The construction of shear and bending moment diagrams can be con- siderably expedited by using the basic di¤erential relationships that exist between the loads, the shears, and the bending moments.
FIG.5.7
SECTION 5.4 Relationships between Loads, Shears, and Bending Moments 173
To derive these relationships, consider a beam subjected to an arbi- trary loading, as shown in Fig. 5.8(a). All the external loads shown in this figure are assumed to be acting in their positive directions. As in- dicated in this figure, the external distributed and concentrated loads acting upward (in the positive y direction) are considered positive; the external couples acting clockwise are also considered to be positive and vice versa. Next, we consider the equilibrium of a di¤erential element of lengthdx, isolated from the beam by passing imaginary sections at dis- tancesxandxþdxfrom the originO, as shown in Fig. 5.8(a). The free- body diagram of the element is shown in Fig. 5.8(b), in whichSandM represent the shear and bending moment, respectively, acting on the left face of the element (i.e., at distancex from the origin O), and dS and dM denote the changes in shear and bending moment, respectively, over the distancedx. As the distance dxis infinitesimally small, the distrib- uted load wacting on the element can be considered to be uniform of magnitudewðxị. In order for the element to be in equilibrium, the forces and couples acting on it must satisfy the two equations of equilibrium, PFyẳ0 and P
Mẳ0. The third equilibrium equation, P
Fxẳ0, is automatically satisfied, since no horizontal forces are acting on the ele- ment. Applying the equilibrium equationP
Fy ẳ0, we obtain þ "P
Fyẳ0 Sỵw dx ðSỵdSị ẳ0
dSẳw dx (5.1)
FIG.5.8
Dividing bydx, we write Eq. (5.1) as dS
dxẳw (5.2)
in which dS=dx represents the slope of the shear diagram. Thus, Eq.
(5.2) can be expressed as
slope of shear diagram
at a point ẳ intensity of distributed
load at that point (5.3) To determine the change in shear between pointsAandBalong the axis of the member (see Fig. 5.8(a)), we integrate Eq. (5.1) from AtoB to obtain
ðB A
dSẳSBSAẳ ðB
A
w dx (5.4)
in which ðSBSAị represents the change in shear between points A and B and ÐB
Aw dx represents the area under the distributed load dia- gram between pointsAandB. Thus, Eq. (5.4) can be stated as
change in shear between
pointsAandB ẳ area under the distributed load
diagram between pointsAandB (5.5) Applying the moment equilibrium equation to the free body of the beam element shown in Fig. 5.8(b), we write
þ’P
Maẳ0 Mỵwðdxịðdx=2ị ðSỵdSịdxỵ ðMỵdMị ẳ0 By neglecting the terms containing second-order di¤erentials, we obtain
dMẳS dx (5.6)
which can also be written as dM
dx ẳS (5.7)
in which dM=dxrepresents the slope of the bending moment diagram.
Thus, Eq. (5.7) can be stated as slope of bending moment
diagram at a point ẳshear at that point (5.8) SECTION 5.4 Relationships between Loads, Shears, and Bending Moments 175
To obtain the change in bending moment between pointsAandB (see Fig. 5.8(a)), we integrate Eq. (5.6) to obtain
ðB A
dMẳMBMAẳ ðB
A
S dx (5.9)
in whichðMBMAịrepresents the change in bending moment between pointsAandBandÐB
AS dxrepresents the area under the shear diagram between pointsAandB. Thus, Eq. (5.9) can be stated as
change in bending moment
between pointsAandB ẳ area under the shear diagram
between pointsAandB (5.10)
Concentrated Loads
The relationships between the loads and shears derived thus far (Eqs.
(5.1) through (5.5)) are not valid at the point of application of concen- trated loads. As we illustrated in Example 5.3, at such a point the shear changes abruptly by an amount equal to the magnitude of the concen- trated load. To verify this relationship, we consider the equilibrium of a di¤erential element that is isolated from the beam of Fig. 5.8(a) by passing imaginary sections at infinitesimal distances to the left and to the right of the point of applicationC of the concentrated loadP. The free-body diagram of this element is shown in Fig. 5.8(c). Applying the equilibrium equationP
Fyẳ0, we obtain þ "P
Fyẳ0 SỵP ðSỵdSị ẳ0
dSẳP (5.11)
which can be stated as
change in shear at the point of
application of a concentrated load ẳ magnitude of
the load (5.12)
The relationships between the shears and bending moments (Eqs.
(5.6) through (5.10)) derived previously remain valid at the points of ap- plication of concentrated loads. Note that because of the abrupt change in the shear diagram at such a point, there will be an abrupt change in the slope of the bending moment diagram at that point.
Couples or Concentrated Moments
Although the relationships between the loads and shears derived thus far (Eqs. (5.1) through (5.5), (5.11), and (5.12)) are valid at the points of application of couples or concentrated moments, the relationships be- tween the shears and bending moments as given by Eqs. (5.6) through (5.10) are not valid at such points. As illustrated in Example 5.3, at the point of application of a couple, the bending moment changes abruptly by an amount equal to the magnitude of the moment of the couple. To derive this relationship, we consider the equilibrium of a di¤erential ele- ment that is isolated from the beam of Fig. 5.8(a) by passing imaginary sections at infinitesimal distances to the left and to the right of the point of applicationDof the coupleM. The free-body diagram of this element is shown in Fig. 5.8(d). Applying the moment equilibrium equation, we write
þ’P Maẳ0 MMỵ ðMỵdMị ẳ0
dM ẳM (5.13)
which can be stated as
change in bending moment at the
point of application of a couple ẳ magnitude of the
moment of the couple (5.14)
Procedure for Analysis
The following step-by-step procedure can be used for constructing the shear and bending moment diagrams for beams by applying the foregoing relationships between the loads, the shears, and the bending moments.
