Project 1

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.

Solution:

`Lundersety`/`Lundersetx`=9/3 =3 which is greater than 2 so it is a one way slab.

Thickness of slab= 0.120m

Self weight of the slab= 25x0.120 =3 KN/m2

Live load= 5 KN/m2

Floor Finish= 3 KN/m2

Total Load= 3+5+3= 11 KN/m2

Load from Slab to Beam= 11x9x1.5x2 

                                   = 297 KN

Converting it to UDL load= 297/9 = 33 KN/m

Self Weight of the beam= 25x0.5x0.75

                                 = 9.375 KN/m

Total Load on the beam= 33+9.375

                                  = 42.375 KN/m

Calculation of Fixed End Moments

`Munderset(FAB)`= `-(wL^2)/12`= `-(42.375*9^2)/12`= -286.031 KN-m

`Munderset(FBA)`= `(wL^2)/12`= `(42.375*9^2)/12`= 286.031 KN-m

`Munderset(FBC)`= `-(wL^2)/12`= `-(42.375*9^2)/12`= -286.031 KN-m

`Munderset(FCB)`= `(wL^2)/12`= `(42.375*9^2)/12`= 286.031 KN-m

`Munderset(FCD)`=`-(wL^2)/12`= `-(42.375*9^2)/12`= -286.031 KN-m

`Munderset(FDC)`= `(wL^2)/12`= `(42.375*9^2)/12`= 286.031 KN-m

`Munderset(FDE)`= `-(wL^2)/12`= `-(42.375*9^2)/12`= -286.031 KN-m

`Munderset(FED)`= `(wL^2)/12`= `(42.375*9^2)/12`= 286.031 KN-m

Calculation of Support End Moments

 `Munderset(AB)`= `Munderset(FAB)`+`4EITunderseta``/L`+ `2EITundersetb``/L`

0= -286.031+ `4EITunderseta``/9`+ `2EITundersetb``/9`

 286.031- `2EITundersetb``/9`= `4EITunderseta``/9`

 `2EITunderseta``/9`= 143.0155 - `EITundersetb``/9`

`Munderset(BA)`= `Munderset(FBA)`+ `4EITundersetB``/L`+ `2EITundersetA``/L`

 `Munderset(BA)`= 286.031+ `4EITundersetB``/9`+ 143.0155 - `EITundersetb``/9`

 `Munderset(BA)`= 429.0465+ `3EITundersetB``/9`

`Munderset(BC)`= `Munderset(FBC)`+ `4EITundersetB``/L`+ `2EITundersetC``/L`

 `Munderset(BC)`= -286.031+ `4EITundersetB``/9`+ `2EITundersetC``/9`

`Munderset(CB)`= `Munderset(FCB)`+ `4EITundersetC``/L`+ `2EITundersetB``/L`

`Munderset(CB)`= 286.031+ `4EITundersetC``/9`+ `2EITundersetB``/9`

 `Munderset(CD)`= `Munderset(FCD)`+ `4EITundersetC``/L`+ `2EITundersetD``/L`

`Munderset(CD)`= -286.031+ `4EITundersetC``/9`+ `2EITundersetD``/9`

`Munderset(DC)`= `Munderset(FDC)`+ `4EITundersetD``/L`+ `2EITundersetC``/L`

`Munderset(DC)`= 286.031+ `4EITundersetD``/9`+ `2EITundersetC``/9`

`Munderset(DE)`= `Munderset(FDE)`+ `4EITundersetD``/L`+ `2EITundersetE``/L`

`Munderset(DE)`= -286.031+ `4EITundersetD``/9`+ `2EITundersetE``/9`

`Munderset(ED)`= `Munderset(FED)`+ `4EITundersetE``/L`+ `2EITundersetD``/L`

0= 286.031+ `4EITundersetE``/9`+ `2EITundersetD``/9`

`2EITundersetE``/9`= -143.0155`- `EITundersetD``/9`

At Joint B

`Munderset(BA)`+  `Munderset(BC)`= 0

429.0465+ `3EITundersetB``/9`-286.031+ `4EITundersetB``/9`+ `2EITundersetC``/9`=0

`7EITundersetB``/9`+`2EITundersetC``/9`= -143.0155

At Joint C

`Munderset(CB)`+ `Munderset(CD)`= 0

286.031+ `4EITundersetC``/9`+ `2EITundersetB``/9`-286.031+ `4EITundersetC``/9`+ `2EITundersetD``/9`=0 

