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Challenge

Project 1

RRajashekar Mandadi17 Nov 2022

Q1]

(a)

AIM: -

Determine the loading on the beam indicated,

Analyze the beams and draw bending moment and shear force diagram

PROCEDURE: -

Given Data

The live load on the floor is 5 kPa

The floor finish and mech loads on the floor is 3 kPa

The floor system is 120 mm thick normal weight concrete floor.

Beams have cross section 500 mmx750 mm.

Modulus of elasticity of concrete is 10 GPa.

Dead Load of Slab = 25*0.12 = 3 kpa

Floor finish load = 3 kpa

Live Load = 5 kpa

Total load = 11 kpa

Load intensity from Slab = 11*3 = 33 kN/m

Self weight of beam = 25*0.5*0.75 = 9.375 kN/m

Total load on beam = 42.375 kN/m

Step 1

Fixed End Moments

`(M^f)_(AB) = -(W*L^2)/12 = - (42.375*9^2)/12 = -286.03`

`(M^f)_(BA) = (W*L^2)/12 = (42.375*9^2)/12 = 286.03`

`(M^f)_(BC) = -(W*L^2)/12 = - (42.375*9^2)/12 = -286.03`

`(M^f)_(CB) = (W*L^2)/12 = (42.375*9^2)/12 = 286.03`

`(M^f)_(CD) = -(W*L^2)/12 = - (42.375*9^2)/12 = -286.03`

`(M^f)_(DC) = (W*L^2)/12 = (42.375*9^2)/12 = 286.03`

`(M^f)_(DE) = -(W*L^2)/12 = - (42.375*9^2)/12 = -286.03`

`(M^f)_(ED) = (W*L^2)/12 = (42.375*9^2)/12 = 286.03`

STEP 2

END MOMENTS

`M_(AB) = (M^f)_(AB)+(2EI)/L(2theta_a+theta_b) = -286.03+(2EI)/9(2theta_a+theta_b)`

`M_(BA) = (M^f)_(BA)+(2EI)/L(2theta_a+theta_b) = 286.03+(2EI)/9(2theta_a+theta_b)`

`M_(BC) = (M^f)_(BC)+(2EI)/L(2theta_b+theta_c) = -286.03+(2EI)/9(2theta_b+theta_c)`

`M_(CB) = (M^f)_(CB)+(2EI)/L(2theta_c+theta_b) = 286.03+(2EI)/9(2theta_c+theta_b)`

`M_(CD) = (M^f)_(CD)+(2EI)/L(2theta_c+theta_d) = -286.03+(2EI)/9(2theta_c+theta_d)`

`M_(DC) = (M^f)_(DC)+(2EI)/L(2theta_d+theta_c) = 286.03+(2EI)/9(2theta_d+theta_c)`

`M_(DE) = (M^f)_(DE)+(2EI)/L(2theta_d+theta_e) = -286.03+(2EI)/9(2theta_d+theta_e)`

`M_(ED) = (M^f)_(ED)+(2EI)/L(2theta_e+theta_d) = 286.03+(2EI)/9(2theta_e+theta_d)`

STEP 3

Kinematic indeterminacy Dk =5

Uknowns = `theta_a,theta_b,theta_c,theta_d,theta_e`

STEP 4

EQUILIBRIUM EQUATIONS

`M_(AB)=0`

`M_(BA)+M_(BC)=0`

`M_(CB)+M_(CD)=0`

`M_(DC)+M_(DE)=0`

`M_(ED)=0`

`M_(AB) = -286.03+(2EI)/9(2theta_a+theta_b) = 0`

`2theta_a+theta_b = 1287.135/(EI)`.......................................................................................(1)

`theta_a+4theta_b+theta_c=0`...............................................................................................(2)

`theta_b+4theta_c+theta_d=0`...............................................................................................(3)

`theta_c+4theta_d+theta_e=0`................................................................................................(4)

`2theta_e+theta_d = -1287.135/(EI)`.......................................................................................(5)