1. Calculate the support reactions.
2. Construct the shear diagram as follows:
a. Determine the shear at the left end of the beam. If no con- centrated load is applied at this point, the shear is zero at this point; go to step 2(b). Otherwise, the ordinate of the shear dia- gram at this point changes abruptly from zero to the magnitude of the concentrated force. Recall that an upward force causes the shear to increase, whereas a downward force causes the shear to decrease.
b. Proceeding from the point at which the shear was computed in the previous step toward the right along the length of the beam, identify the next point at which the numerical value of the or- dinate of the shear diagram is to be determined. Usually, it is necessary to determine such values only at the ends of the beam SECTION 5.4 Relationships between Loads, Shears, and Bending Moments 177
and at points at which the concentrated forces are applied and where the load distributions change.
c. Determine the ordinate of the shear diagram at the point se- lected in step 2(b) (or just to the left of it, if a concentrated load acts at the point) by adding algebraically the area under the load diagram between the previous point and the point currently under consideration to the shear at the previous point (or just to the right of it, if a concentrated force acts at the point). The formulas for the areas of common geometric shapes are listed in Appendix A.
d. Determine the shape of the shear diagram between the previous point and the point currently under consideration by applying Eq. (5.3), which states that the slope of the shear diagram at a point is equal to the load intensity at that point.
e. If no concentrated force is acting at the point under con- sideration, then proceed to step 2(f ). Otherwise, determine the ordinate of the shear diagram just to the right of the point by adding algebraically the magnitude of the concentrated load to the shear just to the left of the point. Thus, the shear diagram at this point changes abruptly by an amount equal to the magni- tude of the concentrated force.
f. If the point under consideration is not located at the right end of the beam, then return to step 2(b). Otherwise, the shear dia- gram has been completed. If the analysis has been carried out correctly, then the value of shear just to the right of the right end of the beam must be zero, except for the round-o¤ errors.
3. Construct the bending moment diagram as follows:
a. Determine the bending moment at the left end of the beam. If no couple is applied at this point, the bending moment is zero at this point; go to step 3(b). Otherwise, the ordinate of the bending moment diagram at this point changes abruptly from zero to the magnitude of the moment of the couple. Recall that a clockwise couple causes the bending moment to increase, whereas a counterclockwise couple causes the bending moment to decrease at its point of application.
b. Proceeding from the point at which the bending moment was computed in the previous step toward the right along the length of the beam, identify the next point at which the numerical value of the ordinate of the bending moment diagram is to be determined. It is usually necessary to determine such values only at the points where the numerical values of shear were computed in step 2, where the couples are applied, and where the maximum and minimum values of bending moment occur. In addition to the points of application of couples, the maximum and minimum values of bending moment occur at points where the shear is zero. At a point of zero shear, if the shear changes
from positive to the left to negative to the right, the slope of the bending moment diagram will change from positive to the left of the point to negative to the right of it; that is, the bending moment will be maximum at this point. Conversely, at a point of zero shear, where the shear changes from negative to the left to positive to the right, the bending moment will be minimum.
For most common loading conditions, such as concentrated loads and uniformly and linearly distributed loads, the points of zero shear can be located by considering the geometry of the shear diagram. However, for some cases of linearly distributed loads, as well as for nonlinearly distributed loads, it becomes necessary to locate the points of zero shear by solving the ex- pressions for shear, as illustrated in Example 5.4.
c. Determine the ordinate of the bending moment diagram at the point selected in step 3(b) (or just to the left of it, if a couple acts at the point) by adding algebraically the area under the shear diagram between the previous point and the point currently under consideration to the bending moment at the previous point (or just to the right of it, if a couple acts at the point).
d. Determine the shape of the bending moment diagram be- tween the previous point and the point currently under con- sideration by applying Eq. (5.8), which states that the slope of the bending moment diagram at a point is equal to the shear at that point.
e. If no couple is acting at the point under consideration, then proceed to step 3(f ). Otherwise, determine the ordinate of the bending moment diagram just to the right of the point by add- ing algebraically the magnitude of the moment of the couple to the bending moment just to the left of the point. Thus, the bending moment diagram at this point changes abruptly by an amount equal to the magnitude of the moment of the couple.
f. If the point under consideration is not located at the right end of the beam, then return to step 3(b). Otherwise, the bending moment diagram has been completed. If the analysis has been carried out correctly, then the value of bending moment just to the right of the right end of the beam must be zero, except for the round-o¤ errors.
The foregoing procedure can be used for constructing the shear and bending moment diagrams by proceeding from the left end of the beam to its right end, as is currently the common practice. However, if we wish to construct these diagrams by proceeding from the right end of the beam toward the left, the procedure essentially remains the same except that downward forces must now be considered to cause increase in shear, counterclockwise couples are now considered to cause increase in bending moment, and vice versa.
SECTION 5.4 Relationships between Loads, Shears, and Bending Moments 179
Example 5.5
Draw the shear and bending moment diagrams and the qualitative deflected shape for the beam shown in Fig. 5.9(a).
FIG.5.9
continued