Structural Design of Multi Storey Building
Short Description
Structural Design of Multi reinforced concrete Building...
Description
MANUAL CONCRETE DESIGN BS 8110:1997 Full structural design of a multi-storey building
Contents: Fundamentals Concepts of Concrete Design Analysis & Design of the Flat slabs Analysis & Design of the Edge beams Analysis & Design of the Columns Analysis & Design of the Staircases Analysis & Design of the Foundations Appendix
T: MUJTABA HADDAD Tel: 0926206662
Section design for moment
Behavior of beams in bending:
Fig -aThe behavior of a cross-section subjected to pure bending is studied be loading a beam at third point as shown in figure above. Initially the beam behaves as a monolithic elastic beam till the stresses at bottom fibers reaches the tensile strength of concrete; because of the very low tensile strength of concrete vertical cracks appear at a fairly low load. As the load increases, cracks lengthen and penetrate deeper towards the compression face, simultaneously the strain in steel also increases. The final failure depends on the amount and yield stress of steel. The three possible modes of failure are:1- Steel yields first: If the tensile force capacity of steel is low then it yields before the strain in concrete at the compression face reaches the maximum permissible value of 0.0035, because steel is a ductile material. Steel elongates while maintaining its strength. The beam continues to deform at constant load and the neutral axis moves up. The beam finally fails when the depth of the compression zone is too small to balance the tensile force in steel. There is always ample warning before failure. 2- Simultaneous yielding of steel and concrete: If the tensile capacity force of steel is moderate, yielding of steel is simultaneously accompanied by the crushing of concrete. 3- Concrete crushes first: If the tensile force capacity of steel is high, then steel doesn’t yield at all before concrete crushes, because concrete is a very brittle material it fails in an explosive manner.
Singly reinforced rectangular beams:
Fig-b- Steel “stress-strain diagrams”
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Fig-c- concrete stress-strain diagrams
Fig-d-
The ultimate moment of resistance of a section is based on the following assumptions:1. 2. 3. 4. 5.
Plane section remains plane. The tensile strength of concrete is ignored. The stresses in the reinforcement is pre-derived from fig(5) with ɤm= 1.05. Lever arm shouldn’t be assumed to be greater than 0.95. Stresses in concrete in compression are derived using either a. The design stress-strain fig(c) with ɤm=1.5 b. The simplified stress where the depth is 0.9 of the depth to the neutral axis fig(e)
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Fig-e(a) Section (b) strain (c) rectangular parabolic stress diagram (d) simplified stress diagram Concrete stress is 0.67fcu/ ɤm= 0.67fcu/1.5=0.447fcu ≈ 0.45 fcu Total compressive force is given by: C=0.447*fcu*b*0.9x= 0.402*b*x*fcu The lever arm z is : Z= d-
= d-0.45x
If M is the applied moment then: M=C*z=0.402*b*x*fcu*(d-0.45x) Multiplied by
for both sides
= 0.402*b*x*fcu Put k =
*d - 0.402*b*x*fcu
*0.45x
and rearranging:
K= 0.402* (1-0.45 ) 0.1809(
2
- 0.402 ) + k=0
Solving for
=
) √
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Put (
=1-0.45 and rearranging: = 0.5+√
Total tensile force in steel is: T = As*fs For internal equilibrium (T) must equal to (C). M=T*Z=As*fs*z
As=
→ (1) Stress fs in steel depends on the strain Єs in steel. Its highly desirable that the final failure is due to yielding of steel rather than crushing of concrete. Its useful therefore to calculate the maximum neutral axis in order to achieve this.
Note
From fig(e) Єs =
*0.0035 → (2)
Strain Єs in steel at a stress of
is given by
Es Єs = Take fy = 460 N/mm2 , ɤm = 1.05 , Es = 200 kN/mm2 =
*1.05 =
Єs = 0.0022 Substitute in (2) Єs =
*0.0035 = 0.0022
= 0.614 In clause (3.4.4.4) Bs 8110 limits the ratio to the maximum of 0.5 in order to ensure that failure is proceeded by steel yielding well before the strain in concrete reaches 0.0035. C = 0.447fcu b *0.9x = 0.402*b*o.5d*fcu = 0.201 Fcub*d Z = d-0.45x = d-0.45*.5d = 0.775d M=C*z= 0.201 Fcu*b*d * 0.775d = 0.156 fcu b*d2 K=
Note
= 0.156
This is the maximum value of the applied moment that the section can resist because it utilises fully the compression capacity of the cross section.
When ( ) ≤ 0.5 steel will always yield
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Then fs =
=0.95 fy
Substitute in (2) M = As*0.95*fy*z As =
Doubly reinforced concrete beams:
Fig-f(a) Section (b) strain diagram (c) stress diagram and internal forces Let (M) be the ultimate design moment, and from the previous calculations for a rectangular sections as singly reinforced the maximum value of the moment that the section can resist is:
Msr = 0.156bd2fcu Put x =0.5d and z = 0.775d (maximum limit) If M > Msr then compression steel is required because concrete cannot provide the necessary compressive resistance alone. Compressive force by concrete = 0.2*fcu*b*d Compressive force by steel = As’ * fs’ From fig(f-b)above: Єsc = 0.0035 If Єsc ≥ Єs steel yields (fs’=0.95 fy) If Єsc ≤ Єs fs must be taken as Es Єsc If fy =460 kN/mm2 , Es = 200 kN/mm2 , then Єs = 0.0022 Therefore steel yield if Єsc ≥ 0.0022 Єsc = 0.0035
=0.0022
≤ 0.37 , if x = 0.5d then If =0.5 ,
≤ 0.186
≤ 0.186
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For equilibrium: T= Cc + Cs 0.95 fy As = 0.2 fcu*b*d +0.95 fy As` (3) Take moment about the tension steel: M = Cc*z1 + Cs*z2 ) = 0.2 fcu*b*d*0.775 *d +0.95 fy As`*(d-d`) = 0.156 fcu bd2 +0.95 fy As` (d-d`)
As`= From (3)
= As =
+ + As` , As =
( )+ As`
Put (d = z/0.775) As = Put k =
+ As`
, k`= 0.156
As` = As =
+As`
Flanged beams:
Fig-gStress distribution in the flange
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In simple slab-beam system shown in the previous page, the slab is designed to span between the beams. The reactions from the slab act as a load on the beam, when a series of beams are used to support a concrete slab the slab acts as the flange of the beams because of the monolithic nature of concrete construction, the end beams become L-beam while the intermediate beams become T-beam. The distribution of the compressive stress on the flange is not uniform (shear leg), but for simplicity in design only part of the flange width is considered to sustain the compressive stress (width=b).
Fig-hT-beam : b = bw+
or the actual flange width if less
L-beam : b = bw+
or the actual flange width if less
Lz: distance between points of zero moments (1.0 L for simple , 0.7 L for continuous) The design procedure for flanged beams depends on the depth of the stress block, two possibilities need to be considered:
Fig-i- (a) Section (b) stresses and internal forces
Stress block within flange:0.9x≤hf If 0.9x≤hf The depth of the flange “same as total depth of the slab” then all concrete below the flange is cracked and the beam may be treated as a rectangular beam with breadth “b” M flange = 0.45 Fcu b h f (d- ) If M M flange C1= 0.45 fcu (b-bw)*hf C2=0.45 fcubw*0.9x Z1=d-hf/2 Z2=d-0.9x/2 Moment of resistance Mr= C1*z1+C2*z2 If the amount of steel provided is sufficient to cause yielding of the steel then fs = 0.95 fy, and with x = 0.5 d Take moment about force C1:
Mr =Tz1-C2(z1-z2) M =0.95fy As (d-0.5hf)-0.2 fcu bw d (0.225d-0.5 hf) M =0.95fy As (d-0.5hf)-0.1 fcu bw d (0.45d- hf)
As =
Notes (2)
-
1. 2. 3.
This equation is given in the code, it gives conservative results for cases where x 267 kN.m ----ok
Final moments:
Shear Forces:
All forces are in kN
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Direction E-W: 123-
No lateral load is subjected, so lateral stability isn`t depend on slab-column connection. Conditions of Clause (3.5.3.3) -single load case- are satisfied. There are 3 equal spans. Since the provision of clause (3.7.2.7) is met, the coefficient in table (3.12) may be used subject to the following provions: 1- Moments at supports maybe reduced by 0.15hc. 2- The limitations of clause (3.7.2.6) need not be checked. Allowance has been made to the coefficient for 20% redistribution.
