dV) - q dx = O
dV -=-q dx (4-1)
The significance of this equation is as follows. The shear force V changes with the distance x measured along the axis of the beam, and Eq. (4-1) shows that the rate of change of V with respect to x is equal to - q. For example, if there is no load on part of the beam (q = O), then dV/dx = O and the shear force is constant in that part of the beam. If q is a constant over part of the beam (uniform load), then dV/dx is constant also and the shear force changes linearly in that part of the beam.
The sign conventions used with Eq. (4-1) are shown in Fig. 4-8a, where both V and q are positive. As explained in Section 4.2, a shear force is positive when it acts upward on the left face and downward on the right face of the element. We have now adopted a sign convention for distributed loads q, namely, a distributed load is positive when it acts downward and negative when it acts upward.
As an illustration of the use of Eq. (4-1), consider the cantilever beam with a linearly varying load that we discussed in Example 2 of the preceding section (see Fig. 4-6). The load on the beam is
which is positive because it acts downward. The shear force is
qox 2 V=---
2L
4.3 Relationships Between Load. Shear Force. and Bending Moment 229
(see Eq. a, Section 4.2). Taking the derivative dV/ dx gives
dV qox -=--=-q dx L
which is in accord with Eq. (4-1). Equation (4-1) can be integrated along the axis of the beam to obtain
a useful equation pertaining to the shear forces acting at two different cross sections. To obtain this relationship, we multiply both sides of Eq. (4-1) by dx and then integrate between two points A and B on the axis of the beam. The result is
f: dV = - f: q dx (a)
The left-hand side of this equation equals the difference v" - v.. of the shear forces at sections B and A. The integral on the right-hand si de represents the area of the load-intensity diagram between A and B, which is equal in magnitude to the resultant of the distributed load between A and B. Thus,
v" - v.. = - f: qdx
= -(area of load-intensity diagram between A and B) (4-2)
The area of the load-intensity diagram may be treated as either positive or negative, depending upon whether q acts downward or upward, respectively.
Because Eq. (4-1) was derived for an element of the beam subjected only to a continuously distributed load (or to no load), we cannot use Eq. (4-1) at a point where a concentrated load is applied. In a similar manner, we cannot use Eq. (4-2) if a concentrated load P acts on the beam between points A and B, because the intensity of load q is undefined for a concentrated load.
Returning to the beam element shown in Fig. 4-8a, let us now consider equilibrium by summing moments about an axis through the left-hand face of the element and perpendicular to the plane of the figure. Taking moments as positive when counterclockwise, we obtain
-M - qdx(d2X) - (V + dV)dx + M + dM = O
Discarding products of differentials because they are negligible compared to the other terms, we obtain the following relationship:
dM = V dx
(4-3)
230 CHAPTER 4 Shear Force and Bending Moment
This equation shows that the rate of change of the bending moment M with respect to x is equal to the shear force. For instance, if the shear force is zero in a region of the beam, then the bending moment is constant in that same region. Equation (4-3) applies only in regions where distributed loads act on the beam. At a point where a concentrated load acts, a sudden change (or discontinuity) in the shear force occurs and the derivative dM / dx is undefined.
Again using the cantilever beam in Fig. 4-6a as an example, we recall that the bending moment (see Eq. b, Section 4.2) is
qox 3 M=---
6L
The derivative dM / dx is
which is the same as the shear force V in the beam (see Eq. a, Section 4.2); thus, Eq. (4-3) is satisfied.
Integrating Eq. (4-3) between two points A and Bon the beam axis glves
f: dM = f: Vdx (b)
The integral on the left-hand side of this equation is equal to the difference Mb - Ma of the bending moments at points A and B. To interpret the integral on the right-hand side, we need to consider V as a function of x and to visualize a diagram showing the manner in which V varies with x. Then we see that the integral on the right represents the area below the diagram between A and B. Therefore, we can express Eq. (b) in the following manner:
Mb - Ma = f: V dx
= area of shear-force diagram between A and B (4-4)
This equation can be used even when concentrated loads are acting on the beam between points A and B. However, it is not valid if a couple acts between A and B, because a couple produces a sudden change in the bending moment and the left-hand side of Eq. (b) cannot be integrated across such a discontinuity.
Now let us consider a concentrated load P acting on the beam element (Fig. 4-8b). In order to have a sign convention for concentrated loads, we assume that a downward load is positive. As before, the stress resultants on the left-hand face are denoted by V and M. On the righthand face, they are denoted by V + VI and M + MI, where VI and MI represent the possible increments in the shear force and bending moment. From equilibrium of forces in the vertical direction, we get
(4-5)
4.4 Shear-Force and Bending-Moment Diagrams 231
which means that an abrupt change in the shear force occurs at any point where a concentrated load acts. As we pass from left to right through the point of load application, the shear force decreases by an amount equal to the magnitude of the downward load. From equilibrium of moments (Fig. 4-8b), we get
- M - P ( d;) - (V + Vd dx + M + MI = ° or MI = p( d; ) + V dx + VI dx
Since the length dx of the element is infinitesimally small, we see from this equation that the increment MI in the bending moment is also infinitesimally small. Thus, we conclude that the bending moment does not change as we pass through the point of application of a concentrated load.