`2EITundersetB``/9`+ `8EITundersetC``/9`+ `2EITundersetD``/9`=0

At Joint D

`Munderset(DC)`+ `Munderset(DE)`= 0

286.01+`4EITundersetd``/9`+`2EITundersetc`/9-286.031+`4EITundersetd``/9`-143.0155-`EITundersetd``/9`=0

 `2EITundersetC``/9`+`7EITundersetD``/9`= 143.0155

Solving the above Three equations we get,

`EITundersetD`= 183.8770

`EITundersetC`= 0

`EITundersetB`= -183.8770

`EITundersetE`= -735.5058

`EITundersetA`= 735.5058

Substituting the above values we get:

 `Munderset(AB)`=0

 `Munderset(BA)`= 367.754 KN-m

 `Munderset(BC)`= -367.754 KN-m

 `Munderset(CB)`= 245.169 KN-m

 `Munderset(CD)`= -245.169 KN-m

 `Munderset(DC)`= 367.757 KN-m

 `Munderset(DE)`= -367.757 KN-m

 `Munderset(ED)`= 0

2. With Load Patterning

Thickness of slab= 0.120m

Self weight of the slab= 25x0.120 =3 KN/m2

Live load= 5 KN/m2

Floor Finish= 3 KN/m2

Total Load= 3+5+3= 11 KN/m2

Load from Slab to Beam= 11x9x1.5x2 

                                   = 297 KN

Converting it to UDL load= 297/9 = 33 KN/m

Self Wight of the beam= 25x0.5x0.75

                                 = 9.375 KN/m

Total Load on the beam= 33+9.375

                                  = 42.375 KN/m

Calculation for Spans without Live Load

Thickness of slab= 0.120m

Self weight of the slab= 25x0.120 =3 KN/m2

Floor Finish= 3 KN/m2

Total Load= 3+3= 6 KN/m2

Load from Slab to Beam= 6x9x1.5x2 

                                   = 162 KN

Converting it to UDL load= 162/9 = 18 KN/m

Self Weight of the beam= 25x0.5x0.75

                                 = 9.375 KN/m

Total Load on the beam= 18+9.375

                                  = 27.375 KN/m

 