Fig-5- Critical strip from direction E-W
Fig-6- coefficients from Table (3.12)
F = 22.16*4.25*5=471 kN h = 0.3568m , L = 4.25m
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F = 471 KN for internal supports, F = (471/2) KN for external supports 0.15 F* hc = o.15*471*0.3568 = 25.2 kN.m 0.15 F* hc = 0.15*(471/2)*0.3568 = 12.6 kN.m
Final bending moments diagrams:
All moments are in K.m
Shear forces diagram:
All forces are in KN
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1.3 Design of flat slab: Direction (N-S): We will consider direction N-S first because in this direction we have greater moments than the other direction. For economic design we have to increase the lever arm as much as possible.
1.3.1 Division of panels (into column and middle strips):
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1.3.2 Check for design moment transferable between slab and column: (clause 3.7.4.2) Mt max = 0.15 be d2 fcu , be = cx+cy < C.S (column strip width) , be = 200+500 =700 < 2125 mm ---ok be: breadth of effective moment transfer strip --- (fig 3.13) Mt max= 0.15 *700 *1672 *25*10-6 = 73.2 kN.m > (57.6/2) kN.m - 167 is the effective depth for the top reinforcement in this direction. - 28.65 KN.m is the column moment from hardy-cross method (moment distribution table). - Mt max should not be less than half the design moment obtained from the equivalent frame analysis if Mt max is calculated to be less than this, the structural arrangements should be changed.
Note
1.3.3 Design of positive moment reinforcement: At span AB , M = 175.3 kN.m , βb = 0.7 K’= 0.402*(βb-0.4)-0.18(βb-0.4)2 K′ = 0.402*(0.7-0.4) – 0.15(0.7-0.4)2 = 0.1044 K’ will be used instead of (k=0.156)
Note
M+ve = 175.3 KN.m/Strip Variation in positive moments are not much (don’t reach the half) so for construction purpose we will design for the maximum moment and generalize it for all zones.
Note
For column strip 55% M = 0.55*175.3 = 96.4 kN.m/column strip width M=
= 45.4 kN.m/m
For middle strip 45% : M = 0.45*175.3 = 79 kN.m / middle strip width M=
= 37.2 kN.m/m
1.3.3.1 Bottom reinforcement at column strip: c.s width = 2.125m , M = 45.4 kN.m/m k=
= 0.064 < k′
=
= 0.5*√ As =
= 0.92 =
= 668.2 mm2
As min =0.13b*h/100=0.13*1000*200/100=260mm2/m
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Spacing : *1000 =
*1000 = 169mm
Use Ø12@150mm c/c Bottom reinforcement at column strips width (N-S)
1.3.3.2 Bottom reinforcement at middle strip: M.s width = 2.125 m , M = 37.2kN.m/m k=
= 0.0525 < k′
=
= 0.5+√ As =
= 0.938 = 537 mm2 > Asmin
=
Spacing : *1000 =
*1000 = 210.4 mm
Use Ø12@200mm c/c Bottom reinforcement at middle strips width (N-S)
1.3.4 Design of negative moment reinforcement: 1.3.4.1 At supports (B, C & D): Take the largest value = 182 kN.m/strip width (At the Cantilever). 3.4.1.1 Top reinforcement at Column strips: M – ve= 0.75 *182 = 136.5 kN.m/column strip width d = 200-25-(16/2) = 167mm = 0.09 < k′
k=
= 2110 mm2/column strip width
= 0.887 , As = Note
According to clause (3.7.3.1) two-third of the moment reinforcement should be placed in a width equal to the half of the column strip.
As = =*2110 = 1406.8 mm2 1406.8mm2 should be placed at (1.0625m) Spacing : *1062.5 =
*1062.5 = 151.8 mm
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1.3.4.1.2 Top Reinforcement at middle strips: M –ve = .25*182 = 45.5kN.m/m.s width M –ve = k=
= 21.4 kN.m/m = 0.03 < k′
=
= 0.5+√ As =
= 0.96 use (0.95) = 305 mm2
=
Spacing : *1000 =
*1000 = 367 mm
Use Ø12@350mm c/c Top reinforcement at middle strips width (N-S) Top reinforcement at support D for moment 182 will be generalized for support (B&C) for moment (153.2&127.1) KN.m respectively cause variation in moment are not much and also for construction simplification.
Note
1.3.4.2 At support A: 1.3.4.2.1 Top reinforcement at column strips:
M = 28.35 kN.m/ strip width For column strip = 0.75*28.35 = 19.85 kN.m/c.s width k=
= 0.013 < k′
=
= 0.5+√ As =
= 0.95 = 285mm2
=
(2/3 *285= 190mm2) Should be placed at 1.0625 m As min = 0.13b*h/100 = 0.13*1062.5*200/100 = 276.25mm2/m --- ok Spacing : *1000 =
*1062.5 = 773.1 mm
Check cracks:3d= 3*167 = 501 mm
Use Ø16@any spacing less than 500 mm match in construction later for column strips width (N-S) Use Ø12@350mm c/c at middle strips width (N-S)
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Direction (E-W):
All moments are in KN.m Consider the Internal strip of the width 4.25m for all other strips. Take M = 150.13 kN.m/strip width for all spans d =h-c-Ø- Ø/2 = 200-25-12-6 = 157 mm
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1.3.5 Design of positive moment reinforcement: 1.3.5.1 Top reinforcement at column strips: M +ve = 0.55*150.13 = 82.6kN.m/c.s width M +ve = k=
= 40.7 kN.m/m = 0.066 < k′
=
= 0.5+√ As =
= 0.92 = 644.8 mm2 > As min
=
Spacing : *1000 =
*1000 = 175.2 mm
Use Ø12@175mm c/c bottom reinforcement for all column strips widths (E-W)
1.3.5.2 Bottom reinforcement at middle strips: M +ve = 0.45*150.13 = 67.56kN.m/m.s width M +ve = k=
= 23.5 kN.m/m = 0.038 < k′
=
= 0.5+√ As =
= 0.955 --- use (0.95) = 360.5 mm2
=
Spacing : *1000 =
*1000 = 313 mm
Use Ø12@300 mm c/c Bottom reinforcement for all middle strips widths (E-W)
1.3.6 Design of Negative moment reinforcement: 1.3.6.1 (At support 2 & 3): M = 147 KN.m/strip width 3.6.1.1 Top reinforcement at column strips: M –ve = 0.75*147 = 110.25kN.m/C.S width k=
=
= 0.091 < k′
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= 0.5+√ As =
= 0.886 = 1885.7 mm2
=
(2/3 *1885.7 = 1257.2) should be placed at 1.0625 Spacing : *1000 =
*1062.5 = 169.8 mm
Use Ø16@150 mm c/c Top Reinforcement for all column strips widths at supports 2&3
1.3.6.1.2 Top reinforcement at middle strips: M –ve = 0.25*147 = 36.75 kN.m/middle strip width M -ve =
= 12.8 kN.m/m
k=
= 0.0218< k′
=
= 0.5+√ As =
= 0.95 = 201.5 mm2 < As min --- use (As min )
=
Spacing : *1000 =
*1000 = 434 mm
Check cracks : 3d = 3*151= 453
Use Ø12@350 mm c/c Top reinforcement at middle strips widths at supports 2&3 (E-W)
1.3.6.2 (At support 1&4): M –ve = 67.3 kN.m 1.3.6.2.1 Top Reinforcement at column strips: M –ve = 0.75*67.3 = 50.5 kN.m/c.s width k=
= 0.041< k′
=
= 0.5+√ As =
= 0.95 =
= 805.6 mm2
(2/3 *806.6 = 537) should be placed at 1.0625
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Spacing : *1062.5 =
*1062.5 = 397 mm
Use Ø16@300 mm c/c Top Reinforcement at column strips widths at supports 1&4 (E-W)
3.6.2.2 Top reinforcement at middle strips: M –ve = 0.25*67.3 = 16.825 kN.m/middle strip width M -ve =
= 5.85 kN.m/m (less than the minimum moment)
Asmin =
= 260 mm2
=
Spacing : *1000 =
*1000 = 434 mm
Check cracks : 3d = 3*151= 453
Use Ø12@350 mm c/c Top reinforcement at middle strips at Supports 1&4 (E-W)
1.4 Checks: 1.4.1 check for shear: Consider the critical internal column
Direction N-S:
Shear force =274.6+239.5 = 514.1 kN “at support B”
Direction E-W:
Shear force =262.6+235.5 = 498.1 kN “at support 2&3” -
Direction N-S is critical The loaded area on column is equal to (5*4.25 =21.25 m2)
1.4.1.1Punshing shear at column face (3.7.6.4): Applied shear stress: V= Veff = Vt Or maybe taken as 1.15 Vt conservative value for internal columns Vt =
(21.25-0.1) = 512 kN
V = *hc = 4*0.3568= 1.427 m (effective parameter)
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= 159 mm (effective depth of the two directions “average’’ )
d=
= 2.59 N/mm2
V= Resistance : 0.8√
= 4 N/mm2
= 0.8*√
5 N/mm2 (whichever lesser) 3.3 N/mm2 < 4 N/mm2
------------satisfied
1.4.1.2 Punching shear at 1.5d from the column face (3.7.7.1)
Applied shear stress : V= Veff = Vt Vt =
(21.25-0.6955) = 497.3 kN
Mt = (from table 1 – moment distribution table) = 18.6 + 18.6 = 37.2 kN.m X = hc+3d = 0.3568+3*0.159 = 0.834 m Veff = 497.2
Mft reduced by 70% -- (clause 3.7.6.2).