Even though the bending moment M does not change at a concentrated load, its rate of change dM / dx undergoes an abrupt change. At the left-hand side of the element (Fig. 4-8b), the rate of change of the bending moment (see Eq. 4-3) is dM / dx = V. At the right-hand side, the rate of change is dM/dx = V + VI ' Therefore, we conclude that at the point of application of a concentrated load P, the rate of change dM / dx decreases abruptly by an amount equal to P.
The last case is a load in the form of a couple M o (Fig. 4-8c). The sign convention for a couple is that a couple acting as a load on a beam is positive when it is counterclockwise. From equilibrium of the element in the vertical direction, we obtain VI = 0, which shows that the shear force does not change at the point of application of a couple. Equilibrium of moments for the element gives
- M + M o - (V + VI) dx + M + MI = ° or, disregarding terms that contain differentials,
(4-6)
This equation shows that there is an abrupt decrease in the bending moment in the beam due to the applied couple M o as we move from left to right through the point of load application.
Equations (4-1) through (4-6) are useful when making a complete investigation of the shear forces and bending moments in a beam, as illustrated in the next section.
4.4 SHEAR-FORCE ANO BENOING-MOMENT OIAGRAMS -The shear forces V and bending moments M in a beam are functions of the distance x measured along the longitudinal axis. When designing a beam, it is desira bie to know the values of V and M at all cross sections.
232 CHAPTER 4 Shear Force and Bending Moment
(a)
Ph/L
l' I O ~·------------+-----,
(bJ Pa/L
Pah/L
'\1~ O ~---------------~----------~--~
(e)
Fig. 4-9 Shear-force and bendingmoment diagrams for a simple beam with a concentrated load
A convenient way to provide this information is to draw graphs showing how V and M vary with x. To plot such a graph, we take the abscissa as the position of the cross section (that is, the distance x), and we take the ordinate as the corresponding value of either the shear force or the bending moment. These graphs are called shear-force and bendingmoment diagrams.
To illustrate the construction of the diagrams, let us consider a simple beam AB carrying a concentrated load P (Fig. 4-9a). The reactions for this beam are
R = Pb a L
Pa Rb =--
L (a)
as found from equilibrium of the entire beam. Next, we cut through the beam to the left of the load P and at distance x from support A. Then we construct a free-body diagram of the left-hand part of the beam, and we find from equilibrium that
Pb V=R =
a L Pbx
M=Rx=a L (b)
These equations show that the shear force is constant from support A to the point of application of the load P and that the bending moment varies linearly with x. The expressions for V and Mare plotted directly beneath the sketch of the beam (Fig. 4-9). In the case of the shear force, the diagram begins at the end of the beam with an abruptjump in the shear force equal to the reaction Ra. Then the shear force remains constant up to x = a. In this same region, the bending-moment diagram is a straight line increasing from M = O at the support to M = Pabl L at x = a.
Next we cut through the beam to the right of the load P (that is, in the region a < x < L), and, from equilibrium of the left-hand part of the beam, we obtain the following expressions:
Pb Pa V=--P=--L L
(c)
Pbx ( x) M = L - P(x - a) = Pa 1 - L (d)
Again we see that the shear force is constant and that the bending moment is a linear function of x. At x = a, the bending moment is equal to PablL; at x = L, it is zero. Equations (c) and (d) for V and Mare plotted in Fig. 4-9.
In deriving Eqs. (c) and (d) for the shear force and bending moment to the right of the load P, we considered the equilibrium of the left-hand part of the beam, which is acted upon by the two forces Ra and P. It would have been slightly simpler in this example to consider the right-hand portion of
qL 2
,
L----fR6
(a)
v~ O ~
(bl - qL 2
qL' 8
'''~ O
(c)
Fig. 4-10 Shear-force and bendingmoment diagrams for a simple beam with a uniform load
4.4 Shear-Force and Bending-Moment Diagrams 233
the beam as a free body. From equilibrium of the right-hand part, we obtain the equations
Pa V= -Rb =--
L
which are the same results as obtained from Eqs. (c) and (d). Let us now observe certain characteristics of the diagrams in Fig. 4-9.
We note first that the slope dVl dx of the shear-force diagram is zero in the regions O < x < a and a < x < L, which is in accord with the equation dVI dx = - q (Eq. 4-1). AIso, in these same regions, the slope dM I dx of the bending moment is equal to V (Eq. 4-3). At the point of application of the load P, there is an abrupt change in the shear-force diagram (equal to P) and a corresponding change in the slope of the bending-moment diagram. To the left of the load P, the slope of the moment diagram is positive and equal to Pb IL; to the right, it is negative and equal to - Pal L.