Calculation For Maximum Bending Moment at D

We calculate the Fixed End Moments for New Loading

`Munderset(FAB)`= `-(wL^2)/12`= `-(42.375*9^2)/12`= -286.031 KN-m

`Munderset(FBA)`= `(wL^2)/12`= `(42.375*9^2)/12`= 286.031 KN-m

`Munderset(FBC)`= `-(wL^2)/12`= `-(27.375*9^2)/12`= -184.78 KN-m

`Munderset(FCB)`= `(wL^2)/12`= `(27.375*9^2)/12`= 184.78 KN-m

`Munderset(FCD)`=`-(wL^2)/12`= `-(42.375*9^2)/12`= -286.031 KN-m

`Munderset(FDC)`= `(wL^2)/12`= `(42.375*9^2)/12`= 286.031 KN-m

`Munderset(FDE)`= `-(wL^2)/12`= `-(42.375*9^2)/12`= -286.031 KN-m

`Munderset(FED)`= `(wL^2)/12`= `(42.375*9^2)/12`= 286.031 KN-m

Calculation For the Support End Moments

`Munderset(AB)`= `Munderset(FAB)`+`4EITunderseta``/L`+ `2EITundersetb``/L`

0= -286.031+ `4EITunderseta``/9`+ `2EITundersetb``/9`

 286.031- `2EITundersetb``/9`= `4EITunderseta``/9`

 `2EITunderseta``/9`= 143.0155 - `EITundersetb``/9`

`Munderset(BA)`= `Munderset(FBA)`+ `4EITundersetB``/L`+ `2EITundersetA``/L`

 `Munderset(BA)`= 286.031+ `4EITundersetB``/9`+ 143.0155 - `EITundersetb``/9`

 `Munderset(BA)`= 429.0465+ `3EITundersetB``/9`

`Munderset(BC)`= `Munderset(FBC)`+ `4EITundersetB``/L`+ `2EITundersetC``/L`

 `Munderset(BC)`= -184.78+ `4EITundersetB``/9`+ `2EITundersetC``/9`

`Munderset(CB)`= `Munderset(FCB)`+ `4EITundersetC``/L`+ `2EITundersetB``/L`

`Munderset(CB)`= 184.78+ `4EITundersetC``/9`+ `2EITundersetB``/9`

 `Munderset(CD)`= `Munderset(FCD)`+ `4EITundersetC``/L`+ `2EITundersetD``/L`

`Munderset(CD)`= -286.031+ `4EITundersetC``/9`+ `2EITundersetD``/9`

`Munderset(DC)`= `Munderset(FDC)`+ `4EITundersetD``/L`+ `2EITundersetC``/L`

`Munderset(DC)`= 286.031+ `4EITundersetD``/9`+ `2EITundersetC``/9`

`Munderset(DE)`= `Munderset(FDE)`+ `4EITundersetD``/L`+ `2EITundersetE``/L`

`Munderset(DE)`= -286.031+ `4EITundersetD``/9`+ `2EITundersetE``/9`

`Munderset(ED)`= `Munderset(FED)`+ `4EITundersetE``/L`+ `2EITundersetD``/L`

0= 286.031+ `4EITundersetE``/9`+ `2EITundersetD``/9`

`2EITundersetE``/9`= -143.0155`- `EITundersetD``/9`

At Joint B

`Munderset(BA)`+  `Munderset(BC)`= 0

429.0465+ `3EITundersetB``/9`-184.78+ `4EITundersetB``/9`+ `2EITundersetC``/9`=0

`7EITundersetB``/9`+ `2EITundersetC``/9`= -244.266

At Joint C

`Munderset(CB)`+ `Munderset(CD)`= 0

184.78+ `4EITundersetC``/9`+ `2EITundersetB``/9`-286.031+ `4EITundersetC``/9`+ `2EITundersetD``/9`=0

`2EITundersetB``/9`+ `8EITundersetC``/9`+`2EITundersetD``/9`= 101.251

At Joint D

`Munderset(DC)`+ `Munderset(DE)`= 0

286.031+ `4EITundersetD``/9`+ `2EITundersetC``/9`-286.031+ `4EITundersetD``/9`+ `2EITundersetE``/9`=0

286.031+ `4EITundersetD``/9`+ `2EITundersetC``/9`-286.031+ `4EITundersetD``/9`-143.0155`- `EITundersetD``/9`=0

`2EITundersetC``/9`+ `7EITundersetD``/9`= 143.0155

Solving the above three Joint Equations:

`EITundersetB`= -362.873

`EITundersetC`= 170.861

`EITundersetD`= 135.058

Now Substituting the above values we get

`Munderset(DC)`= 384.025 KN-m

`Munderset(DE)`= -384.025 KN-m

Calculation Max Bending Moment at B

Since the Loading and structure is symmetrical so we can say that

`Munderset(BA)`= 384.025 KN-m

`Munderset(BC)`=-384.025 KN-m

Calculation for Maximum Bending Moment At C

We calculate the Fixed End Moments for New Loading

`Munderset(FAB)`= `-(wL^2)/12`= `-(27.375*9^2)/12`= -184.781 KN-m`

`Munderset(FBA)`= `(wL^2)/12`= `(42.375*9^2)/12`= 184.781 KN-m

`Munderset(FBC)`= `-(wL^2)/12`= `-(42.375*9^2)/12`=-286.031 KN-m

`Munderset(FCB)`= `(wL^2)/12`= `(42.375*9^2)/12`= 286.031 KN-m

`Munderset(FCD)`=`-(wL^2)/12`= `-(42.375*9^2)/12`= -286.031 KN-m

`Munderset(FDC)`= `(wL^2)/12`= `(42.375*9^2)/12`= 286.031 KN-m

`Munderset(FDE)`= `-(wL^2)/12`= `-(27.375*9^2)/12`=-184.781 KN-m

`Munderset(FED)`= `(wL^2)/12`= `(27.375*9^2)/12`= 184.781 KN-m

Calculation For Support End Moments:

