Draft

5.3Shear& Moment Diagrams 107

3. If we needto determinethemaximummoment alongBC, we know that

dMBC

dx

= 0

at thepoint whereVBC = 0, thatisVBC(x) = 64: 06 3 x= 0)x=

64 : 06

3

=

25 : 0 ft. In otherwords,maximummoment occurswheretheshearis zero.

ThusM

max

BC

= 432 + 64:06(25:0) 3

(25:0)

2

2

= 432 + 1; 601 : 5 937 :5 = 232k.ft

4. FinallyalongCD, themoment variesquadratically(sincewe hada linearshear),

themoment rstincreases(positive shear),andthendecreases(negative shear).The

moment alongCDis givenby

MCD = MC+

R

x

0

VCD(x)dx= 139:8 +

R

x

0

(13: 22 3 x)dx

= 139 :8 + 13: 22 x 3

x

2

2

which is a parabola.

Substitutingforx= 15,we obtainat nodeC MC = 139:8 + 13:22(15) 3

15

2

2

=

139 :8 + 198: 3 337 :5 = 0

p

Example5-7: FrameShearandMoment Diagram;HydrostaticLoad

Theframeshownbelow is thestructuralsupportof a
ume.Assumingthattheframesare

spaced2 ft apartalongthelengthof the
ume,

1. Determineallinternalmember endactions


2. Draw theshearandmoment diagrams


3. Locateandcomputemaximuminternalbendingmoments


4. If thisis a reinforcedconcreteframe,show thelocationof thereinforcement.

Solution:

Thehydrostaticpressurecauseslateralforcesontheverticalmemberswhich canbe treated

as cantileversxedat thelower end.