Project 1

Given below is a floor plan of a building. The red lines running East-west are beams supported by the girders spanning North-South between columns. The floor system is 120 mm thick normal weight concrete floor. The floor finish and mech loads on the floor is 3 kPa. The live load on the floor is 5 kPa. All beams have cross section 500 mmx750 mm. The beams are equally spaced. The rectangles in cyan color are columns. Modulus of elasticity of concrete is 10 GPa.

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.
2. For the same beam, consider load patterning to determine maximum positive and negative moments in each span.

GIVEN DATA=

FLOOR FINISH AND MECH = 3 KPA = 3 KN/M2

LIVE LOD = 5KA = 5KN/M2

SLAB THICK = 120MM

BEAM CROSS SECTION =500mm X 750 mm

ELASTICITY OF CONCRETE = 10 GPA =`10^7`KN/M2

DEAD LOAD = 0.12X 25 = 3 KN/M2

TOTAL LOAD ON THE FLOOR = 3+5 + 3 = 11 KN/M2

SELF WT O BEAM = 0.5X 0.75 X25= 9.375 KN/M

SINCE THE SLAB IS - ONE WAY SLAB

LENGTH OF TOTAL FLOOR = 36 M

LOAD CARRIED BY THE BEAM IN THE FORM OF UDL BY ONE PARTIAL PARTS OF FOOR = W= w X LX/2

11 kn/m^2 X1.5 /2 = 8.25 KN/M

= 2 X8.25 = 16.5 KN /M

16.5 + SELF WT OF BEAM= 16.5+9.375 = 25.875 KN/M

TOTAL WT OF BEAM = 25.875 KN/M

FXED END MOMENT FOR SPANS

MF AB = - 25.87 X9 X9 /12 = -174 KN M

MF BA = + 174KN M

MF BC = -174 KN M

MFCB = + 174KNM

MF CD = -174 KN M

MF DC = + 174 KNM

MF DE = -174 KN M

MFED = +174 KNM

FOR END MOMENT FOR SPAN AB

MAB = MFAB + 2EI/L( 2TA + TB) EI = 10^7 X0.5 X 0.75^3/12 = 3,12,500 KN M2

= -174+ K( 2TA +TB)

MBA= MFBA + K( 2TB + TA )

= +174 + K (2TB + TA)

FOR SPAN BC

MBC = MFBC + 2EI/L( 2TB + TC)

-174 + K( 2TB +TC)

MCB = MFCB + 2EI/L( 2TC+ TB)

+174 + K( 2TC +TB)

FOR SPAN CD

MCD = MFCD + 2EI/L( 2TC + TD)

= -174 + K( 2TC +TD)

MDC = MFDC + 2EI/L( 2TD+ TC)

= +174 + K( 2TD +TC)

FOR SPAN DE

MDE = MFDE + 2EI/L( 2TD+ TE)

= -174 + K( 2TD +TE)

MED = MFED + 2EI/L( 2TE + TD)

= +174 + K( 2TE +TD)

FROM MOMENT EQUIIBRIUM EQUN TAKE VALUE OF K =1

MB =0

174 + K (2TB + TA) + -174 + K( 2TB +TC)= 0

4TB + TA+ TC =0 --------------------------------------------1)

MC =0

+174 + K( 2TC +TB) -174 + K( 2TC +TD)

= 4TC+ TB+ TD =0 ------------------------------------------------(2)

MD=0

+174 + K( 2TD +TC) + -174 + K( 2TD +TE) =0

4TD+ TC+TE =0 --------------------------------------------------(3)

MAB=0

-174 + K( 2TA +TB) = 0

2TA +TB = 174 ------------------------------------------------4)

MED=0

+174 + K( 2TE +TD) =0

2TE +TD = -174 -------------------------------------------------- 5)