`Munderset(AB)`= `Munderset(FAB)`+`4EITunderseta``/L`+ `2EITundersetb``/L`

0= -184.781+ `4EITunderseta``/9`+ `2EITundersetb``/9`

`2EITunderseta``/9`= 92.3905-`EITundersetb``/9`

`Munderset(BA)`= `Munderset(FBA)`+ `4EITundersetB``/L`+ `2EITundersetA``/L`

`Munderset(BA)`= 184.781+ `4EITundersetB``/9`+ 92.3905-`EITundersetb``/9`

`Munderset(BA)`= 277.172+`3EITundersetB``/9`

`Munderset(BC)`= `Munderset(FBC)`+ `4EITundersetB``/L`+ `2EITundersetC``/L`

`Munderset(BC)`=-286.031+ `4EITundersetB``/9`+ `2EITundersetC``/9`

`Munderset(CB)`= `Munderset(FCB)`+ `4EITundersetC``/L`+ `2EITundersetB``/L`

`Munderset(CB)`= 286.031+`4EITundersetC``/9`+ `2EITundersetB``/9`

`Munderset(CD)`= `Munderset(FCD)`+ `4EITundersetC``/L`+ `2EITundersetD``/L`

`Munderset(CD)`= -286.031+`4EITundersetC``/9`+ `2EITundersetD``/9`

`Munderset(DC)`= `Munderset(FDC)`+ `4EITundersetD``/L`+ `2EITundersetC``/L`

`Munderset(DC)`= 286.031+ `4EITundersetD``/9`+ `2EITundersetC``/9`

`Munderset(DE)`= `Munderset(FDE)`+ `4EITundersetD``/L`+ `2EITundersetE``/L`

`Munderset(DE)`= -184.781+ `4EITundersetD``/9`+ `2EITundersetE``/9`

`Munderset(ED)`= `Munderset(FED)`+ `4EITundersetE``/L`+ `2EITundersetD``/L`

0= 184.781+ `4EITundersetE``/9`+ `2EITundersetD``/9`

`2EITundersetE``/9`= -92.3905`- `EITundersetD``/9`=0

At Joint B

`Munderset(BA)`+  `Munderset(BC)`= 0

277.172+`3EITundersetB``/9`+ -286.031+ `4EITundersetB``/9`+ `2EITundersetC``/9`=0

`7EITundersetB``/9`+ `2EITundersetC``/9`= 8.859

At Joint C

`Munderset(CB)`+ `Munderset(CD)`= 0

286.031+`4EITundersetC``/9`+ `2EITundersetB``/9`-286.031+`4EITundersetC``/9`+ `2EITundersetD``/9`=0

`2EITundersetB``/9`+ `4EITundersetC``/9`+ `2EITundersetD``/9`=0

At Joint D

`Munderset(DC)`+ `Munderset(DE)`= 0

286.031+ `4EITundersetD``/9`+ `2EITundersetC``/9`-184.781+ `4EITundersetD``/9`+ `2EITundersetE``/9`=0

286.031+ `4EITundersetD``/9`+ `2EITundersetC``/9`-184.781+ `4EITundersetD``/9`-92.3905`- `EITundersetD``/9`=0

`2EITundersetC``/9`+  `7EITundersetD``/9`= -8.86

Solving the above Joint Equations:

