Project 1

Ans 1. Determine the loading on the beam indicated, analyze the beams and draw bending moment and shear force diagram. Use slope deflection method. Do not consider any load patterning. Also, do not consider any girder deflections for this analysis purpose.

GIVEN-----

Self weight of floor= `0.120*25=3(KN)/m^2`

FF & ME= 3KN/M^2

LL= 3KN/M^2

TOTAL = 11KN/M^2

Total load on the beam is

11*3=33kn/m

and self weight of the beam

0.5*0.75*25=9.375 KN/m

Then the total load =42.375 KN/m

total reaction forces

STEP -2

CALculate the fixed end moment

1. `M_(AB)^F=-(42.375*9*9)/12=-286.031KN/m`
2. `M_(BA)^F=+286.031 (KN)/m`
3. `M_(BC)^F=-286.031 KN/m`
4. `M_(CB)^F=+286.031 KN/m`
5. `M_(CD)^F=-286.031 KN/m`
6. `M_(DC)^F=+286.031 KN/m`
7. `M_(DE)^F=-286.031 KN/m`
8. `M_(ED)^F=286.031 KN/m`

STEP 3

CALCULATE THE MOMENT USING SLOP DIFLECTION METHODhe

NOW FIND THE UNKNOWNS

`theta_A, theta_B, theta_C , theta_D, theta _E`

GIVEN

E=10^7 KPa

`M_(AB)=M_(AB)^F+(2EI)/L[2theta_A+theta_B]`

`I=[B(D)^3]/12`=0.0175 M^4

EI= 1.75*10^5 KNM^2

after puting the EI value in the equvation

1. `M_(AB=-286.031+0.78*10^5theta_A+0.39*10^5theta_B`
2. `M_(BA=286.031+39000theta_B+75000theta_A`
3. `M_(BC=-286.031+75000theta_B+39000theta_C`
4. `M_(CB=286.031+39000theta_C+75000theta_B`
5. `M_(CD=-286.031+75000theta_C+39000theta_D`
6. `M_(DC=286.031+39000theta_C+75000theta_D`
7. `M_(DE=-286.031+75000theta_D+39000theta_E`
8. `M_(ED=286.031+39000theta_D+75000theta_E`

here total no of equliberium equvation is

1. `M_(AB=0`
2. `M_(BA)+M_(BC)=0`
3. `M_(CB)+M_(CD)=0`
4. `M_(DC)+M_(DE)=0`
5. `M_(ED)=0`

NOW PUTING THE VALUES IN THESE EQUVATION AND FIND THE UNKNOWNS

1. `75000theta_A+39000theta_B=286.031`
2. `150000theta_B+39000theta_C+39000theta_D=0`
3. `39000theta_B+150000theta_C+39000theta_D=0`
4. `39000theta_C+150000theta_D+39000theta_E=0`
5. `39000theta_D+75000theta_E=-286.031`

AFTER SOLVING THESE EQUVATION

1. `theta_A=0.00419`
2. `theta_B=-0.00105`
3. `theta_C=0.00`
4. `theta_D=0.00105`
5. `theta_E=-0.00419`

AFTER PUTTING THE VALUES IN THE EQUVATION 1 TO 8 WE HAVE

1. `M_(AB)=0`
2. `M_(BA)=367.54 KNm`
3. `M_(BC)=-367.54 KNm`
4. `M_(CB)=245.08 KNm`
5. `M_(CD)=-245.08KNm`
6. `M_(DC)=367.54KNm`
7. `M_(DE)=-367.54KNm`
8. `M_(ED)=0`

NOW FIND THE REACTION FORCES

• `-R_(B_1)*9+367.5+42.35*9*4.5=0`
• `R_(B_1)=231.41 KN`
• `R_A+R_(B_1)=42.35*9`
• `R_A=149.74KN`

• `-R_(C_1)-367.5+245.08+42.35*9*4.5=0`
• `R_(C_1)=176.9KN`
• `R_(B_2)+R_(C_1)=42.35*9`
• `R_(B_2)=204.25KN`

TOTAL Rb=435.66KN

SIMILARLY FIND THE ALL REACTION AND WE HAVE

Rc=353.94 KN

Rd=435.59 KN

Re=149.75 KN

SOLUTION -2

GIVEN-

DEAD LOAD OF BEAM=9.375 KN/m

ME& FF=3KN/m^2

SLAB DL= 3KN/m^2

LL= 5 KN/m^2

FOR THIS SPAN SLAB WIDTH IS 3m

SO,

ME AND FF=3*3=9 KN/m

SLAB DL= 3*3=9 KN/m

BEAM LOAD =9.375KN/m

TOTAL LOAD =27.375 KN/M

THIS IS THE COMMOM UDL FOR HOLE SPAN

AND LL= 5*3=15 KN/m

AFTER CALCULATING FEM FOR THIS BEAM

WE HAVE

1. `M_(AB)^F=-286.031 KNm`
2. `M_(BA)^F=+286.031KNm`
3. `M_(BC)^F=-184.78KNm`
4. `M_(CB)^F=+184.78KNm`
5. `M_(CD)^F=-286.031KNm`
6. `M_(DC)^F=+286.031KNm`
7. `M_(DE)^F=-184.78KNm`
8. `M_(ED)^F=+184.78KNm`

SLOP DIFLECTION METHOD

`M_(AB)=-286.031+(2*1.75*10^5)/9[2theta_A+theta_B]`

1. `M_(AB)=-286.031+75000theta_A+39000theta_B`
2. `M_(BA)=286.031+39000theta_A+75000theta_B`
3. `M_(BC)=-184.78+75000theta_B+39000theta_C`
4. `M_(CB)=184.78+39000theta_B+75000theta_C`
5. `M_(CD)=-286.031+75000theta_C+39000theta_D`
6. `M_(DC )=286.031+39000theta_C+75000theta_D`
7. `M_(DE)=-184.78+75000theta_D+39000theta_E`
8. `M_(ED)=184.75+39000theta_D+75000theta_E`

now using of the equiliberium equation

• `M_(AB)=0`
• `M_(BA)+M_(BC)=0`

`39000theta_A+150000theta_B+39000theta_C=-101.251`

• `M_(CD)+M_(CB)=0`

`39000theta_B+150000theta_C+39000theta_D=101.251`

• `M_(DC)+M_(DE)=0`

`39000theta_C+150000theta_D+39000theta_E=-101.251`

• `M_(ED)=0`

AFTER SOLVING THESE EQUATION WE HAVE

1. `theta_A=0.00473`
2. `theta_B=-0.00213`
3. `theta_C=0.00119`
4. `theta_D=-0.00003`
5. `theta_E=-0.00365`

after puting these theta values in equvations we have

1. `M_(AB)=0`
2. `M_(BA)=304.5knm`
3. `M_(BC)=-304.5knm`
4. `M_(CB)=194.53knm`
5. `M_(CD)=-194.53knm`
6. `M_(DC)=330.101knm`
7. `M_(DE)=-330.101knm`
8. `M_(ED)=0`

now calculating reaction forces we have

1. `R_A=156.875KN`
2. `R_B=378KN`
3. `R_C=289.5KN`
4. `R_D=362.7KN`
5. `R_E=86.5KN`