Consider next the area of the shear-force diagram (see Eq. 4-4). As we move from x = O to x = a, the area of the shear-force diagram is (PbIL)a, or PablL. This quantity represents the increase in bending moment between these same two points. From x = a to x = L, the area of the shear-force dia gram is - Pab I L, which means that in this region the bending moment decreases by that amount. Thus, the bending moment is zero at end B of the beam, as expected. If we consider the entire shearforce diagram and if we note that M = O at both ends of the beam, then Eq. (4-4) requires that the area of the diagram between the ends of the beam be zero. This conclusion would not apply if the beam were subjected to a load in the form of a couple.
The maximum or minimum values of shear forces and bending moments are needed when design ing beams. For a simple beam with a single concentrated load, the maximum shear force occurs at the end of the beam nearest to the concentrated load, and the maximum bending moment occurs under the load itself.
To further illustrate the construction of shear-force and bendingmoment diagrams, let us consider a simple beam with a uniformly distributed load (Fig. 4-10a). In this case, each of the reactions Ra and Rb is equal to qL12; hence at a cross section at distance x from the left-hand end A, we obtain
qL V=--qx
2 qLx qx 2
M=---2 2
(e)
The first of these equations shows that the shear-force diagram consists of an inclined straight line having ordinates qLI2 and - qLI2 at x = O and x = L, respectively (Fig. 4-lOb). The slope of this line is -q, as expected from the equation dVl dx = -q (Eq. 4-1). The bending-moment diagram is a parabolic curve that is symmetric about the middle of the beam (Fig. 4-1 Oc). At each cross section, the slope of the bending-moment
234 CHAPTER 4 Shear Force and Bending Moment
PJ IlJ r ll~ r>! Il , V'
;& ~ RJ
x
tR. L
(a)
R.
vi I}P,
lip, O p,(J (b) - R.
M,
Mkt:& O
(e)
Fig. 4-11 Shear-force and bendingmoment diagrams for a simple beam with several concentrated loads
diagram is equal to the shear force (see Eq. 4-3):
dM = ~(qLX _ qX 2 ) = qL _ qx = V dx dx 2 2 2
The maximum value of the bending moment occurs at the point where dM / dx = O (that is, at the cross section where the shear force is zero). This section is at the middle of the beam in our example; hence we substitute x = L/2 into the expression for M to obtain
qL2
M max = -8-
as shown in the bending-moment diagram.
(f)
The diagram of load intensity (Fig. 4-lOa) has area qL. In accord with Eq. (4-2), the shear force V decreases by this amount as we move along the beam from A to B. The area of the shear-force diagram between x = O and x = L/2 is qL 2 /8, and this area represents the increase in the bending moment between those same two points. In a similar manner, the bending moment decreases by qL 2 /8 as we proceed from x = L/2 to x = L.
If several concentrated loads act on a simple beam (Fig. 4-11a), expressions for V and M may be determined for each region of the beam between the points of load application. Again measuring distance x from end A of the beam, we obtain for the first region of the beam (O < x < ad the following equations:
(g)
For the second region (al < x < a2), we get
(h)
For the third region of the beam (a2 < x < a3), it is advantageous to consider the right-hand part of the beam rather than the left, because fewer loads act on the corresponding free body. Hence, we obtain
(i)
and (j)
Finally, for the fourth portion of the beam, we obtain
(k)
From Eqs. (g) through (k) we see that in each region of the beam the shear force remains constant; hence the shear-force diagram has the shape shown in Fig. 4-11 b. AIso, the bending moment in each part of the beam is a linear function of x; therefore, the corresponding part of the diagram is represented by an inclined straight line. To assist in drawing these lines, we obtain the bending moments under the concentrated loads by substituting x = al' x = a2, and x = a3 into Eqs. (g), (h), and (k), respectively. In this
4.4 Shear-Force and Bending-Moment Diagrams 235
manner we obtain for the bending moments the values
(1)
From these values, we can readily construct the bending-moment diagram (Fig. 4-11c) because the diagram consists of straight lines between points of loading.
Note that, at each point where a concentrated load acts, the shearforce diagram changes abruptly by an amount equal to the load. Furthermore, at each such discontinuity in the shear force, there is a corresponding change in the slope dM / dx of the bending-moment diagram. Also, the change in bending moment between two load points equals the area of the shear-force diagram between those same two points (see Eq. 4-4). For example, the change in bending moment between loads PI and P2 is M 2 - MI. Substitut ing from Eqs. (1), we get
which is the area of the rectangular shear-force diagram between x = al andx=a2 •
The maximum bending moment in a beam with only concentrated loads must occur under one of the loads or at a reaction. From the equation dM/dx = V, we know that the slope of the bending-moment diagram is equal to the shear force. It follows that the bending moment has a maximum or minimum value at cross sections where the shear force changes sign, which occurs only under a load. If, as we proceed along the x axis, the shear force changes from a positive to a negative value (as in Fig. 4-11 b), then the slope in the bending-moment diagram also changes from positive to negative. Therefore, we must have a maximum bending moment at this cross section. Conversely, a change in shear force from a negative to a positive value indicates a minimum bending moment. The shear-force diagram can (but usually does not) intersect the horizontal axis at several points. Corresponding to each such intersection point, there is a local maximum or minimum in the bending-moment diagram. The values of all local maximums and minimums must be calculated in order to tind the absolute maximum positive and negative bending moments in the beam for use in design.