By solving the above Equations, we get

EI = 175781.25 kN-m2

`theta_a = 735.51/(EI) = 735.51/175781.25 = 0.0042`

`theta_b = -183.88/(EI)= -183.88/175781.25 = -0.00105`

`theta_c = 0`

`theta_d = 183.88/(EI)= 183.88/175781.25 = 0.00105`

`theta_e = -735.51/(EI) = -735.51/175781.25 = -0.0042`

STEP 5

Substitute these values in equations and get the End moment values

`M_(AB) = 0`

`M_(BA) = 419.22 kN-m`

`M_(BC) = -419.22 kN-m`

`M_(CB) = 255.77 kN-m`

`M_(CD) = -255.77 kN-m`

`M_(DC) = 419.22 kN-m`

`M_(DE) = -419.22 kN-m`

`M_(ED) = 0`

STEP 6

F.B.D

Ra+Rb = 42.375*9 = 381.375

Moment @ B = 0

Ra*9 - 42.375*9*(9/2) + 419.22 = 0

Ra = 144.1 kN

Rb1 = 381.375 - 144.1 = 237.27 kN

Rb2+Rc1 = 42.375*9 = 381.375

Moment @ C = 0

Rb2*9 - 42.375*9*(9/2) - 419.22 + 255.77= 0

Rb2 = 208.85 kN

Rc1 = 381.375 - 208.85 = 172.53 kN

Rc2+Rd1 = 42.375*9 = 381.375

Moment @ C = 0

Rc2*9 - 42.375*9*(9/2) + 419.22 - 255.77= 0

Rc2 = 172.53 kN

Rd1 = 381.375 - 172.53= 208.85 kN

Rd2+Re = 42.375*9 = 381.375

Moment @ E = 0

Rd2*9 - 42.375*9*(9/2) - 419.22 = 0

Rd2 = 237.27 kN

Re = 381.375 - 237.27 = 144.1 kN

STEP 7

FINAL REACTIONS

Ra = 144.1 kN

Rb = Rb1 +Rb2 = 237.27 + 208.85 = 446.12 kN

Rc = Rc1 + Rc2 = 172.53 +172.53 = 345.06 kN

Rd = Rd1 +Rd2 = 208.85 + 237.27= 446.12 kN

Re = 144.1 kN

(b)

For the same beam, Consider Load patterning to Determine Maximum Positive Moments in each span.

Dead Load of Slab = 25*0.12 = 3 kpa

Floor finish load = 3 kpa

Total Dead load = 6 kpa

Load intensity from Slab = 6*3 = 18 kN/m

Self weight of beam = 25*0.5*0.75 = 9.375 kN/m

Total Dead load on beam = 27.375 kN/m

Live Load = 5 kpa

Live load on beam = 5*3 = 15 kN/m

FOR SPAN AB & CD

Distribute the Dead load on all the Spans,

Place the Live load on span AB and CD

Step 1

Fixed End Moments

`(M^f)_(AB) = -(W*L^2)/12 = - (42.375*9^2)/12 = -286.03`

`(M^f)_(BA) = (W*L^2)/12 = (42.375*9^2)/12 = 286.03`

`(M^f)_(BC) = -(W*L^2)/12 = - (27.375*9^2)/12 = -184.78`

`(M^f)_(CB) = (W*L^2)/12 = (27.375*9^2)/12 = 184.78`

`(M^f)_(CD) = -(W*L^2)/12 = - (42.375*9^2)/12 = -286.03`

`(M^f)_(DC) = (W*L^2)/12 = (42.375*9^2)/12 = 286.03`

`(M^f)_(DE) = -(W*L^2)/12 = - (27.375*9^2)/12 = -184.78`

`(M^f)_(ED) = (W*L^2)/12 = (27.375*9^2)/12 = 184.78`

STEP 2

END MOMENTS

`M_(AB) = (M^f)_(AB)+(2EI)/L(2theta_a+theta_b) = -286.03+(2EI)/9(2theta_a+theta_b)`

`M_(BA) = (M^f)_(BA)+(2EI)/L(2theta_a+theta_b) = 286.03+(2EI)/9(2theta_a+theta_b)`

`M_(BC) = (M^f)_(BC)+(2EI)/L(2theta_b+theta_c) = -184.78+(2EI)/9(2theta_b+theta_c)`

`M_(CB) = (M^f)_(CB)+(2EI)/L(2theta_c+theta_b) = 184.78+(2EI)/9(2theta_c+theta_b)`

`M_(CD) = (M^f)_(CD)+(2EI)/L(2theta_c+theta_d) = -286.03+(2EI)/9(2theta_c+theta_d)`

`M_(DC) = (M^f)_(DC)+(2EI)/L(2theta_d+theta_c) = 286.03+(2EI)/9(2theta_d+theta_c)`

`M_(DE) = (M^f)_(DE)+(2EI)/L(2theta_d+theta_e) = -184.78+(2EI)/9(2theta_d+theta_e)`

