Shear and bending moment diagrams depict the variations of these quantities along the length of the member. Such diagrams can be con- structed by using the method of sections described in the preceding sec- tion. Proceeding from one end of the member to the other (usually from left to right), sections are passed, after each successive change in loading, along the length of the member to determine the equations ex- pressing the shear and bending moment in terms of the distance of the section from a fixed origin. The values of shear and bending moments determined from these equations are then plotted as ordinates against the position with respect to a member end as abscissa to obtain the shear and bending moment diagrams. This procedure is illustrated by the following examples.
SECTION 5.2 Shear and Bending Moment Diagrams 167
Example 5.3
Draw the shear and bending moment diagrams for the beam shown in Fig. 5.5(a).
Solution
Reactions See Fig. 5.5(b).
þ !P
Fxẳ0 Axẳ0 þ’P
MDẳ0 Ayð9ị ỵ265ð6ị ỵ245ỵ30ð6ịð0ị ẳ0
Ayẳ203:89 kN"
þ "P Fyẳ0 203:8926530ð6ị ỵDyẳ0
Dyẳ241:11 kN"
FIG.5.5
continued
Shear Diagram To determine the equation for shear in segmentABof the beam, we pass a sectionaaat a distance xfrom supportA, as shown in Fig. 5.5(b). Considering the free body to the left of this section, we obtain
Sẳ203:89 kN for 0<x<3 m
As this equation indicates, the shear is constant at 203.89 kN from an infinitesimal distance to the right of pointAto an infinitesimal distance to the left of pointB. At pointA, the shear increases abruptly from 0 to 203.89 kN, so a vertical line is drawn from 0 to 203.89 on the shear diagram (Fig. 5.5(c)) atAto indicate this change. This is followed by a horizontal line fromAtoBto indicate that the shear remains constant in this segment.
Next, by using sectionbb(Fig. 5.5(b)), we determine the equation for shear in segmentBCas Sẳ203:89265ẳ 61:11 kN for 3 m<xa6 m
The abrupt change in shear from 203.89 kN at an infinitesimal distance to the left ofBto61.11 kN at an infinitesimal distance to the right ofBis shown on the shear diagram (Fig. 5.5(c)) by a vertical line fromþ203.89 to61.11. A horizontal line at61.11 is then drawn fromBtoCto indicate that the shear remains constant at this value throughout this segment.
To determine the equations for shear in the right half of the beam, it is convenient to use another coordinate,x1, directed to the left from the endE of the beam, as shown in Fig. 5.5(b). The equations for shear in segmentsEDand DCare obtained by considering the free bodies to the right of sectionsdd andcc, respectively. Thus,
Sẳ30x1 for 0ax1<3 m and
Sẳ30x1241:11 for 3 m<x1a6 m
These equations indicate that the shear increases linearly from zero atEtoþ90 kN at an infinitesimal distance to the right ofD; it then drops abruptly to151.11 kN at an infinitesimal distance to the left ofD; and from there it increases linearly to61.11 kN atC. This information is plotted on the shear diagram, as shown in Fig. 5.5(c). Ans.
Bending Moment Diagram Using the same sections and coordinates employed previously for computing shear, we determine the following equations for bending moment in the four segments of the beam. For segmentAB:
Mẳ203:89x for 0axa3 m For segmentBC:
Mẳ203:89x265ðx3ị ẳ 61:11xỵ795 for 3 max<6 m For segmentED:
Mẳ 30x1
x1
2 ẳ 15x12 for 0ax1a3 m For segmentDC:
Mẳ 15x12ỵ241:11ðx13ị ẳ 15x12ỵ241:11x1723:33 for 3 max1<6 m
The first two equations, for the left half of the beam, indicate that the bending moment increases linearly from 0 atAto 611.67 kN-m atB; it then decreases linearly to 428.34 kN-m atC, as shown on the bending moment diagram in Fig.
5.5(d). The last two equations for the right half of the beam are quadratic inx1. The values ofMcomputed from these equations are plotted on the bending moment diagram shown in Fig. 5.5(d). It can be seen thatMdecreases from 0 atE to135 kN-m atD, and it then increases toþ183.33 kN-m at an infinitesimal distance to the right ofC. Note that atC, the bending moment drops abruptly by an amount 428:34183:33ẳ245 kN-m, which is equal to the magnitude of the moment of the counterclockwise external couple acting at this point.
continued
SECTION 5.2 Shear and Bending Moment Diagrams 169
A point at which the bending moment is zero is termed the point of inflection. To determine the location of the point of inflectionF (Fig. 5.5(d)), we setMẳ0 in the equation for bending moment in segmentDCto obtain
Mẳ 15x12ỵ241:11x1723:33ẳ0
from whichx1ẳ3:99 m; that is, pointF is located at a distance of 3.99 m from endE, or 123:99ẳ8:01 m from
supportAof the beam, as shown in Fig. 5.5(d). Ans.
Example 5.4
Draw the shear and bending moment diagrams for the beam shown in Fig. 5.6(a).
Solution
Reactions See Fig. 5.6(b).
þ !P
Fxẳ0 Bxẳ0 þ’P
Mcẳ0 1
2 ð9ịð27ị 9
3 Byð6ị ẳ0 Byẳ60:75 kN"
þ "P Fyẳ0 1
2 ð9ịð27ị ỵ60:75ỵCyẳ0 Cyẳ60:75 kN"
Shear Diagram To determine the equations for shear in segmentsABandBCof the beam, we pass sectionsaaand bbthrough the beam, as shown in Fig. 5.6(b). Considering the free bodies to the left of these sections and realizing that the load intensity,wðxị, at a point at a distance x from endAiswðxị ẳ 279 xẳ3x kN/m, we obtain the following equations for shear in segmentsABandBC, respectively:
Sẳ 1
2 ðxịð3xị ẳ 3x2
2 for 0ax<3 m Sẳ 3x2
2
þ60:75 for 3 m<x<9 m
The values ofScomputed from these equations are plotted to obtain the shear diagram shown in Fig. 5.6(c). The point Dat which the shear is zero is obtained from the equation
Sẳ 3x2 2
ỵ60:75ẳ0
from whichxẳ6:36 m. Ans.
continued
Bending Moment Diagram Using the same sections employed previously for computing shear, we determine the following equations for bending moment in segmentsABandBC, respectively:
Mẳ 1
2 ðxịð3xị x
3 ẳ x3
2 for 0axa3 m Mẳ x3
2 ỵ60:75ðx3ị for 3 maxa9 m
continued FIG.5.6
SECTION 5.2 Shear and Bending Moment Diagrams 171
The values of M computed from these equations are plotted to obtain the bending moment diagram shown in Fig. 5.6(d). To locate the point at which the bending moment is maximum, we di¤erentiate the equation for M in segmentBCwith respect toxand set the derivativedM=dxequal to zero; that is,
dM
dx ẳ 3x2 2
ỵ60:75ẳ0
from whichxẳ6:36 m. This indicates that the maximum bending moment occurs at the same point at which the shear is zero. Also, a comparison of the expressions fordM=dxand Sin segment BCindicates that the two equations are identical; that is, the slope of the bending moment diagram at a point is equal to the shear at that point. (This rela- tionship, which is generally valid, is discussed in detail in a subsequent section.)
Finally, the magnitude of the maximum moment is determined by substitutingxẳ6:36 m into the equation forM in segmentBC:
Mmaxẳ ð6:36ị3 2
" #
ỵ60:75ð6:363ị ẳ75:5 kNm Ans.