In general, the maximum positive or negative bending moments in a beam may occur at a concentrated load (provided the shear force changes sign at the load), at a reaction, at a cross section where the shear force equals zero (see Fig. 4-10), or at a section where a couple is applied. The discussions and examples in this section illustrate ali of these possibilities.
When several loads act on a beam, the shear-force and bendingmoment diagrams can be obtained by superposition (or summation) of the diagrams obtained from each of the loads acting separately. For instance, the shear-force diagram of Fig. 4-11 b is actually the sum of three separate diagrams, each of the type shown in Fig. 4-9b for a single
236 CHAPTER 4 Shear Force and Bending Moment
'/
(a)
.Wm.l"
M~ O
(e)
Fig. 4-12 Example 1
concentrated load. We can make an analogous comment for the bendingmoment diagram of Fig. 4-11c. These conclusions concern ing superposition of shear-force and bending-moment diagrams follow from the fact that shear forces and bending moments are linear functions of the applied loads.
Example I
Construct the shear-force and bending-moment diagrams for a simple beam with a uniform load of constant intensity q acting over a part of the span (Fig. 4-12a).
We begin the analysis by determining the reactions for the beam:
R = qb(c +~) • L 2
(m)
To obtain the shear forces and bending moments, we consider separately the three regions of the beam. For the left-hand part of the beam (O < x < a), we tind
V= R. M =R.x (n)
For a cross section within the loaded portion of the beam, the shear force is obtained by subtracting from the reaction R. the load q(x - a) acting on the beam to the left of the cross section. The bending moment in this same region is obtained by subtracting the moment of the load to the left of the cross section from the moment of the reaction R •. In this manner, we tind
v = R. - q(x - a)
q(x - a)2 M = R x - -'--'---___=_---'-
• 2
For the unloaded portion of the beam at the right-hand end, we get
(o)
(p)
(q)
Using Eqs. (n) through (q), we can readily construct the shear-force and bendingmoment diagrams.
The shear-force diagram (Fig. 4-12b) consists of horizontal straight li nes in the unloaded regions of the beam and an inc1ined straight line in the loaded region, as expected from the equation dVjdx = -q. The bending-moment diagram (Fig. 4-12c) consists of two inc1ined straight lines in the unloaded portions of the beam and a parabolic curve in the loaded portion. The inc1ined lines have slopes equal to R. and - Rb' respectively (see Eq. 4-3). Each of these lines is tangent to the parabolic curve at the point where it meets the curve. This conc1usion follows from the fact that there are no abrupt changes in the magnitude of the shear force at these points. Hence, from Eq. (4-3), we see that the slope of the bending-moment diagram cannot change abruptly.
The maximum bending moment Mmax occurs where the shear force equals zero.lts value can be found by setting the shear force V (from Eq. o) equal to zero, solving for x, and then substituting this value of x into the expres sion for the
Fig. 4-13 Example 2
4.4 Shear-Force and Bending-Moment Diagrams 237
bending moment (Eq. pl. The result is
qb M max = 8L2 (b + 2c)(4aL + 2bc + b2 ) (r)
This moment always occurs at a cross section within the region of the uniform load and is always positive. As a special case, consider a beam with a uniform load extending over the entire span (see Fig. 4-10). Then we substitute a = c = O and b = L into Eq. (r) and obtain Mmax = qL2/8 (see Eq. f).
Example 2
Construct the shear-force and bending-moment diagrams for the cantilever beam with two concentrated loads shown in Fig. 4-13a.
(a)
v O ... 1 _____ --,
- P,
(b)
O
(e)
Again measuring distance X from the left-hand end of the beam, and considering first the region O < x < a, we obtain
(s)
For the right-hand portion of the beam (a < x < L), we obtain
(t)
The corresponding shear-force and bending-moment diagrams are shown in Figs. 4-13b and c. The shear force is constant between the loads, and it reaches its maximum numerical value at the support. The bending-moment diagram
238 CHAPTER 4 Shear Force and Bending Moment
Fig. 4-14 Example 3
consists of two inclined straight lines, the slopes of which are equal to the shear forces in the corresponding portions of the cantilever beam. The maximum bending moment occurs at the support (x = L), and it is equal to the area of the shearforce diagram, as expected from Eq. (4-4).