`EITundersetB`= 11.244

`EITundersetC`= 0.5109

`EITundersetD`=- 13.288

Substituting the above values we get

`Munderset(CB)`= 288.757 KN-m

`Munderset(DE)`= -288.757 KN-m

Calculation For Bending Moment at Mid Span of Beam

Calculation For Fixed End Moments

`Munderset(FAB)`= `-(wL^2)/12`= `-(42.375*9^2)/12`= -286.031 KN-m

`Munderset(FBA)`= `(wL^2)/12`= `(42.375*9^2)/12`= 286.031 KN-m

`Munderset(FBC)`= `-(wL^2)/12`= `-(27.375*9^2)/12`= -184.78 KN-m

`Munderset(FCB)`= `(wL^2)/12`= `(27.375*9^2)/12`= 184.78 KN-m

`Munderset(FCD)`=`-(wL^2)/12`= `-(42.375*9^2)/12`= -286.031 KN-m

`Munderset(FDC)`= `(wL^2)/12`= `(42.375*9^2)/12`= 286.031 KN-m

`Munderset(FDE)`= `-(wL^2)/12`= `-(42.375*9^2)/12`= -184.781 KN-m

`Munderset(FED)`= `(wL^2)/12`= `(42.375*9^2)/12`= 184.781 KN-m

Calculation For Support End Moments

`Munderset(AB)`= `Munderset(FAB)`+`4EITunderseta``/L`+ `2EITundersetb``/L`

0= -286.031+ `4EITunderseta``/9`+ `2EITundersetb``/9`

 286.031- `2EITundersetb``/9`= `4EITunderseta``/9`

 `2EITunderseta``/9`= 143.0155 - `EITundersetb``/9`

`Munderset(BA)`= `Munderset(FBA)`+ `4EITundersetB``/L`+ `2EITundersetA``/L`

 `Munderset(BA)`= 286.031+ `4EITundersetB``/9`+ 143.0155 - `EITundersetb``/9`

 `Munderset(BA)`= 429.0465+ `3EITundersetB``/9`

`Munderset(BC)`= `Munderset(FBC)`+ `4EITundersetB``/L`+ `2EITundersetC``/L`

 `Munderset(BC)`= -184.78+ `4EITundersetB``/9`+ `2EITundersetC``/9`

`Munderset(CB)`= `Munderset(FCB)`+ `4EITundersetC``/L`+ `2EITundersetB``/L`

`Munderset(CB)`= 184.78+ `4EITundersetC``/9`+ `2EITundersetB``/9`

 `Munderset(CD)`= `Munderset(FCD)`+ `4EITundersetC``/L`+ `2EITundersetD``/L`

`Munderset(CD)`= -286.031+ `4EITundersetC``/9`+ `2EITundersetD``/9`

`Munderset(DC)`= `Munderset(FDC)`+ `4EITundersetD``/L`+ `2EITundersetC``/L`

`Munderset(DC)`= 286.031+ `4EITundersetD``/9`+ `2EITundersetC``/9`

`Munderset(DE)`= `Munderset(FDE)`+ `4EITundersetD``/L`+ `2EITundersetE``/L`

`Munderset(DE)`= -184.781+ `4EITundersetD``/9`+ `2EITundersetE``/9`

`Munderset(ED)`= `Munderset(FED)`+ `4EITundersetE``/L`+ `2EITundersetD``/L`

0= 184.781+ `4EITundersetE``/9`+ `2EITundersetD``/9`

`2EITundersetE``/9`= -92.3905`- `EITundersetD``/9`

At Joint B

`Munderset(BA)`+  `Munderset(BC)`= 0

429.0465+ `3EITundersetB``/9`-184.78+ `4EITundersetB``/9`+ `2EITundersetC``/9`=0

`7EITundersetB``/9`+ `2EITundersetC``/9`= -244.266

At Joint C

`Munderset(CB)`+ `Munderset(CD)`= 0

184.78+ `4EITundersetC``/9`+ `2EITundersetB``/9`-286.031+ `4EITundersetC``/9`+ `2EITundersetD``/9`=0

`2EITundersetB``/9`+ `8EITundersetC``/9`+`2EITundersetD``/9`= 101.251

At Joint D

`Munderset(DC)`+ `Munderset(DE)`= 0

286.031+ `4EITundersetD``/9`+ `2EITundersetC``/9`-184.781+ `4EITundersetD``/9`+ `2EITundersetE``/9`=0

286.031+ `4EITundersetD``/9`+ `2EITundersetC``/9`-184.781+ `4EITundersetD``/9`-92.3905 - `EITundersetD``/9=0`

`2EITundersetC``/9`+  `7EITundersetD``/9`= -8.8595

Solving the above Joint Equations

`EITundersetB`= -377.273

`EITundersetC`= 221.175

`EITundersetD`= -51.802

Calculation For Reaction at `RundersetA`

 `Munderset(BA)`= 429.0465+ `3EITundersetB``/9`

                          = 429.0465- 125.757

                          = 303.288 KN-m

 Now taking Span AB

Taking Moment About Point B

`RundersetA`x9+303.288-42.375x9x4.5=0

`RundersetA`= 156.988 KN

Calculation for the Maximum Positive Bending Moment For Every Span

So the maximum Bending moment occurs where shear force is zero.

`RundersetA`- `w*X`=0

156.988-42.375`*X`=0

156.988=42.375`*X`

X=3.7m

`Munderset(3.7m)`from A

=`RundersetA`x(3.7)-42.375x3.7x(3.7)/2

= 156.988x(3.7)-42.375x3.7x(3.7)/2

= 290.798 KN-m

Other Spans will also ahve the same value=290.798 KN-m