Note
V=
= 544.03 kN
= 0.97 N/mm2
=
Resistance: Vc =
(
As =(
)1/3 ( +
=
)1/4
)1/3 )/2 = 1340 mm2
= 0.843 < 3 -------ok
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(
)1/4 = (
)1/4 = 1.26 > 0.67 ------ok
)1/3 =
)1/3 = 1
Note
m = an allowance for ( ( clause 3.7.6.4)
m =1.25
has been made in the equation of Veff , so it will be omitted here
Vc = 0.79*0.8431/3 *1.25* 1 = 0.94 N/mm2 < 0.97 N/mm2 --------- not satisfied One of the following options may be used: 12345678-
Use a drop panel. Use a column head. Use shear reinforcement. Increase the slab depth. Increase column dimensions. Increase top reinforcement. Increase characteristic strength of concrete. Decrease the cover if possible.
Fcu = 30 N/mm2 will be used instead of 25 N/mm2 (Optimum solution) Vc = 0.79*0.8431/3 *1.26*
)1/3 = 0.999N/mm2 > 0.97 N/mm2 ------------satisfied
No need to recalculate after changing in f cu
Note
1.4.2 Check for deflection: )actual ≤
)limit
)limit =
)Basic * Mft*0.9
)actual = 5000/169 =29.58 mm )Basic = 26
(Table 3.9) )≤2
Mft = 0.55+ fs =
)*fy
)=
M/bd2 =
)*460
) = 280.36 N/mm2
= 1.444
Mft = 0.55+
) = 1.249 < 2 ------------ok
)limit = 26*1.249*0.9 = 29.2 mm ----- Not satisfied - In punching shear calculation we changed fcu to be 30 N/mm2, this will lead to minimize As required in column strip & middle strip. At column strip : K=
= 0.053 ,
= 0.937
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As required = M/0.95fy z = 45.4*10 6/(0.95*460*0.937*169) = 656 mm2 At middle strip : K=
= 0.0439
= 0.948 As required = M/0.95fy z = 37.2*10 6/(0.95*460*0.937*169) =531 mm2 Fs new =
)*460
) = 276.1 N/mm2
Mft = 1.264 )limit = 26*1.264*0.9 = 29.57 mm ----- also not satisfied
Use Ø12@175mm c/c Bottom reinforcement at middle strip width with 645.7mm 2 provided area. Fs =
)*460
Mft = 0.55+
) = 260.2 N/mm2 ) = 1.32
)limit = 26*1.32*0.9 = 30.88 mm The slab is finally satisfied with respect to deflection.
1.4.3 Check for Cracks ( Clause 3.12.11.2.7 ): - Maximum spacing = 350 mm d= (167+151)/2 = 159 mm 3d = 477 mm . 350 mm < 3d < 750 mm ------ ok Grade 460 N/mm2 is used, slab depth not exceed 200 mm. No further checks required.
1.5 Curtailment of reinforcement (clause 3.12.10): Top reinforcement: According to the simplified detailing in figure (3.25) of the code, all bars need to continue up to 0.15L from the face of the column on either side of the column and then only 50% of the steel need to continue to a distance of 0.3L from column face. Bottom reinforcement: According to the same figure only 40% of the steel need to go over the support. This can be achieved by curtailing alternate bars at 0.2L from the center of the column, but at edges spans curtail bars at 0.1L from the center of the column.
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Note
- The simplified detailing rules is only used when t (3-12) is used, otherwise its important to figure out the point of counter flexure and pass the reinforcement distance d after it in both directions. - The arrangement of panels division (column and middle strips) will be used to simplify curtailment as follows: Half C.S = Lx/4 = 0.25L (from the centre) > 0.15L (from the column face) . Hence Its convenient to continue all the bars to distance 0.25L from the centre of the column for construction purposes. - For bottom bars its more convenient to pass 50% of the bars instead of 40% for construction simplification. - The spacing of reinforcement should be arranged for continuity and practical purposes.
1.6 Drawing details: - Figure “1” Bottom reinforcement “direction N-S” - Figure “2” Bottom reinforcement “direction” E-W” - Figure “3” Top reinforcement “direction N-S” - Figure “4” Top reinforcement “direction” E-W”
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2. Design of beams: 2.1 Dimension requirements: According to architectural design the breadth of the beam will be taken as 0.2 m. The depth of the beam is generally be controlled by consideration of deflection, but a value of D = (1.6-2.5)B will be in a suitable range to control deflection. Take D = 2.5B =25*0.2 = 0.5 m
2.2 Estimation of loads: Consider the western beam (the critical one)
2.2.1 Beam self-weight: S.W = Density *section dimensions = 24*0.2*0.3 = 1.44 kN/m
2.2.2 Partition weight: Weight = density*breadth*height = 19.2*0.2*3 = 11.52 kN/m Ultimate design load = 1.4(1.44+1.2*11.52) =21.37 kN/m Note
20% increase in partition weight for transverse partitions.
2.2.3 Slab loads: Slab loads excluding partitions = 22.16-1.4*5 = 15.16 kN/m Slab loads per meter = 15.16 *2.125 = 32.22 kN/m Total weight = 21.37 + 32.22 = 53.6 kN/m
2.3 Analysis:
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Provisions of clause 3.4.3: a) Imposed load may not exceed dead load. b) Load should be uniformly distributed over three or more spans. c) Variation in span length should not exceed 15% of the longest. Since the provisions of (clause 3.4.3) are met, table 3.5 may be used to obtain bending moments and shear forces. Coefficient from (Table 3.5):
F = 53.6*5 = 268 kN , L = 5 m
Bending moments diagram:
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Shear Forces diagram:
Note
- No redistribution of moments from this table should be made. - The design will be generalized for all other beams because it gives the greatest values in the analysis for bending moments and shear forces.