Example 3
Construct shear-force and bending-moment diagrams for the beam with an overhang shown in Fig. 4-14a. The beam is subjected to a uniform load of constant intensity q = 1.0 kN/m on the overhang and a counterclockwise couple Mo = 12.0 kN·m acting midway between the supports.
q = 1.0 kN/ m
Al" ti: Mo = 12.0 kN·m
q)
. .t---L/2 = 8 m--b = 4 m Rb
L/2 = 8 m-fR, (a)
V + 1.25 (kN)
o ~'\to--(b)
M (kN·m) + 20
°'i7<1V - 10.0
(e)
We can readily calculate the reactions Rb and Re, and we find that Rb is upward and Re is downward, as shown in the figure. Their numerical values are as follows:
Rb = 5.25 kN Re = 1.25 kN
Using the techniques already described, we now draw the shear-force diagram (Fig. 4-14b). Note that the shear force does not change at the point of application of the couple Mo. The bending-moment diagram has the shape shown in Fig. 4-14c. At B, the moment is
qb 2 1 Mb = -2 = -"2(1.0 kN/m)(4 m)2 = -8.0 kN·m
which is also equal to the area of the shear-force diagram between A and B. The slope of the bending-moment diagram from B to C is 1.25 kN (that is, the slope
Fig. 4-15 Example 4
4.4 Shear-Force and Bending-Moment Diagrams 239
equals the shear force), but the bending moment changes abruptly due to the couple M o. This change is equal to Mo (see Eq. 4-6). Maximum Of minimum values of the bending moment occur where the shear force changes sign and where the couple is applied.
Example 4
A cantilever beam AB supporting a uniform load of constant intensity q is shown in Fig. 4-15a. Draw the shear-force and bending-moment diagrams for this beam.
The reactions Rb and Mb , obtained by considering the equilibrium of the
q I Mh
A cI '='=:::'='='=' :::::' :::::' =' :::' ,='='=:::'=1='=1 :::::' ::::=! ;, )
x
r--- tRI'
~------------ L--------------~·I (a)
(b)
M O __ -======---------------,
~-"L' 2
(e)
entire beam, are
(u)
The shear force V and bending moment M at any cross section located at distance x from end Acan easily be found with the aid of a free-body diagram similar to the one shown in Example 2 of Section 4.2 (see Fig. 4-6b). By this means we obtain
V= -qx (v)
Now we can plot these equations to obtain the shear-force and bending-moment diagrams (Figs. 4-15b and c). The values of V and M at support B are consistent with the values of Rb and M b previously determined (see Eqs. u). We also observe
240 CHAPTER 4 Shear Force and Bending Moment
PROBlEMS / CHAPTER"
that the slope of the shear force diagram is equal to -q and that the slope of the bending-moment diagram at any point is equal to the shear force V.
In this example it is also feasible to determine V and M by integrating the differential relationships between load, shear force, and bending moment. The shear force is obtained from Eq. (4-2), which is repeated here:
v" - v" = - f q dx (w)
For our purposes, we take A at the left-hand end of the beam where the shear force is zero (v" = O), and we take H at distance x from A. Then, from Eq. (w), we get
V = v" = - J: qdx = -qx (x)
which agrees with the previous result. The bending moment is obtained from Eq. (4-4):
Mb-Ma = f Vdx
Since the bending moment at end A is zero, we obtain
(y)
(z)
Integrating the differential reIationships is quite simple in this example because the loading pattern is continuous and there are no concentrated loads or couples. If concentrated loads or couples are present, discontinuities in the V and M diagrams will exist, and we cannot integrate Eqs. (w) and (y) across such discontinuities, as explained in Section 4.3.
4.2-1 Determine the shear force V and bending moment M at the middle of the simple beam AH shown in the figure.
I 18 kN 15 kN/ m
~ i IIIII!!!! ~ ~1.5m- 6m 3m '
Prob.4.2-1
4.2-2 Calculate the shear force V and bending moment M at a section located 0.5 m from the fixed support A of the cantilever beam shown in the figure.
4.2.J What are the shear force V and bending moment M at the midpoint of the overhanging beam shown in the figure?
~lm+lm-I- 2m Prob.4.2-2
------ b-----+--a---
Prob.4.2·3
4.2.4 Find the shear force V and bending moment M at the midpoint of the simple beam AH shown in the figure (page 241).
~ ~ ~~~I_.5_m ________ 6m ______ I_.5_m_~~
Prob.4.2-4
4.2-5 Find the shear force V and bending moment M at a section located 5 m from the left-hand end A of the beam with an overhang shown in the figure.
8kN/m 5 kN/m
;4; '10 1 B
, I ~ 3m--l_3m~2m
rTm •
2m
Prob.4.2-5
4.2.6 The beam shown in the figure is simply supported at A and B. The loads consist of a couple M o = 4 kN . m at A and a concentrated load P = 9 kN at the end of the overhang. Determine the shear force V and bending moment M at a section 2 m from the left-hand support.
