Calculate the reactions at the supports A B and C for the be

Calculate the reactions at the supports A, B, and C for the beam in figure 4 and then draw the shear force and bending moment diagrams. At A and B there are simple supports while at C there is a pin joint. If the cross section of the beam is rectangular, with dimensions b=10 nun and h=24 nun what is maximum bending stress in the beam?

Solution

solution:

1)here it is continuous beam hence we have to solve it for support reaction by three moment principle as follows

MaL1+2Mb(L1+L2)+McL2=6a1x1/L1+6a2x2/L2

2)here for concentrated load is acting centrally hence

L1=4 m

4m=4 m another section ratio

m=1

here moment at a,b,c are

Ma=-[3PL1/16][2+3m/3+3m]=-7.5 kNm

Mb=-[3PL1/16][2/3+3m]=-3 kNm

Mc=[3PL1/16][1/3+3m]=1.5 kNm

here for second concentrated load moment are

L2=4

4m=4

m=1

here again

Ma=[wL2^2/8][1/3+3m]=3 kNm

Mb=-[wL2^2/8][2/3+3m]=-6 kNm

Mc=[-wL2^2/8][2+3m/3+3m]=-13.5 kNm

hence resultant moment are

Ma=-7.5+3=-4.5 KNm

Mb=-3-6=-9 KNm

Mc=1.5-15=-13.5 KNm

4)hence support reaction are

Ra=Rb=12/2=6 KN

Ra\'=Mb/L1=-9/4=-2.25 KN

Ra1=Ra-Ra\'=6+2.25=8.25 KNm

Rb2=Rb+Ra\'=6-2.25=3.75 KN

6)for second span

Rb=Rc=36/2=18 KN

Rc\'=Mb/L2=-9/4=-2.25 KN

Rb2=Rb-Rc\'=18+2.25=20.25 KN

Rc=Rb+Rc\'=18-2.25=15.75 KN

hence net reaction for beam are

Ra=8.25 KN

Rb=3.75+20.25=24 KN

Rc=15.75 KN

7)now shear force diagram as follows

Ral=0 KN

Rar=8.25 KN

R12l=8.25 KN

R12r=8.25-12=-3.75 KN

Rbl=-3.75 KN

Rbr=-3.75+24=20.25 KN

Rcl=20.25-36=-15.75 KN

RCr=-15.75+15.75=0

8)bending moment diagram is

Ma=-4.5 KNm

M12=-4.5+8.25*2=12 KNm

Mbl=-4.5+8.25*4-12*2=4.5 KNm

Mbr=4.5-9=-4.5 KNm

Mcl=-9+24*4-72=15 KNm

MCr=15-15=0 KNm