Draft

5.3Shear& Moment Diagrams 109

1. BaseatBtheshearforcewas determinedearlierandwas equalto 2: 246 k. Based

ontheorientationof thexyaxis,thisis a negative shear.

2. TheverticalshearatBis zero(neglectingtheweight ofAB)


3. Theshearto theleftofCisV = 0 + (:749)(3)= 2 : 246 k.


4. Theshearto theright ofCisV = 2 : 246 + 5:99 = 3: 744 k

Momentdiagrams

1. At thebase:B M= 4: 493 k.ftas determinedabove.


2. At thesupportC,Mc= 4 : 493 + (:749)(3)(

3

2

) = 7 : 864 k.ft

3. Themaximummoment is equaltoMmax= 7 : 864 + (:749)(5)(

5

2

) = 1: 50 k.ft

Design: Reinforcement shouldbe placedalongthe berswhich areundertension,thatis on

thesideof thenegative moment

7

. The gurebelow schematicallyillustratesthelocation

of the
exural

8

reinforcement.

Example5-8: ShearMoment DiagramsforFrame

7

Thatis why in mostEuropeancountries,thesignconventionfordesignmoments is theoppositeof theone

commonlyusedin theU.S.A.;Reinforcement shouldbe placedwherethemoment is \postive".

8

Shearreinforcement is madeof a seriesof verticalstirrups.