Mo = 4k ·m
'" B
~ I ~ ... -------- 4 m ------1-1 m--
Prob.4.2-6
4.2-7 A lumberjack weighing 100 kg stands at the midpoint of a floating log that is 5 m long. What is the maximum bending moment in the log?
4.2·8 A simply supported beam AB supports a trapezoidal load (see figure). The intensity of the distributed load varies linearly from 50 kN/m at A to 30 kN/m at B. Calculate the shear force V and bending moment M at the midpoint of the beam.
4.2-9 A curved bar ABC is subjected to loads in the form of two equal and opposite forces P, as shown in the figure. The axis of the bar forms a semicircle of radius r. Determine the axial force N, shear force V, and bending moment M acting at a cross section defined by the angle () (see figure).
Problems / Chapter 4 241
50 kNlm
~--------2m----------~
Prob.4.2-8
~ ~N Ll/~~ :i't\
A O C A
Prob.4.2-9
4.2-10 The beam ABC shown in the figure is attached to a pin support at C and a roller support at A. A uniform load of intensity q acts on part AB and a triangular load of maximum intensity 2q acts on part Be. (a) Obtain an expres sion for the bending moment M in part AB at distance x from support A. (b) From this expression, determine the maximum bending moment Mmax in part AB.
Prob. 4.2-10
I ~l
4.2-11 A beam ABC with a vertical arm BD is supported as a simple beam at A and C (see figure on page 242). A cable passes over a small pulley that is attached to the arm at D. One end of the cable is attached to the beam at E. What is the force P in the cable if the bending moment in the beam just to the left of B is equal numerically to 7.5 kN· m?
242 CHAPTER 4 Shear Force and Bending Moment
DJ':\--___ _ -:_p ..... ~
ablc
B c
, ~2m~2m+2m~ Prob. 4.2-1 1
4.2-12 The beam ABCD is loaded by a force W = 30 kN by the arrangement shown in the figure. The cable passes over a small frictionless pulley at B and is attached at E to the vertical arm. Calculate the axial force N, shear force V, and bending moment M at section C, which is just to the left of the vertical arm.
Prob. 4.2-12
4.2-13 The beam ABCD is held in equilibrium by uniformly distributed loads of intensities q 1 and q2 as shown in the figure. Find the shear force V and bending moment M at the foIlowing cross sections of the beam: (a) cross section at point B, and (b) cross section at midpoint of the beam. (As sume a = 1 m, b = 3 m, and ql = 40 kN/m.)
ti,
Prob. 4.2-13
4.2-14 The beam ABCD shown in the figure has overhanging ends and carries a distributed load of Iinearly varying
intensity. For what ratio a/L will the shear force Valways be zero at the midpoint of the beam?
~ A~D
II-I-L~I-II Prob. 4.2-14
*4.2-15 A beam ABC with an overhang (see figure) supports a concentrated load P = 40 kN at the left-hand end and a uniform load of intensity q throughout its entire length. It is known that the shear force and bending moment at section x-x are equal to -80 kN and -480 kN·m, respectively. Find the shear force V and bending moment M at section Y-Y.
p
Prob. 4.2-15
*4.2-16 The centrifuge shown in the figure rotates in a horizontal plane (the xy plane) on a smooth surface about
w
Prob. 4.2-16
the z axis (which is vertical) with an angular acce1eration (J..
Each of the two arms has weight w per unit length and supports weight W = lOwL at its end. Derive formulas for the maximum shear force and maximum bending moment in the arms, assuming b = L/9 and c = L/lO.
When solving Problems 4.4-1 through 4.4-38, draw approximately to scale the shear-force and bending-moment diagrams and labeI ali critical ordinates, includ ing the maximum and minimum values.
4.4.1 Construct the shear-force and bending-moment diagrams for a simple beam supporting two equal concentrated loads (see figure).
r I ~ a4
1------------ L------------1~
Prob.4.4-1
4.4-2 Construct the shear-force and bending-moment diagrams for a cantilever beam carrying a uniform load of intensity q over one-half of the length (see figure).
q
t-----LI2-----t----- Ll2 --+1
Prob.4.4-2
4.4·3 A simple beam AH is subjected to a couple .of moment Mo acting at distance a from the left-hand support (see figure). Draw the shear-force and bending-moment diagrams for this beam.
(/ -------1
I------------ L----------~
Prob.4.4-3
4.4·4 The simple beam AH shown in the figure is subjected to a concentrated load P and a couple MI = P L/ 4 acting at
Problems / Chapter 4 243
the positions indicated. Draw the shear-force and bendingmoment diagrams for this beam.
PL '\/ 1 = 4
0) L 3 --~4-- L 3-1
Prob.4.4-4
L3
4.4-5 A simply supported beam AHC supports a vertical load P by means of a bracket HDE (see figure). Draw the shear-force and bending-moment diagrams for the beam.