2.4 Design of the Edge beams: 2.4.1 Design of moment reinforcement: 2.4.1.1 Design of positive moments: - Maximum positive moment = 121.1 kN.m - The beam will be designed as L section The effective width of the flange of the L-beam is given by lesser of : a. Actual width = 2125 mm b. bw +Lz/10 = 200 + 0.7*5000/10 = 550 mm Lz = distance between points of zero moment (clause 3.4.1.5) So (b) will be taken as 550 mm From table 3.3 and 3.4 the value of the cover is 25 mm, assuming Ø 16 bars, the effective depth d is estimated as : d = h-c- Ølink- Ø/2 = 500-25-8-16/2 = 459 mm
All dimensions are in (m)
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Check for depth of the stress block: The moment of resistance of the section when the block is equal to the slab depth is given by: Mflange = 0.45fcu*b*hf*(d-hf/2) Mflange = 0.45*30*550*200*(459-200/2)*10-6 = 536 kN.m Applied moment = 121.1 kN.m < 536 kN.m The stress block is inside the slab, the beam can be designed as a rectangular section with (b=550 mm) K=
=
= 0.0348
= 0.5+√
=0.959
As = =
-- use 0.95 = 635.5 mm2
=
= 0.36 < 0.9 and for Fy = 460 mm2:
Minimum tension steel = 0.18% (see Appendix) Asmin =
= 180 mm2 < AS --- ok
=
Number of bars =
=
= 3.16 bars
If 4 bars is used in one row spacing will be equal to: (500-25*2-8*2-4*16) /3 = 23.3 mm < (Hagg + 5 mm ) --- not acceptable 4 bars will be used in two rows with 25mm vertical spacing Effective depth for the upper row = 500-25-8-16-25-8 = 418 mm The average depth will be = (459+418)/2 = 438.5 mm K ==
= 0.038 , = 0.5+√
=0.95
= 665.2 mm2
As =
Number of bars = 665.2 /201 = 3.3 Use 4T16 with 804 mm2 provided Area bottom reinforcement for all beams
2.4.1.2 Design of negative moments: Maximum negative moment = 147.4 kN.m Section will be designed as rectangular one. K=
=
= 0.128 < 0.156 ,
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= 929 mm2 > Asmin
As = - Number of bars =
= 4.62
- clear horizontal spacing = (200-25*2-8*2-3*16)/2 = 43 mm > ( hagg+5 mm) ----ok Use 5T16 with 1005mm2 provided Area Top Reinforcement for all beams
2.5 Design for Torsional moment: According to clause 3.4.5.13 calculations for torsional moment are not necessary in framed construction Torsional cracking will be adequately controlled by shear reinforcement.
2.6 Checks: 2.6.1 Check for shear at column face: Maximum shear force = 160.8 kN Maximum shear stress =
= 1.833 N/mm2 < (0.8√
=
= 4.38N/mm2) ---- Satisfied
2.6.2 check for shear at 1.0 d from the column face : The shear at distance (d= 438.5mm) from the column face = 160.8-53.6*(0.4385+0.5/2) = 123.9 kN The enhancement of shear strength near the support using the simplified approach given in clause ( (3.4.5.10) is taken into account.
Note
The Applied shear: V = v/(bd) = 123.9*103/(200*438.5) = 1.41 N/mm2 The resistance : Vc=
(
)1/3 (
)1/4
)1/3
The effective area of steel at d from the face of the column is 5T16 with provided Area = 1005 mm2 (
)=(
(
)1/4 = (
Vc =
) = 1.146 Vc not ok (shear reinforcement required) .
2.7 Design of shear reinforcement: (Vc + 0.4) < V < 0.8√
, (0.702+0.4) < 1.41 < 4.38
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From table 3.7 (see Appendix c ) ≥ ≥ 200(1.41 - 0.702)/(0.95*250) using 8 mm two legs links spacing required is: ≥ Sv ≤ 168.5 mm < (0.75d = 328.8 mm) Use Ø 8 mm link @150 mm c/c The top layer of bars are curtailed at (0.15L = 750mm) from the column face only 3 bars will be continued to a distance (0.25L = 1250 mm ). (see figure 3.24-Appendix c ) Vc with 3T16: Vc =
)1/3 (1)1/4
(
)1/3 = 0.592N/mm2
Shear force at distance 0.75+d from the column face: F = 160.8 – 53.6(0.75+0.5/2 + 0.4385) = 83.7 KN V = v/bd = 83.7*103/200*438.5 = 0.954 N/mm2 From (Table 3.7) : 0.5Vc < v < Vc + 0.4 0.5*0.592 < 0.954 < 0.952 + 0.4 , minimum links may be provided. ≥
,
≥
Sv ≤ 298.3 mm < (0.75d = 328.8mm) Use Ø 8 mm link with 275 mm c/c after distance 0.75 m from the column face (see drawing details).
2.8 Deflection check: The deflection of the beam is checked using the rules given in (clause 3.4.6) referring to table 3.4 of the code: 200 mm , b 550 mm = 0.3
( )basic = 20.8
= 0.3636
( )basic = x
= 1.0
( )basic = 26 , x = 20.8 +
( )actual ≤ ( )limit ,
(26-20.8) = 21.27
( )limit = ( )basic *Mft*Mfc
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The modification factor for tension reinforcement Mft : ≤ 2
Mft = 0.55+ fs (Service stress) = 2/3 *fy * M/bd2 =
= 2/3 *460 *
= 242.4 N/mm2
= 1.145
Mft = 0.55+
= 1.5 < 2 ---- ok
Modification factor for compression reinforcement Mfc : 2 bars will be continued along the beam for construction purpose Mfc = 1 +(
/(3 +
)) Formula in Table 3.11
As’ prov = 2T16 = 402 mm2 =
= 0.166
Mfc =1+(0.166/(3+0.166)) = 1.052 < 1.5 Allowable Ratio: ( )limit = 21.27*1.5*1.052 = 33.56 mm ( )actual = (5000/438.5) = 11.4 mm < 33.56 mm Hence the beam is very satisfactory with respect to deflection .
2.9 Check for Cracking: 2.9.1 Minimum spacing: The clear distance between bars on the tension face is = 43 mm > (hagg +5 mm) 2.9.2 Maximum spacing: 43 mm doesn’t exceed 155 mm as per table (3.28) The distance from the side or bottom to the nearest longitudinal bar is 25+8+16/2 = 41 mm The clear distance from the corner to the nearest longitudinal bars: √
-16/2 = 66 mm < (155/2 = 77.5 mm )
This beam is satisfactory with regard to cracking.
2.10 Curtailment of Reinforcement: See figure 3.24 -simplified detailing rules for beams- (Appendix C).
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2.11 Beams detailing: (All bars are Ø 16) Edge beams parallel to Direction (N-S):
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3. Analysis & Design of the Columns: Design Information: Structure Type: Braced (No lateral load ). End condition: The end of column is connected monolithically to slab on either side at interior columns, and by edge beam at edge and corner columns on the upper side and to slab at the bottom side. Suggested section: (0.2*0.5)m (the minimum section for the slab to be adequate in punching shear). Number of stories: Four stories. Load on floors: 22.16 kN/m2 Storey height: 3 m (clear height). Characteristic strength of reinforcement: 460 N/mm2 Characteristic strength of concrete: 30N/mm2 Exposure condition: moderate. Fire resistance period: 1.5 hours. Maximum aggregate size: 25 mm
3.1 Design of the interior columns: (C1) The Figure below is the largest loaded area on an interior column. Maximum loaded area on a column is: 4.25*5 = 21.25 m2
Plan
Elevation
3.1.1 Estimation of loads: Load on floors= 22.16 kN/m2 Self-weight of the column = 24*(0.2*0.5*3)*1.4 = 10 KN. Load at position N1 : (see figure above) N1 = 22.16*3*21.25+10*3 = 1442.7 kN Load at position N2: N2 = 22.16*21.25*5+10*5 = 2404.5 kN
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3.1.2 Design of the column: Column classification: From table (3.19) End connection at top: -------- “condition 2” End connection at bottom: ---- “condition 2” Value of
= 0.85
=
= 5.1 < 15 , short in this direction.
=
= 12.75 < 15 , short in this direction also.
The column is short Note
inc
Since the spans are equal in each direction, equation 38 in clause 3.8.4.3 may be used. The equation includes a reduction equal to 10% due to nominal eccentricity that may happen in construction.
N = 0.4fcuAc+0.8Asc fy
(Equation 38)
3.1.2.1 Design for the maximum load at the base ( N2 = 2404.5 kN ). Substitute in equation above: 2404.5*103 = 0.4*30(300*500 - Asc) + 0.8 *460 *Asc Asc = 3383.4 mm2 Asmin = Asmax =
= 400 mm2 --------ok
=
= 6000 mm2 --------ok
By using Ø 16 with As = 201 mm2 Number of bars required =
= 16.8 bars (Too much bars in this section ).
Increase the column dimensions: Try (200*600) mm2 2404500 = 0.4*30*(200*600 - Asc) + 0.8*460*Asc Asc = 2709.3 mm2 Number of bars =
= 13.4 bars, Use 14T16 .