`M_(ED) = (M^f)_(ED)+(2EI)/L(2theta_e+theta_d) = 184.78+(2EI)/9(2theta_e+theta_d)`

STEP 3

Kinematic indeterminacy Dk =5

Uknowns = `theta_a,theta_b,theta_c,theta_d,theta_e`

STEP 4

EQUILIBRIUM EQUATIONS

`M_(AB)=0`

`M_(BA)+M_(BC)=0`

`M_(CB)+M_(CD)=0`

`M_(DC)+M_(DE)=0`

`M_(ED)=0`

`M_(AB) = -286.03+(2EI)/9(2theta_a+theta_b) = 0`

`2theta_a+theta_b = 1287.135/(EI)`.......................................................................................(1)

`theta_a+4theta_b+theta_c=-455.625/(EI)`..............................................................................(2)

`theta_b+4theta_c+theta_d=455.625/(EI).........................................................................................(3)`

`theta_c+4theta_d+theta_e=-455.625/(EI)`................................................................................................(4)`

`2theta_e+theta_d = -831.51/(EI)`.......................................................................................(5)`

By solving the above Equations, we get

EI = 175781.25 kN-m2

`theta_a =833.14/(EI)`

`theta_b = -379.14/(EI)`

`theta_c = 227.81`

`theta_d = -76.48/(EI)`

`theta_e = -377.51/(EI)`

STEP 5

Substitute these values in equations and get the End moment values

`M_(AB) = 0`

`M_(BA) = 302.66 kN-m`

`M_(BC) = -302.66 kN-m`

`M_(CB) = 201.77 kN-m`

`M_(CD) = -201.77 kN-m`

`M_(DC) = 302.66 kN-m`

`M_(DE) = -302.66 kN-m`

`M_(ED) = 0`

STEP 6

F.B.D

Ra+Rb = 42.375*9 = 381.375

Moment @ B = 0

Ra*9 - 42.375*9*(9/2) + 302.66 = 0

Ra = 157.06 kN

Rb1 = 381.375 - 157.06 = 224.32 kN

Rb2+Rc1 = 27.375*9 = 246.375

Moment @ C = 0

Rb2*9 - 27.375*9*(9/2) - 302.66 + 201.77= 0

Rb2 = 134.4 kN

Rc1 = 246.375 - 134.4 = 111.98 kN

Rc2+Rd1 = 42.375*9 = 381.375

Moment @ D = 0

Rc2*9 - 42.375*9*(9/2) + 302.66 - 201.77= 0

Rc2 = 179.48 kN

Rd1 = 381.375 - 179.48= 201.9 kN

Rd2+Re = 27.375*9 = 246.375

Moment @ E = 0

Rd2*9 - 27.375*9*(9/2) - 302.66 = 0

Rd2 = 156.82 kN

Re = 246.375 - 156.82 = 89.55 kN

STEP 7

FINAL REACTIONS

Ra = 157.06

Rb = Rb1 +Rb2 = 224.32 + 134.4 = 358.72 kN

Rc = Rc1 + Rc2 = 111.98 + 179.48 = 291.46 kN

Rd = Rd1 +Rd2 = 201.9 + 156.82 = 358.72 kN

Re = 89.55 kN

MAXIMUM POSITIVE BENDING MOMENT IN SPAN BC & DE

Distribute the Dead load on all the Spans,

Place the Live load on span BC and DE.

Step 1

Fixed End Moments

`(M^f)_(AB) = -(W*L^2)/12 = - (27.375*9^2)/12 = -184.78`

`(M^f)_(BA) = (W*L^2)/12 = (27.375*9^2)/12 = 184.78`

`(M^f)_(BC) = -(W*L^2)/12 = - (42.375*9^2)/12 = -286.03`

`(M^f)_(CB) = (W*L^2)/12 = (42.375*9^2)/12 = 286.03`

`(M^f)_(CD) = -(W*L^2)/12 = - (27.375*9^2)/12 = -184.78`

`(M^f)_(DC) = (W*L^2)/12 = (27.375*9^2)/12 = 184.78`

`(M^f)_(DE) = -(W*L^2)/12 = - (42.375*9^2)/12 = -286.03`

`(M^f)_(ED) = (W*L^2)/12 = (42.375*9^2)/12 = 286.03`

STEP 2

END MOMENTS

`M_(AB) = (M^f)_(AB)+(2EI)/L(2theta_a+theta_b) = -184.78+(2EI)/9(2theta_a+theta_b)`