A B C
A II v ~ t:
L4 Li4 L2
L
Prob.4.4-5
4.4-6 A simple beam AH supports a uniform load of intensity q = 6.0 kN/m over a portion of the span (see figure). Assuming that L = 10 m, a = 4 m, and b = 2 m, draw the shear-force and bending-moment diagrams for this beam.
li = 6.0 kN m
10H'H I
a = 4m
Prob.4.4-6
4.4-7 A simple beam AH subjected to two bending couples is shown in the figure. Construct the shear-force and bending-moment diagrams for this beam.
~ I
Prob.4.4-7
,\1/ 1 2M I
244 CHAPTER 4 Shear Force and Bending Moment
4.4-8 Draw the shear-force and bending-moment diagrams for the beam ABC loaded as shown in the figure.
r r fiI
~ ~ l. l~ C
~8
"~ 11_ 11 _\_11-Prob.4.4-8
4.4-9 The cantilever beam AB supports a concentrated load and a couple as shown in the figure. Construct shearforce and bending-moment diagrams for the beam.
4kN
3kN·m
-:') ~ 8 ~ A
I 2 m ----I~-- 2 m
Prob.4.4-9
4.4-10 Construct shear-force and bending-moment diagrams for the beam ABC loaded as shown in the figure. The cable passes over a small frictionless pulley at C and supports a weight W = 5.0 kN.
W = SkN
Prob. 4.4-10
4.4-11 Draw the shear-force and bending-moment diagrams for the simple beam of Problem 4.2-1.
4.4-12 Draw the shear-force and bending-moment diagrams for the cantilever beam of Problem 4.2-2.
4.4-13 Draw the shear-force and bending-moment diagrams for the overhanging beam of Problem 4.2-3.
4.4-14 Draw the shear-force and bending-moment diagrams for the simple beam shown in Problem 4.2-4.
4.4-15 Draw the shear-force and bending-moment diagrams for the beam with an overhang shown in Problem 4.2-5.
4.4-16 Draw the shear-force and bending-moment diagrams for the beam with an overhang shown in Problem 4.2-6.
4.4-17 Draw the shear-force and bending-moment diagrams for a cantilever beam with a linearly varying load of maximum intensity qo (see figure).
II"
A e=====================~
I---L---Prob. 4.4-17
4.4-18 A beam with two equal overhangs (see figure), supporting a uniform load of intensity q, has totallength L. Find the distance a between supports A and B so that the maximum bending moment in the beam has the smallest possible numerical value. Draw the shear-force and bending-moment diagrams for this condition.
q
I J • J J J J J J J J J + + + + + l I i A~ ~8 i
t----a .1 1.------- L ------~
Prob. 4.4-18
4.4-19 Draw the shear-force and bending-moment diagrams for the beam with an overhang shown in the figure on page 245.
4.4-20 through 4.4-30 Construct the shear-force and bending-moment diagrams for the beams shown in the figures on pages 245-6.
1 I • c
Prob. 4.4-19
-l k 101 5 kN I ~~=tJ=J=J=J='='=J =I ======~ţ 8 IA 2m-~~--201--~
Prob. 4.4-20
~--2m-~~--2m--~
Prob. 4.4-21
</0
Prob. 4.4-22
30 kN/ 01
Prob. 4.4-23
Problems / Chapter 4 245
'I .. tiu
Prob. 4.4-24
JP = 150
+ q = 900 /01
IlOllll O,B B
~ ~ I L/ 2= 1 m-I- L/2 = Im i
Prob. 4.4-25
Sk i In IOk
,Abf ' , I _-..r~ ~ 1 ni" .
IOk
201 201~-2m-l-lm Prob. 4.4-26
5kN/ 01
Prob. 4.4-27
1.201
Prob. 4.4-28
246 CHAPTER 4 Shear Force and Bending Moment
ISk /m 1 45kN
1 1++,+,++++ 1 B t ~i=:::::::::::::::::::::::::::::::::::::::::::::~=:;:::==' c
I ~ 1- -- 6m - -2m 2m
Prob. 4.4-29
'1 = 3 kN 111
3 111 3m
Prob. 4.4-30
4.4-31 Beam ABCD has two equal overhangs and supports a symmetric triangular load (see figure). Determine the distance b between the supports so that the bending moment at the midpoint of the beam is zero. Draw the shear-force and bending-moment diagrams for this condition.
{f"
/. %
f-h-I I---- - --L
Prob. 4.4-31
4.4-32 The beam ABC shown in the figure consists of a cantilever part AB attached to a simple span BC by a pin at B. The pin can transmit a shear force but not a bending
Prob. 4.4-32
moment. Draw the shear-force and bending-moment diagrams for the beam.
4.4·33 The beam ABCD supports a distributed load of Iinearly varying intensity (see figure). Draw the shear-force and bending-moment diagrams for this beam.