Design of links: Ølink = ¼ Ømax but not less than 6 mm Ølink = 16/4 = 4 mm, use Ø6 mm . Spacing = 12Ømin = 12*16 =192 mm , Use Ø6 @ 175 mm c/c
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Section details:
section 1-1
3.1.2.2 Design of the section at (N1 = 1442.7 KN ) Substitute in the main equation 1442.7*103 = 0.4*30*( 200*500 – Asc ) + 0.8*460*Asc Asc = 681.7 mm2 > Asmin Number of bars =
= 3.4 bars Use 4T16
Design of links: Ølink = 16/4 = 4 mm, use Ø6 mm Spacing = 12*16 =192 mm Use Ø 6mm @ 175 mm c/c Section details:
section 2-2
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3.2 Design of the edge column: (C2) Consider the critical edge column:
Plan
Elevation
Maximum loaded area = 4.25*2.5 = 10.625 mm2 Note
The spans in direction E-W are equal. The column is subjected to Axial load and moment about minor axis.
3.2.1 Estimations of loads and moments: Loads on floors = 22.16 kN/mm2 Self-weight of the column =(0.2*0.5*2.7)*24*1.4 = 9 kN Self-weight of the beam =(0.2*0.3*4.25)*24*1.4 = 8.57 kN
3.2.2 Obtain values of Ns: N1 = load on floor *loaded area by column -
N1 = 22.16*10.625 = 235.45 kN N2 = N1 +beam s.w +s.w of the column = 235.45+8.57+9 = 253.02 KN N3 = N2+slab weight =253.02 + 235.45 = 488.47 kN N4 = 488.47 +8.57+9 = 506.04 kN N5 = 506.04 +235.45 = 741.49 kN N6 = 741.49 +8.57+9 = 759.06 kN N7 = 506.04 +235.45 = 994.51 kN N8 = 994.51 +8.57+9 = 1012.08 kN N9 = 1012.08 +235.45 = 1247.53 kN N10 = 1247.53+8.57+9 = 1265.1 kN
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3.2.3 Obtain moments: Individual sub-frames will be used to obtain moments
At Roof:
Individual Sub-frame
Relative stiffnesses: *0.5 = 2.833*10-4
Ks = 0.5 =
= 1.111*10-4
Kc= = Note
The beam is being ignored for safer and more simple analysis.
∑ K = Ks+Kc = 2.833+1.111 = 3.944 Distribution factors: Dc = Kc/∑K = 1.111/3.944 = 0.282 Fixed end moments: M=
=
= 196.25 kN.m
The column moment = 0.282*196.25 = 55.4 kN
At lower points:
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Relative stiffnesses: Ks = 2.833*10-4 , Kupper column =1.111*10-4 , Klower column =1.111*10-4 , ∑k = 5.055 Distribution factors: Dupper column = 1.111/5.055 = 0.22 Fixed end moment = 196.25 kN.m Mupper column = Mlower column = 0.22*196.25 = 43.2 kN.m
At ground floor level:
Relative stiffnesses: Ks = 2.833*10-4 , Kupper column =1.111*10-4 , Klower column =
= 1.67*10-4
∑k = 2.833+1.111+1.67 = 5.614 Distribution factors: Dupper column = 1.111/5.614 = 0.2 Dlower column = 1.67/5.614 = 0.3 Fixed end moment = 196.25 kN.m Mupper column = = 0.2*196.25 = 39.25 kN Mlower column = 0.3*196.25 = 58.9 kN.m M at base = 0.5(M lower column) = 0.5*58.9 = 29.45 kN.m
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Final Axial loads and moments:
3.2.4 Design of the columns sections: Column classification: From (table 3.19) End connection at top: (direction x-x “condition 1” – direction y-y ‘‘condition 2’’) End connection at bottom: (direction x-x “condition 1” – direction y-y ‘‘condition 2’’) Value of
1
= 0.75 ,
= =
2
= 0.85
= 4.06 < 15 , short in this direction = 12.75 < 15 short in this direction also.
The column is short Note
The column will be designed to withstand an axial load and moment about its minor axis.
Fcu = 30 N/mm2 , Fy = 460 N/mm2 , c = 30 mm , d = 200-30-6-8 = 156 , = 0.78 Note
Interpolation from chart No. 26 and chart No.27 will be made.
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For section 1-1:
Take N1 = 235.45 kN , M1 = 55.4 kN.m =
= 2.35 ,
=
From chart No. 26:
=1.8
From chart No. 27:
=1.6
= 2.8
Interpolation between the two values: = 1.6 + Asc =
=
Number of bars =
*(1.8-1.6) = 1.72 = 1720 mm2 = 8.55 bars Use 10T16 with Ø6@175mmc/c
Note
At lower point N2 increase by 17.57 kN but the moment decreases by 12.2 kN.m., no need to check the section here and 10T16 will be satisfied
section 1-1 For section 2-2
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Take N8 = 1012.08 kN , M = 43.2 kN.m =
= 9.94 ,
From charts: Asc =
=
= 2.16
= 0.84 = 840 mm2
=
Number of bars =
= 4.18 bars Use 6T16 with Ø6 @ 175mmc/c
Section details:
section 2-2 For section 3-3:
Upper values are critical Take N9 = 1247.53 kN , M = 58.9 kN.m =
= 12.47
=
= 2.95
From charts Asc =
=
= 2.2 = 2200 mm2
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Number of bars =
= 10.9 bars Use 12T16 with Ø6@175mmc/c
Section Details:
section 3-3
5.3 Design of the corner columns: (C3)
Note
The corner column is subjected to axial load and moments about both axis ( Biaxial Bending ).
5.3.1 Estimation of loads: Loads on floors = 22.16 kN/m2 Self-weight of columns = (0.2*0.5*2.7)*24*1.4 = 9 kN Self-weight of the beam = (0.2*0.3*(2.5+2.125)*24*1.4 =9.33 kN 5.3.1.1 Obtain values of Ns: -
N1 = 22.16*5.3 + 9.33 = 126.8 kN
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-
N2 = 126.8 + 9 = 135.8 KN N3 = 135.8 + 126.8 + 9.33 = 271.93 kN N4 = 271.93 + 9 = 280.93 kN N5 = 280.93 + 126.8 + 9.33 = 417.06 kN N6 = 417.06 + 9 = 426.06 kN N7 = 426.06 +126.8 + 9.33 = 562.19 kN N8 = 562.19 + 9 = 571.19 kN N9 = 571.19 +126.8 + 9.33 = 707.32 kN N10 = 707.32 + 9 = 716.32 kN
5.3.2 Analysis: 5.3.2.1 Obtain moments: Note
Individual sub-frames will be used to obtain moments in each direction.
First: moments about the major axis: At roof:
w = 53.6 kN/m ( calculated before in beams ). Note
In general loads at roof is lesser than at floors, but for special purpose (say client requirements) the roof will be designed to withstand additional loads (same loads as floor).
Relative stiffnesses: Ks = 0.5 = Note
*0.5 = 2.083*10-4 , Kc = =
= 7.72*10-4
It is more accurate to obtain the stiffness for a T section , but we considered the beam stiffness only, a part of slab is ignored for simpler and safer analysis.
