Project 1
1) Requirement #1
The beam under consideration has 4 spans & supports on 5 girder support. Assuming the supports are A,B,C,D & E from the left.
a) Loads on Beam
i) Dead loads
Assume the unit weight of concrete = 25 kN/m^3
Own weight =B*H*concrete unit weight = 0.5*0.75*25 = 9.375 kN/m'
Load from slabs
All slabs' dimensions are 9m*3m with aspect ratios 9/3 > 2 so that they are all one-way slabs with load distribution in the shortest spans.
Own weight from slabs = t*concrete unit weight*Spacing between beams= 0.12*25*3= 9.0 kN/m'
Finishes & Mechanicals = Area load*Spacing between beams = 3*3= 9.0 kN/m'
The total unfactored dead loads = 9.375+9+9= 27.375 kN/m'
ii) Live loads
The total unfactored live loads = Area load*Spacing between beams = 5*3= 15.0 kN/m'
iii) Total unfactored loads (D+L)
The total unfactored loads (D+L) = 27.375 + 15.0 = 42.375 kN/m' (uniformly distributed)
b) Structural Analysis
By using the slope deflection method, the beam is kinematically indeterminate from the 5th degree
So we have 5 unknown rotations at the 5 supports Ta, Tb, Tc, Td & Te respectively from the left
Forming the slope deflection equation given that the fixed end moment for the uniformly distributed load = WL^2/12
Considering the clockwise moments & rotations as +ve & the counterclockwise moments & rotations as -ve
Also, the beam has equal spans with the same uniform load & constant cross-section, so for all spans:
FEM=±42.375*9^2/12=±286.2; EI=10000000*0.5*0.75^3/12=175781.25; L=9;
4EI/L=4*175781.25/9=703125; 2EI/L=2*175781.25/9=351562.5
For span AB:
Ma = FEMAB + (4EI/L)Ta + (2EI/L)Tb
Ma=0 (hinged support) then:
0 = -286.2 + (703125)Ta + (351562.5)Tb ---> 1
Mb = FEMBA + (4EI/L)Tb + (2EI/L)Ta
Mb = 286.2 + (703125)Tb + (351562.5)Ta ---> 2
For span BC:
Mb = FEMBC + (4EI/L)Tb + (2EI/L)Tc
Mb = -286.2 + (703125)Tb + (351562.5)Tc ---> 3
Mc = FEMCB + (4EI/L)Tc + (2EI/L)Tb
Mc = 286.2 + (703125)Tc + (351562.5)Tb ---> 4
For span CD:
Mc = FEMCD + (4EI/L)Tc + (2EI/L)Td
Mc = -286.2 + (703125)Tc + (351562.5)Td ---> 5
Md = FEMDC + (4EI/L)Td + (2EI/L)Tc
Md = 286.2 + (703125)Td + (351562.5)Tc ---> 6
For span DE:
Md = FEMDE + (4EI/L)Td + (2EI/L)Te
Md = -286.2 + (703125)Td + (351562.5)Te ---> 7
Me = FEMED + (4EI/L)Te + (2EI/L)Td; Me=0
0 = 286.2 + (703125)Te + (351562.5)Td ---> 8
Equilibrium Conditions:
Mb from span AB = -Mb from span BC
Substitute from 3 in 2
Also due to symmetry in geometry & loads, Tc must be zero
Tc=0
Final slope deflection equations:
(703125)Ta + (351562.5)Tb - 286.2 = 0 ---> I (rearranging equation 1)
Mb from span AB = 286.2 + (703125)Tb + (351562.5)Ta = - [-286.2 + (703125)Tb + (351562.5)Tc]
(351562.5)Ta + (1406250)Tb = 0 ---> II (Substituting from 3 in 2; Tc=0)
Solving the equations I & II to find Ta & Tb:
Ta = 4.65E-04; Tb = -1.16E-04; Tc = 0 rad.
Also due to symmetry Td=-Tb & Te=-Ta therefore:
Td=1.16E-04; Te=-4.65E-04
Substitute with Ta & Tb in equation 2 to find Mb
Mb = 286.2 + (703125)Tb + (351562.5)Ta ---> 2
Mb = 367.9 kN.m (clockwise for span AB)
Substitute with Tb & Tc=0 in equation 4 to find Mc
Mc = 286.2 + (703125)Tc + (351562.5)Tb ---> 4
Mc = 245.4 kN.m (clockwise for span BC)
And due to symmetry Md=Mb=367.9 kN.m (clockwise for span CD); Me=0 (hinged support)
Then the beam became statically determinate & we can separate each span & study its equilibrium (∑M=0 & ∑Y=0)
And the SFD & BMD will be as the following:
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2) Requirement #2
Using the previous unfactored load (service) & considering the load patterning, therefore:
Wu= 1.2D +1.6L ---> ASCE
Wu=1.2(27.375) +1.6(15) = 56.85 kN/m'
The ratio between the ultimate load (Wu) to the service load (Ws) = 56.85/42.375 = 1.34
Therefore to bring the ultimate moments we will multiply (magnify) the service moments by 1.34 (due to load superposition property) then:
Mub = 367.9*1.34= 493 kN.m (negative)
Muc = 245.4*1.34= 328.8 kN.m (negative)
We can separate each span & study its equilibrium (∑M=0 & ∑Y=0) as follows:
For span AB:
Taking the moment B to bring the reaction at A (Ra)
Ra=[56.85*9*4.5-493]/9= 201 kN (upward)
Rb=W*L-Ra=56.85*9-201= 310.65 kN (upward)
Assuming the point of zero shear (max moment) at distance x from the left support A then:
Ra = Wu*x
x= 201/56.85 = 3.54 m
Taking the left moment at distance x to bring the max. moment in span AB then:
Mumax(AB)=Ra*x-Wu*x*x/2=201*3.54-56.85*3.54*3.54/2= 355.3 kN.m (positive)
For span BC:
Similarly, taking the moment at C to bring the reaction at B (Rb)
Rb=[56.85*9*4.5+493-328.8]/9= 274.1 kN (upward)
Rc=W*L-Rb=56.85*9-274.1= 237.55 kN (upward)
Assuming the point of zero shear (max moment) at distance x from the left support B then:
Rb = Wu*x
x= 274.1/56.85 = 4.82 m
Taking the left moment at distance x to bring the max. moment in span BC then:
Mumax(BC)=Rb*x-Wu*x*x/2-493=274.1*4.82-56.85*4.82*4.82/2-493= 167.7 kN.m (positive)
And due to the symmetry in geometry & loads, span DC is a mirror of span BC & span ED is a mirror of span AB
And the SFD & BMD will be as the following:
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21-10-2022