The.second.and.the.third.equations.lead.to.the.following.differential.and.
integral.relations
.
dV
dx =– ( ),q x V= −∫ q dxfor distributed loads. (4.1a) V= −Pfor concentrated loads. (4.1b)
.
dM
dx =V M, = ∫V dx. (4.2a)
ΔM = Mo for.concentrated.moments. (4.2b) Note.that.we.have.replaced.the.differential.operator.d.with.the.symbol.Δ.in.
Equation.4.1b.and.Equation.4.2b.to.signify.the.fact.that.there.will.be.a.sudden.
change.across.a.section.when.there.is.a.concentrated.load.or.a.concentrated.
moment.externally.applied.at.the.location.of.the.section.
Differentiating. Equation. 4.2a. once. with. respect. to.x and. eliminating.V.
using.Equation.4.1a,.we.arrive.at
.
d M
dx2 2 =– ,q M= ∫∫– q dxdx. (4.3) The.preceding.equations.reveal.the.following.important.features.of.shear.
and.moment.variation.along.the.length.of.a.beam.
. 1..The. shear. and. moment. change. along. the. length. of. the. beam. as. a.
function.of.x..The.shear.and.moment.functions,.V(x).and.M(x),.are.
called. shear. and. moment. diagrams,. respectively,. when. plotted.
against.x.
. 2..According.to.Equation.4.1a,.the.slope.of.the.shear.diagram.is.equal.
to.the.negative.value.of.the.intensity.of.the.distributed.load,.and.
the. integration. of. the. negative. load. intensity. function. gives. the.
shear.diagram.
. 3..According.to.Equation.4.1b,.wherever.there.is.a.concentrated.load,.
the.shear.value.changes.by.an.amount.equal.to.the.negative.value.of.
the.load.
. 4..According. to. Equation. 4.2a,. the. slope. of. the. moment. diagram. is.
equal.to.the.value.of.the.shear,.and.the.integration.of.the.shear.func- tion.gives.the.moment.diagram.
. 5..According. to. Equation. 4.2b,. wherever. there. is. a. concentrated.
moment,.the.moment.value.changes.by.an.amount.equal.to.the.value.
of.the.concentrated.moment.
. 6..According.to.Equation.4.3,.the.moment.function.and.load.intensity.
are.related.by.twice.differentiation/integration.
Furthermore,.integrating.once.from.Equation.4.1a.and.Equation.4.2a.leads.to
Vb Va q dx
a b
∫
= + − . (4.4)
and
Mb Ma V dx
a b
∫
= + . (4.5)
where.a.and.b.are.two.points.on.a.beam.
Equations. 4.4. and. 4.5. reveal. practical. guides. to. drawing. the. shear. and.
moment.diagrams:
. 1..When.drawing.a.shear.diagram.starting.from.the.leftmost.point.
on. a. beam,. the. shear. diagram. between. any. two. points. is. flat. if.
there.are.no.loads.applied.between.the.two.points.(q =.0)..If.there.
is.an.applied.load.(q ≠ 0),.the.direction.of.change.of.the.shear.dia- gram.follows.the.direction.of.the.load.and.the.rate.of.change.is.
equal.to.the.intensity.of.the.load..If.a.concentrated.load.is.encoun- tered,. the. shear. diagram,. going. from. left. to. right,. moves. up. or.
down.by.the.amount.of.the.concentrated.load.in.the.direction.of.
the.load.(Equation.4.1b)..These.practical.rules.are.illustrated.in.the.
figure below.
a b
a b
V
Vb
Va
a b
a b
V
Vb Va
q
1 q
a b
V
Vb
Va
P
a b P
Shear.diagram.rules.for.different.loads.
. 2..When.drawing.a.moment.diagram.starting.from.the.leftmost.point.
on.a.beam,.the.moment.diagram.between.any.two.points.is.(a).linear.
if. the. shear. is. constant,. (b). parabolic. if. the. shear. is. linear,. and. so.
forth..The.moment.diagram.has.a.zero.slope.at.the.point.where.the.
shear.is.zero..If.a.concentrated.moment.is.encountered,.the.moment.
diagram,.going.from.left.to.right,.moves.up.or.down.by.the.amount.
of.the.concentrated.moment.if.the.moment.is.counterclockwise.or.
clockwise.(Equation.4.2b)..These.practical.rules.are.illustrated.in.the.
following.figure.
a b
V
a b
M
a b
M
Mo
Mo
a b
a b
M
Mb Ma
Va
1 Va
a b
V
Moment.diagram.rules.for.different.shear.diagrams.and.loads.
Example 4.4
Draw the shear and moment diagrams of the loaded beam shown.
3 kN/m
2 m 3 m 3 m
6 kN
Example for shear and moment diagrams of a beam.
Solution
We shall give a detailed step-by-step solution.
1. Find reactions. The first step in shear and moment diagram con- struction is to find the reactions. Readers are encouraged to verify the reaction values shown in the following figure, which is the FBD of the beam with all the forces shown.
3 kN/m
2 m 3 m 3 m
6 kN
10 kN 2 kN
FBD of the beam showing applied and reaction forces.
2. Draw the shear diagram from left to right.
V
Linear1 m –6 kN 3 kN
2 m
V
–6 kN
4 kN Flat
10 kN
V
–6 kN –2 kN Flat 4 kN
6 kN
Drawing the shear diagram from left to right.
3. Draw the moment diagram from left to right.
–6 kN-m
–6 kN-m
–6 kN-m Parabolic M
Linear
6 kN-m
6 kN-m M
M Linear
4 kN 1 m 2 m
Drawing the moment diagram from left to right.
Example 4.5
Draw the shear and moment diagrams of the loaded beam shown.
3 kN/m
2 m 6 m
6 kN
2 m 6 kN
Example for shear and moment diagrams of a beam.
Solution
We shall draw the shear and moment diagrams directly.
1. Find reactions.
3 kN/m
2 m 3 m 3 m
6 kN
2 m 6 kN
15 kN 15 kN
FBD showing all forces.
2. Draw the shear diagram from left to right.
3 kN/m
2 m 3 m 3 m
6 kN
2 m 6 kN
15 kN
15 kN
6 kN
–6 kN 9 kN
–9 kN V
V
Drawing shear diagram from left to right.
3. Draw the moment diagram from left to right.
2 m 3 m 3 m 2 m
15 kN
13.5 kN-m M
–6 kN/m 6 kN/m
1.5 kN M
–12 kN/m –12 kN/m
Drawing moment diagram from left to right.
Example 4.6
Draw the shear and moment diagrams of the loaded beam shown next.
30 kN-m
2 m
6 kN 2 m 6 kN
3 m 3 m
Example for shear and moment diagrams of a beam.
Solution
1. Find reactions.
30 kN/m
2 m 3 m 3 m
6 kN
2 m 6 kN
5 kN 5 kN
FBD showing all forces.
2. Draw the shear diagram from left to right.
2 m 3 m 3 m
6 kN 2 m
V 6 kN
V
5 kN 5 kN
6 kN 1 kN
6 kN
Drawing shear diagram from left to right.
3. Draw the moment diagram from left to right.
2 m 3 m 3 m 2 m
30 kN-m M
6 kN/m
6 kN/m
M
1 kN/m
1 kN/m
15 kN-m 12 kN-m
–12 kN-m –15 kN-m
Drawing moment diagram from left to right.