TA= 99.42

TB= -24.85

TC = 0.00

TD= 24.85

TE = -99.42

MAB= -174+ K( 2TA +TB) = -0.01 KNM

MBA= +174 + K (2TB + TA) = 223 KNM

MBC = -174 + K( 2TB +TC) = -223 KNM

MCB = +174 + K( 2TC +TB) = 149.15 KNM

MCD= -174 + K( 2TC +TD) = -149.15 KNM

MDC=+174 + K( 2TD +TC) =223.7 KN M

MDE= -174 + K( 2TD +TE) = -223.7 KNM

MED =+174 + K( 2TE +TD) = 0.01

FOR SPAN AB

MOMENT ABOUT B

RA X 9 - 0-25.87X9X4.5 + 223=0

RA= +91.5

RB= 25.87X9 -91.5 =141.33

SPAN BC

RB X9 -223 -25.87X9 X4.5+149.15 =0

RB = 124 KN

RC = 25.87 X9 -( 124) = 108.83 KN

SPAN CD

RC X 9 -149.15 -25.87X9 X4.5+223.7 =0

RC = =108 KN

RD = 124.83

SPAN DE

RD X9 - 223.7) -25.87X 9X 4.5 =0

RD = 141

RE = 91.83 KN

RA = 91.5 KN

RB = 141.33 KN

RC = 249.28 KN

RD = 216.35

RE = 257.68 KN

B M DIAGRAM

2

SPAN AB AND CD LOADS= 25.75 KN/M

SPAN BC AND DE LOADS= W* L/2 =6* 1.5 /2= 4.5 KN/M + SELF WT OF BEAM ( 9.375KN/M) =

= 2x 4.5 +9.375 = 18.375

MFAB =-174 KN M

MFBA=+174 KN M

MFCD =-174 KN M

MFDC=+174 KN M

MFBC = - WL2/12= - 18.875 X 9 X9/12 = -124 knm

MFCB = +124 KN M

MFDE = -124KN M

MFED=+ 124 KN M

END MOMENTS OF SPANS

MAB = MFAB + 2EI/L( 2TA+TB)

= -174 + K ( 2TA +TB)

MBA = +174 + K (2TB + TA)

MBC = MF BC+ 2EI/L(2TB +TC)

-124 + K (2TB+ TC)

MCB = +124+K( 2TC+ TB)

MCD = -174 + K( 2TC +TD)

MDC= +174 + K( 2TD +TC)

MDE = -124 + K( 2TD +TE)

MED = +124 + K( 2TE+TD)

FRO MOMENT EQUILIBRIUM EQUN

MB=0

MBA + MBC =0

+174 + K (2TB + TA) + -124 + K (2TB+ TC) =0

50 + K( 4TB +TA+TC ) =0

4TB +TA+TC = -50/K = - 0.0014 --------------------------------------------1

MC =0

M CB + MCD =0

+124 +K( 2TC+ TB) + -174 + K( 2TC +TD) =0

(4TC+ TB+ +TD) = 50/K

4TC +TB+TD = 0.0014 -----------------------------------------------------2

MD=0

MDC +MDE=0

+174 + K( 2TD +TC) + -124 + K( 2TD +TE) =0

K(4TD+ TC +TE) ) = -50

4TD+ TC +TE = -0.0014 ------------------------------------------3

MA =0

-174 + K ( 2TA +TB) =0

2TA+TB =174/K = 0.0050 --------------------------------------------4

ME =0

+124 + K( 2TE+TD) =0

2TE+ TD = -124/ 34,722 = -0.0035

2TE+ TD =-0.0035 ----------------------------------------------------5

TA= 0.0033 / 0.00316

TB = -0.0016 /-0.00132

TC = 0.0011/ 0.00070

TD= - 0.0004 /-0.00010

TE=- 0.0018 /-0.00170

MAB = -174 + K ( 2TA +TB) = -174 +34722 X( 2* 0.0031 -0.00113)= 3.860 KN M

MBA = +174 + K (2TB + TA)= +17 4+34,722 X(2*-0.0013 +0.0031)= 191 KN-M

MBC = -124 + K (2TB+ TC)= -93.65 +34722X(2*-0.0013+0.0007)= 190 KN M

MCB = +124 +K( 2TC+ TB) = 124+ 34722 X(2*0007-0.0013)=127 KN M

MCD= -174 + K( 2TC +TD)= -174 + 34722X(2*0.00070-0.00010)= -128 KN M

MDC= +174 + K( 2TD +TC) =+174 +34722 X(2*-00010+0.00070) = 191 KNM

MDE = -93.65 + K( 2TD +TE)= -124 + 34722X (2*-0.0001 -0.0017)= -189 KN M

MED= +93.65 + K( 2TE+TD) =124+ 34722 X(2*-0.0017 -0.0001) = +2.473

FOR SPAN AB

RA=91.5 KN

RB = 141.33 KN

FOR SPAN BC