`M_(BA) = (M^f)_(BA)+(2EI)/L(2theta_a+theta_b) = 184.78+(2EI)/9(2theta_a+theta_b)`

`M_(BC) = (M^f)_(BC)+(2EI)/L(2theta_b+theta_c) = -286.03+(2EI)/9(2theta_b+theta_c)`

`M_(CB) = (M^f)_(CB)+(2EI)/L(2theta_c+theta_b) = 286.03+(2EI)/9(2theta_c+theta_b)`

`M_(CD) = (M^f)_(CD)+(2EI)/L(2theta_c+theta_d) = -184.78+(2EI)/9(2theta_c+theta_d)`

`M_(DC) = (M^f)_(DC)+(2EI)/L(2theta_d+theta_c) = 184.78+(2EI)/9(2theta_d+theta_c)`

`M_(DE) = (M^f)_(DE)+(2EI)/L(2theta_d+theta_e) = -286.03+(2EI)/9(2theta_d+theta_e)`

`M_(ED) = (M^f)_(ED)+(2EI)/L(2theta_e+theta_d) = 286.03+(2EI)/9(2theta_e+theta_d)`

STEP 3

Kinematic indeterminacy Dk =5

Uknowns = `theta_a,theta_b,theta_c,theta_d,theta_e`

STEP 4

EQUILIBRIUM EQUATIONS

`M_(AB)=0`

`M_(BA)+M_(BC)=0`

`M_(CB)+M_(CD)=0`

`M_(DC)+M_(DE)=0`

`M_(ED)=0`

`M_(AB) = -184.78+(2EI)/9(2theta_a+theta_b) = 0`

`2theta_a+theta_b = 831.51/(EI)`.......................................................................................(1)`

`theta_a+4theta_b+theta_c=455.625/(EI)`..............................................................................(2)`

`theta_b+4theta_c+theta_d=-455.625/(EI).........................................................................................(3)`

`theta_c+4theta_d+theta_e=455.625/(EI)`................................................................................................(4)`

`2theta_e+theta_d = -1287.135/(EI)`.......................................................................................(5)`

By solving the above Equations, we get

EI = 175781.25 kN-m2

`theta_a =377.51/(EI)`

`theta_b =76.48/(EI)`

`theta_c = 227.81/(EI)`

`theta_d = 379.14/(EI)`

`theta_e = -833.14/(EI)`

STEP 5

Substitute these values in equations and get the End moment values

`M_(AB) = 0`

`M_(BA) = 302.66 kN-m`

`M_(BC) = -302.66 kN-m`

`M_(CB) = 201.77 kN-m`

`M_(CD) = -201.77 kN-m`

`M_(DC) = 302.66 kN-m`

`M_(DE) = -302.66 kN-m`

`M_(ED) = 0`

STEP 6

F.B.D

Ra+Rb = 27.375*9 = 246.375

Moment @ B = 0

Ra*9 - 27.375*9*(9/2) + 302.66 = 0

Ra = 89.55 kN

Rb1 = 381.375 - 157.06 = 156.82 kN

Rb2+Rc1 = 42.375*9 = 381.375

Moment @ C = 0

Rb2*9 - 42.375*9*(9/2) - 302.66 + 201.77= 0

Rb2 =201.90 kN

Rc1 = 381.375 - 201.90 = 179.48 kN

Rc2+Rd1 = 27.375*9 = 246.375

Moment @ D = 0

Rc2*9 - 27.375*9*(9/2) + 302.66 - 201.77= 0

Rc2 = 111.98 kN

Rd1 = 246.375 - 111.98= 134.40 kN

Rd2+Re = 42.375*9 = 381.375

Moment @ E = 0

Rd2*9 - 42.375*9*(9/2) - 302.66 = 0

Rd2 = 224.32 kN

Re = 381.375-224.32 = 157.06

STEP 7

FINAL REACTIONS

Ra = 89.55 kN

Rb = Rb1 +Rb2 = 224.32 + 134.4 = 358.72 kN

Rc = Rc1 + Rc2 = 111.98 + 179.48 = 291.46 kN

Rd = Rd1 +Rd2 = 201.9 + 156.82 = 358.72 kN

Re = 157.06 kN

RESULT: -

(a)

Determined the loading on the beam indicated,

Analyzed the beams and drawn Bending moment and Shear force diagram.

Used slope deflection method.

(b)

Considered load patterning and Determined maximum positive moments in each span.

SPAN AB & CD

SPAN BC & DE

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