∑ K = Ks+Kc = 2.083+7.72 = 9.803 Distribution factors: Dc = Kc/∑K = 7.72 /9.803= 0.79 Fixed end moments: M=
=
= 111.67 kN.m
The column moment = 0.79*111.67 = 88.32 kN
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At lower points :
Relative stiffnesses: Ks = 2.083*10-4 , Kupper column = Klower column = 7.72*10-4 ∑k = 7.72*2+ 2.083 = 17.523 Distribution factors: Dupper column = 7.72/17.523= 0.44 Fixed end moment = 111.67 kN.m Mupper column = Mlower column = 0.44*111.67 = 49.13 kN.m
At ground floor level:
Relative stiffnesses: Ks = 2.083*10-4 , Kupper column =
= 7.716*10-4 , Klower column =
= 12.255*10-4
∑k = 22.054 Distribution factors: Dupper column = 7.716/20.054= 0.35 , Dlower column = 12.255/20.054= 0.555 Fixed end moment = 111.67 kN.m
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Mupper column = = 0.35*111.67 = 39.1 kN Mlower column = 0.555*111.67 = 62 kN.m M at base = 0.5(M lower column) = 0.5*62 = 31 kN.m
Second: moment obtained about the minor axis: Load acting on the frame = ( Beam self-weight + partition weight + slab loads ) Load acting on the frame = ( 24*0.2*0.3*1.4 + 1.4*0.2*2.7*19.2*1.2 + (22.16 – 5*1.4) = 57.3 KN/m At roof:
Relative stiffnesses: Ks = 0.5
=
*0.5 = 2.45*10-4 = 1.235*10-4
Kc = =
∑ K = Ks+Kc = 2.45+1.235 = 3.685 Distribution factors: Dc = Kc/∑K = 1.235 /3.685= 0.335 Fixed end moments: M=
=
= 86.3 kN.m
The column moment = 0.335*112.74 = 28.91 kN.m
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At lower points:
Relative stiffnesses: Ks = 2.45*10-4 , Kupper column = Klower column =
= 1.234*10-4
∑k = 1.234*2+ 2.45 = 4.918 Distribution factors: Dupper column = Dlower column = 1.234/4.918= 0.251 Fixed end moment = 86.3 kN.m Mupper column = Mlower column = 0.251*86.3 = 21.7 kN.m
At ground floor level:
Relative stiffnesses: Ks = 2.45*10-4 Kupper column =
= 1.234*10-4 , Klower column =
= 1.96*10-4
∑k = 2.45+1.234+1.96 = 5.644 Distribution factors: Dupper column = 1.9234/5.644= 0.219 , Dlower column = 1.96/5.644= 0.347 Fixed end moment =
= 86.33 kN.m
Mupper column = = 0.219*86.33 = 18.91 kN.m Mlower column = 0.347*86.33 = 29.96 kN.m
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M at base = 0.5(M lower column) = 0.5*29.96 = 14.98 kN.m
Final results:
5.3.3 Design of the corner column: Column classification: From (table 3.19) End connection at top:
“condition 1”
End connection at bottom:
“condition 1”
Value of
= 0.8
=
=
= 4.06 < 15 , short in this direction.
=
=
= 10.12 < 15 short in this direction also.
The column is short
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The column is subjected to axial load and biaxial bending ( Clause 3.8.4.5 ), if: ≥
, Mx′ = Mx+
My
<
, My′ = My+
Mx
Obtain the value of h' and b' ( figure 3.22 – see Appendix ) h′ = 500-30-6-16/2 = 456 mm b′ = 200-30-6-16/2 = 156 mm For section 1-1:
Take N1 = 126.8 kN , Mx = 88.32 kN.m , My = 28.91 kN.m Obtain the value of coefficient β (Table 3.22 ) =
= 0.042
Interpolation between two values Β = 0.88 +
* ( 1 – 0.88 ) = 0.95
Substitute on the equations: =
= 0.1936 ,
=
, Mx′ = Mx+
> =
= 0.1853 My = 88.32 +0.95(456/156)*28.91 = 168.6 KN.m
= 1.268 ,
=
= 3.37
Fy = 460 N/mm2 , fcu = 30 N/mm2 , = From chart: Asc =
=
= 0.9 ( chart No 29 will be used ).
= 1.6 =
Number of bars =
= 1600 mm2 = 7.9 bars Use 8 T16 with Ø 6mm link @175mmc/c
Section at the bottom: =
= 0.1077 ,
=
= 0.139
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, My′ = My+
<
Mx
My′ = 21.7 + 0.95(156/456)*49.13 = 37.67 KN.m =
= 1.358 ,
=
= 1.88
Fy = 460 N/mm2 , fcu = 30 N/mm2 =
=
= 0.78 ( Interpolation between charts 26& 27 will be made )
From charts: Asc =
= 1.1
=
Note
= 1100 mm2 , Number of bars =
= 5.47 bars
Section at top is critical and it will govern the column design. But we have to make sure that there is at least 3 bars in each side at the longer dimension for the section to be adequate at bottom.
Use 10 T16 with Ø 6mm link @175mmc/c Section details:
Section 1-1 For section 2-2:
Take N9 = 707.32 kN , Mx = 62 kN.m , My = 29.8 KN.m Obtain the value of coefficient β (Table 3.22 ) =
= 0.236
Interpolation between two values
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Β = 0.65 + = <
* ( 0.77 – 0.65 ) = 0.727
= 0.135 ,
=
, My′ = My+
=
= 7.07 ,
= 0.191 Mx = 29.8 + 0.727(156/456)*62 = 45.22 KN.m =
= 2.26
Fy = 460 N/mm2 , fcu = 30 N/mm2 , = From charts: Asc =
Note
=
= 0.78 ( Interpolation between charts 26& 27 will be made )
= 0.9
=
= 900 mm2 , Number of bars =
= 4.47 bars
-No need to check the section at the base values of moment are exactly the half. -This section will be generalized at all stories except the upper one ( 3 rd floor column ).
Use 6 T16 with Ø 6mm link @175mmc/c Section details:
Section 2-2
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5.3.4 Drawing Details:
Column/Level
Short Column
Ground Floor
1st floor
2nd floor
3rd floor
C1
C2
C3
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4. Design of the staircase: Design information:
Type of stair: spanning longitudinally Support condition: stair is not constructed monolithically with the supports, Type of supports: stair is supported on the landing slab on side and on the flat slab on the other one. Thickness of both slabs: 200 mm Characteristic strength of concrete: 30 N/mm2 Characteristic strength of steel reinforcement: 460 N/mm2 Exposure condition: moderate. Fire resistance: 1.5 hours Dimension of the stair: Riser 160 mm , Tread 300 mm, waist: 200 mm, breadth: 1000 mm ( see figure below )
4.1 Estimation of loads: 4.1.1 Dead load: -
Weight of waist = 24*0.2*3.4*1 = 16.32 kN Weight of steps = 24*0.5*0.16*0.3*10*1 = 5.76 kN Weight of tiles = 25*(3+1.6)*0.02 *1 = 2.3 kN Weight of plaster = 20*3.4*0.02 *1 = 1.36 kN
Total dead load = 25.74 kN
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4.1.2 Live loads: -
Live load = 3*3*1 = 9 KN
Ultimate Design load = 1.4*D.L+1.6*L.L = 1.4*25.74+1.6*9 = 50.44 KN The dead load is calculated using the slope length while the imposed load acts on the plan length.
Note
4.2 moment reinforcement: 4.2.1 Longitudinal Reinforcement: With no effective end restrain: M= The Effective span = distance between supports + effective depth (clause 3.10.1.4) When we design the slabs to act as a support to the stair, we must take into account the additional moment that may happen due to settlement (Deflection) at them.
Note
Effective depth = 200 – 25 – 12/2 = 169 mm The Effective span = 3000 + 169 =3169 mm M= K=
= 20 KN.m =
= 0.0233
= 0.5+√
= 0.97, use 0.95
As = Asmin =
= 285 mm2
=
= 260 mm2
=
Spacing =
* 1000 =
*1000= 396.4 mm
Number of bars = (1000/333) + 1 = 4 bars Use Ø12 @ 333 mm c/c (longitudinal Reinforcement). 4.2.2 Transverse distribution steel: Use Asmin = 260 mm2 Spacing =
=
*1000= 434 mm Use Ø12 @ 350 mm c/c (Transverse Reinforcement).
4.3 Check for deflection: )actual ≤
)limit
)actual = 3200/169 = 18.9 mm
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)limit =
)Basic * Mft
)Basic = 20 simply supported span ( Table 3.19 ) Mft = 0.55+ fs =
)
)*fy
)=
M/bd2 =
)*460
) = 195.4 N/mm2
= 0.7
Mft = 0.55+ Note
) = 2.017 use 2 Since the stair flight occupies more than 60% of the span, a further increase of 15% is permitted.
)limit = 20*2*1.15 = 46 mm The stair is very satisfactory with respect to deflection.
4.4 check for shear: Applied V = v = w/2 = 50.44/2 = 25.22 kN = 0.148 N/mm2
= Vc =
(
Vc =
(
)1/3 ( )1/3 (
)1/4
)1/3 )1/4
)1/3 = 0.536 N/mm2
The stair is satisfactory with regard to shear.