'1"
A <--__ ..--___________ ~--..J D
l U4'--+-I, -L-----'"-~ J Prob. 4.4-33
4.4-34 The beam ABCDE shown in the figure has simple supports at A, C, and E and a hinge (or pin) at D. A load of 4 kN acts at the end of the bracket that extends from the beam at B, and a load of 2 kN acts at the midpoint of part DE. Draw the shear-force and bending-moment diagrams for the beam. (Note that the pin at D can transmit a shear force but not a bending moment.)
Prob. 4.4-34
4.4·35 The shear-force diagram for a simple beam is shown in the figure. Determine the loading on the beam and draw
+ 20
(~N)~ O~--------------------~~-----r----~
- 4m I - 20 -2 m--2m-!
Prob. 4.4-35
the bending-moment diagram, assuming that no couples act as loads on the beam. (Note that the shear force has units of kilonewtons.)
4.4-36 The shear-force diagram for a beam is shown in the figure. Assuming that no couples act as loads on the beam, draw the bending-moment diagram. (Note that the shear force has units of kilonewtons.)
+22.0
+ 17.5 V (kN)
+ 1.5 O
-1 6.5
---'!I.sm -11----- 6.0 m 1.5 m
Prob. 4.4-36
4.4·37 Construct the shear-force and bending-moment diagrams for a simple beam with a triangular load (maximum intensity qo) that extends for three-fourths of the span length L (see figure).
~qB"n
;A .....----1 _1 -L 3U4 . : I
Prob. 4.4-37
4.4·38 A beam AB is subjected to concentrated loads P and 2P as shown in the figure. The beam rests on a foundation
~l-~----- L----_2_~l· U41 A 8
Prob. 4.4-38
Problems / Chapter 4 247
that exerts a continuously distributed reaction against the beam. Assuming that the distributed reaction varies Iinearly in intensity from A to B, determine the intensities q. and qb of the reaction at the ends A and B, respectively. Also, construct the shear-force and bending-moment diagrams for the beam.
4.4-39 Two equalloads Pare separated by a fixed distance d (see figure). This load combination may be located at any distance x from the left-hand support of the simple beam AB. (a) For what distances x will the shear force in the beam be a maximum? Also, derive a formula for the maximum shear force Vmax • (b) Derive a formula for the distance x so as to produce the maximum bending moment Mmax in the beam; also, obtain an expression for Mmax .
x
~ --------- L--------~· I
Prob. 4.4-39
4.4-40 A simple beam AB carries two connected wheel loads P and 2P that are distance d apart (see figure). The loads may be placed at any distance x from the left-hand support of the beam. Determine the distance x for (a) maximum shear force in the beam and (b) maximum bending moment in the beam if P = 6 kN, d = 1.6 m, and L = 8 m. Also, determine the maximum shear force Vmax
and maximum bending moment Mmax .
P 2P
I ,~-~-d-A
~ ;& 1 L I
Prob. 4.4-40
4.4·41 Three wheelloads W" W2 , and W3 move across a simply supported beam as shown in the figure on page 248. Determine the position of the wheels, as defined by the distance x from end A, so as to produce the maximum bending moment in the beam, assuming that W1 = 20 kN and W2 = W3 = 80 kN. Also, determine the maximum bending moment Mmax .
248 CHAPTER 4 Shear Force and Bending Moment
ii -----24 m ---- -.11
Prob. 4.4-41
4.4-42 A small retaining wall 2.4 m high is constructed of vertical wood planks ABC, as shown in the figure. The planks are 125 mm wide and are simply supported at B and C by a horizontal steel beam and a concrete foundation wall, respectively. The steel beam is 0.6 m from the top of the wall. The soil pressure against the wall is assumed to vary linearly from zero at the top (point A) to p = 18 kPa at the bottom (point C). Calculate the maximum bending moment Mmax in the planks.
Wood plank.
Slcel beam
* 4.4-43 The loads on a simple beam consist of n equally spaced forces (see figure). The total applied load is P; therefore, each force is equal to Pin. The length of the beam is L; hence the spacing between loads is LI(n + 1). (a) Derive general formulas for the maximum bending moment in the beam. (b) From these formulas, determine the
maximum bending moment for several successive values of n (n = 1,2,3,4, ... ). (c) Compare these results with the maximum bending moment in the beam due to a uniformly distributed load of intensity q such that qL = P.
~-J I ~ ~I!_~--I_~_I_~_I--I--L "· I~~
Prob. 4.4-43
Top vicw
Sidc vicw
Prob. 4.4-42
*4.4-44 A beam ABC of length L supports a triangularly distributed load of maximum intensity qo as shown in the figure. The beam is simply supported at points B and C, with Cat the right-hand end and B at any desired distance a
from the left-hand end. (a) Determine the ratio a/L so that the numerically largest bending moment in the beam is a minimum. (b) What is the maximum bending moment in the beam under this condition?