4.5 Stair Details:
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5. Design of the foundations: Design information:
Soil investigation report: shallow isolated or combined footing may be used Safe bearing pressure: 200 kN/m2 Characteristic strength of concrete:35 N/mm2(minimum grade of concrete to be used on foundation Characteristic strength of steel reinforcement: 460 N/mm2 Exposure condition: moderate Cast statement: footing constructed on a blind layer of concrete Columns dimension: (200*600)mm2 for interior columns and (200*500)mm2 for others
5.1 design footing for interior columns: Suggested footing: pad footing. 5.1.1 Estimation of loads: For ultimate limit state: Consider the critical interior column Column design axial load = 2404.5 kN For serviceability limit state: Calculating the un factored load: D.L+L.L = 12.4+3 = 15.4 kN/m2 Loaded area carried by the column: 21.25 m2 Self-weight of the column = 8 kN Un factored load = 15.4 *21.25 + 8*5 = 1676.25 kN Assume weight for the base = 110 kN Total un factored load = 1676.25+110 = 1786.25 kN
5.1.2 Calculate the plan size for the footing using the permissible bearing pressure and the loading arrangement for the serviceability limit state: Required base area =
=
= 8.93 m2
Use a square footing B=h=√
=√
= 2.98 m
Provide a base 3m square with provided area 9 m2 Ultimate earth pressure = Note
= 271.2 kN/m2
271.2Kn/m2 > 200 kN/m2 settlement takes place during construction and working life of the structure..
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5.1.3 Assume a suitable thickness (h) and check the shear stress of the column face is not exceed 5 N/mm2 or 0.8√ whichever smaller Assume 600 mm thick footing d average = (d1+d2)/2 d1 = h-c-Ø/2 , d2 = h-c- Ø -Ø/2 c =50 mm ( footing constructed on a blinding layer of concrete) Assume Ø16 mm in design later d1 = 600-50-8 = 542 mm , d2 = 600-50-16-8 = 526 mm d average = (542+526)/2 = 534 mm Check the adequacy of the depth : Check for shear at column face = 3.22 N/mm2 < (4.73N/mm2 = 0.8√
V = N/ud =
)
Initial check for punching shear at 1.5 d from the column face: Punching shear stress: V=
=
u = critical parameter = (0.2+3*0.534)*2+ (0.6+3*0.534 )*2 = 8.01 m
Punching shear force v = maximum pressure (Base area-area within parameter) v = 271.2(3*3- (0.2+3.0534)*(0.6+3.0534)) = 1364.67 KN V = 1364.67*103/8010*534 = 0.32 N/mm2
Note
-This value is not excessive and it can be easily controlled even if the punching shear failed therefore h = 600 mm will be suitable. -The shear resistance Vc depends on the value of As which is not obtained yet, so we will make a final check for punching shear after calculating As.
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5.1.4 Bending reinforcement: (direction x-x) At the column face which is the critical section
M = 271.2*3*1.42 / 2 = 797.33 kN.m K=
=
= 0.0258
= 0.5 + √ As = Asmin =
= 0.97 take 0.95 = 3543.5 mm2
= =
= 2340 mm2 -- ok
Number of bars required = 3543.5/201 = 17.6 bars Provide 18T16 bar with 3618 mm2 The distribution of reinforcement is determined to satisfy the rule. (c1+3d1) = (600+3*542) = 1669.5 mm Lc1 = 3000/2 = 1500 mm Lc1 ≯ (c1+3d1) Reinforcement may be distributed uniformly in direction x-x
5.1.5 Bending reinforcement: (direction y-y)
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M = 271.2*3*1.22 / 2 = 585.8 kN.m K=
=
= 0.02
= 0.5 + √ As =
= 0.977 take 0.95 = 2682.6 mm2
=
Asmin =
= 2340 mm2 -- ok
=
(c2+3d2) = (200+3*526) = 1333.5 mm Lc2 = 3000/2 = 1500 mm Lc2 > (c2+3d2) 2/3 As should be placed at a distance equal to (c 2+3d2) Note
The bending moment tends to be somewhat higher towards the column than the other way from it .
*2682.6 = 1788.4 mm2 , 1788.4 should be placed at 1778 mm Spacing =
*1778 = 199.8 mm say 200 Provide Ø16 @200 mm c/c distribute uniformly in direction y-y
Number of bars required = 3000/200 = 15 bars As provided = 15 *201 = 3015 mm2
5.1.6 Vertical shear at 1.0 d from column face: Check direction x-x first
V = 271.6*3*0.858 = 699 kN v=
= Note
= 0.43 N/mm2 The bars extended to a distance more than (d) beyond the critical section, so the steel is effective in increasing the shear stress.
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obtain vc: = (
= 0.222 < 3 -----ok
)1/4 = (
)1/4 = 0.926 > 0.67 ------ok ( member without shear reinforcement )
)1/3 =
)1/3 = 1.119
(0.222)1/3 *0.926*
Vc =
= 0.396 N/mm2
v > vc ---- section fails As required to satisfy the section in shear 0.43 =
(
)1/3 *0.926*
As required = 4603.1 mm2 Number of bars 4603.1/201 = 22.9 Provide 23T 16 with 4623 mm2 provided area Vc =
(
)1/3 *0.926*
= 0.431
And hence the shear stress is satisfactory Check direction y-y:
V = 271.6*3*0.674 = 549.2 kN v=
= 0.348 N/mm2
=
Resistance:
(
=
= 0.191 < 3 -----ok
)1/4 = (
)1/4 = 0.9338 > 0.67 ------ok
)1/3 =
)1/3 = 1.119 , (fcu < 40)
Vc =
(0.191)1/3 *0.9338*
= 0.38 N/mm2
vc > v ------- the section is adequate in shear.
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5.1.7 Final check for punching shear: Punching shear stress was 0.32 N/mm2 Resistance stress: Vc =
(
Vc =
(
)1/3 (
)1/4
)1/3
)1/3 (
)1/4
)1/3 = 0.407 N/mm2
5.1.8 Cracking: The bar spacing does not exceed 750 mm and the flexural reinforcement supplied is 0.28 which is less than 0.3 % , No further checks required.
5.1.9 Drawing details:
-F1-
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5.2 Design of pad footing subjected to moment for edge columns: 5.2.1 Estimation of loads and moment : For the ultimate limit state : Column design axial load = 1265.1 kN Ultimate moment (My) = 29.45 kN.m For serviceability limit state calculating the un factored load: D.L+L.L = 12.4+3 = 15.4 kN/m2 Loaded area carried by the column = 10.625 m2 Average s.w of columns = 7 kN Un factored load = 15.4*10.625*5 +7*5 = 853.2 KN
5.2.2 Calculating the size of footing: Calculating the eccentricity e= M/N e=
= 23.3mm
B should be greater than or equal to (6e) (middle third rule) 6*e = 6*23.3= 139.8 mm Assume weight for the base = 80 kN Total un factored load = 80+853.2 = 933.2 kN Area required = 933.2/200 = 4.67 m2 , Use a square footing : B=h=√ Note
T
=√
= 2.16 m > (6e = 0.1398m)
When using b greater than (6e) we thus insure that all the footing is under compression. (middle third rule).
Provide a base 2.2 m square with provided area 4.84 m2
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Pressure on soil: Maximum pressure =
+
Minimum pressure =
-
= =
+
= 261.3 + 16.6 = 278 kN/m2
-
= 261.4 - 16.6 = 244.8 kN/m2
Pressure at column face q = 244.8 +
*(278 – 244.8) = 263 kN/m2
5.2.3 Design of moment reinforcement: First: the direction parallel to the moment: Moment at face of the column: ( critical ) M = 263*2.2*12/2+0.5(278.9-263)*2.2*0.666*1 = 300.3 kN.m assume 500 thickness for the base , d1 = h –c – Ø/2 = 500 – 50 – 8 = 442 mm K=
=
= 0.02 < 0.156 ----ok
= 0.5 +√ As = Asmin =
= 0.97 , take 0.95 = 1636.6 mm2
= =
= 1300 mm2
Number of bar = 1636.6/201 = 8.14 , use 9T16 Arrangement of bars: Lc1 = 2000/2 = 1000 (c1+3d1) = (500+3*442) = 1369.5 mm Lc1 <
(c1+3d1) --- Reinforcement may distributed uniformly
Second: The other span moment steel : Average pressure = 0.5 (278 +244.8) = 261.4 kN/m2 M = 261.4*2.2*0.852/2 = 207.7 kN.m
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d2 = h – c – Ø - Ø/2 = 500 – 50 –16 - 8 = 426 mm k = .015 , z/d = 0.98 (use 0.95) As =
= 1174.4 mm2
=
= 1430 mm2 -----(use Asmin)
Asmin =
Number of bars = 1430/201 = 7.1 ,use 9T16 Lc2 = 2000/2 = 1000 (c2+3d2) = (500+3*426) = 1333.5 mm , Lc2 <
(c2+3d2) ----- ok
5.2.4 Check for punching shear at column face: V = N/u*davg = 1265.1*103/((200*2+500*2)(426+442)0.5 = 2.08 N/MM2 < 4.73 N/mm2 ------- ok
5.2.5 Check for punching shear at 1.5d from column face: Applied stress Use the average pressure to obtain the applied force qavg = 0.5(278+244.8) = 261.4 N/mm2 v = 261.4(2.2*2.2-((0.2+3*0.434)+(0.5+3*0.434))= 557.6 KN u = (0.2+3*0.434)*2 + (0.5+3*0.434)*2 = 6.608 = 0.194 N/mm2
v= Vc =
(
(
)
)1/3 (
)1/4
)1/3 = 0.398 N/mm2
Section is satisfactory with respect to punching shear.
5.2.6 Check for vertical shear at 0.1d from the column face:
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*(278 -244.8) = 271.3 kN/m2
qs = 244.8 +
v = 271.3*2.2*0.442 + 0.5(278-271.3)*2.2*0.442 = 267.1 KN = 0.302 N/mm2
v = Resistance: Vc =
(
Vc =
(
)1/3 (
)1/4
)1/3 (
)1/3 )1/4
)1/3 =0.4 N/mm2 ---- ok
5.2.7 Drawing Details:
-F2-
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5.3 Design of combined footing:
5.3.1 Estimation of loads: For ultimate limit state: For the western column: Loaded Area = 5*(
) =15.375 m2
Factored load = 22.16*15.375*5 +12*5 = 1763.55 kN Un factored load = 15.4*15.375*5 + 8*5 = 1223.9 kN For the Eastern column: Loaded area without stair zone = ( 2.35/2 + 1.9/2 )*2.8 + ( 1.9/2 * 2.2 ) = 8.04m2 Factored load = 22.16*8.04*5+12*5+25.22*4 = 1051.7 KN. Unfactored load = 15.4*8.04+8*5+(25.22/1.5)*4 = 726.3 KN.
5.3.2 Calculate the plan size of the footing: for serviceability limit state: Assume 150 kN self-weight of the footing Total Unfactored load = 1223.9+726.3+150 = 2100.2 kN Required Area = 2100.2/200 = 10.5m2 Provide a rectangular base (2.8*3.8)m2 , with 10.64 m2 provided area .
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5.3.3 Resultant of column loads and centroid of the base: Take moments about the Eastern column: (1223.9+726.3) x` = 1223.9*1.9
Note
, x` = 1.1924m from C2
The base is arranged so that the centre of gravity of the loads coincides with the centreline of the base, in which case the soil pressure will be uniform. This arrangement will be made for the maximum ultimate loads.
5.3.4 Bearing pressure at the ultimate limit state Column loads = 1763.55+1051.7 = 2815.25 KN Earth pressure = 2815.25 / (2.8*3.8) = 264.6 kN/m2
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5.3.5 Design of moment reinforcement: 5.3.5.1 Longitudinal bending: - Mid span between the columns: the moment is positive in this zone = 2184 mm2
use Asmin = spacing =
*2800 = 257 mm
Use Ø16 @ 250 mm c/c Top Reinforcement (parallel to the length of the footing). - At the column face (Western column): M = 740.88*(1.2-0.1)2 /2 = 448.23 kN.m d = 600 -50 – 8 = 542 mm K=
=
= 0.016
= 0.5+√
= 0.98 (use 0.95)
As =
= 2109.2 mm2 < As min
=
Use Ø16@250 mm c/c Bottom Reinforcement (normal to the length of the footing) 5.3.5.2 Transverse bending: M = 264.6*1*1.12 /2 = 160.1 kN.m , d = 600-50-16-8= 526 mm K= AS =
=
= 0.016 ,
= 0.5+√
= 0.98 (use 0.95)
= 733mm2/m
=
= 780mm2/m (use As min )
As min =
Spacing = (201/780)*1000 = 257.7 mm
use Ø16@250 mm c/c Bottom Reinforcement (normal to the length of the footing) 5.3.6 Check for punching shear: At the column face: =
= 2.06 N/mm2 < (0.8√
= 4.73N/mm2)
At 1.5d from column face: v = 264.6((0.2+3*0.534) * (0.6 +3*0.534) - 0.2*0.6) =10018.2 kN v=(
)
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= 0.238 N/mm2
Page 68
Vc =
(
)1/3 (
)1/4
)1/3 = 0.346 N/mm2 , Section is satisfactory with respect to punching shear
5.3.7 Drawing details:
-F3-
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Appendix-A: Symbols(General): ᵞm d.l l.l fcu fy
partial safety factor for strength of materials. characteristic dead load. characteristic imposed load. characteristic strength of concrete. characteristic strength of reinforcement.
Symbols (beams): AS As´ b bw d d' hf M X Z βb Ac Asv bv fyv N SV V v vc As prov As' prov fs
area of tension reinforcement. area of compression reinforcement. width or effective width of the section or flange in the compression zone. average web width of a flanged beam. effective depth of the tension reinforcement depth to the compression reinforcement. thickness of the flange. design ultimate moment. depth to the neutral axis. lever arm. (moment at the section after redistribution)/ (moment at the section before redistribution). area of concrete section. total cross-section of links at the neutral axis, at a section. breadth of section. characteristic strength of links (not to be taken as more than 500 N/mm2). design axial force. spacing of links along the member. design shear force due to ultimate loads. design shear stress at a cross-section. design concrete shear stress. area of tension reinforcement provided at mid-span (at support for cantilever). area of compression reinforcement. estimated design service stress in the tension reinforcement.
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Symbols(Flat slab): av be CX Cy dh F hc l l1 l2 lc lh lx ly Mt n u uo Vt Veff x
distance from the edge of the loaded area to the perimeter considered. breadth of effective moment transfer strip. plan dimensions of column. depth of the head. total design ultimate load on the full width of panel between adjacent bay centre lines. effective diameter of a column or column head. given in Table 3.12 should be taken as the full panel length in the direction of span. panel length parallel to span, measured from centres of columns. panel width, measured from centres of columns lh. dimensions of the column measured in the same direction as lh. effective dimension of a head. shorter span of flat slab panel. longer span of flat slab panel. design moment transferred between slab and column. design ultimate load per unit area (= 1.4gk + 1.6qk). effective length of the outer perimeter of the zone. effective length of the perimeter which touches a loaded area. design shear transferred to column. design effective shear including allowance for moment transfer. dimension of a shear perimeter parallel to axis of bending.
Symbols(Columns): Ac Asc b h le lex ley lo lc Mx Mx' My My' Madd N
net cross-sectional area of concrete in a column. area of vertical reinforcement. width of a column (dimension of cross-section perpendicular to h). depth of cross-section measured in the plane under consideration. effective height of a column in the plane of bending considered. effective height in respect of the major axis. effective height in respect of the minor axis. clear height between end restraints. height of column measured between centres of restraints. design ultimate moment about the x-axis. effective uniaxial design ultimate moment about the x-axis. design ultimate moment about the y-axis. effective uniaxial design ultimate moment about the y-axis. additional design ultimate moment induced by deflection of column. design ultimate axial load on a column.
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Symols(Bases): av c cx cy d h lc lx ly v vc
distance from the face of a column to the critical shear section. column width. horizontal dimension of a column, parallel to lx horizontal dimension of a column, parallel to ly. effective depth of a pad footing or pile cap. thickness of pad footing or pile cap. half the spacing between column centres (if more than one) or the distance to the edge of the pad (whichever is the greater). length of the longer side of a base. length of the shorter side of a base. design shear stress at a section. design concrete shear stress.
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Appendix-B: Word Behavior Section Subjected Pure Bending Load Figure Initially Monolithic Elastic Strength Appear Occupies Increased Cracks Lengthen Penetrate Simultaneously Stress Strain Failure Yield Capacity Compression Ductile Permissible Elongate Maintaining Deform Constant Ample Moderate Accompanied Brittle Material Manner Ultimate Resistance Ignore Comprise Illustrated Resultant
Mujtaba Haddad [0926206662]
Translation سلوك مقطع معرض نقي عزم حمل شكل مبدئيا وحدة متراصة مرن مقاومة يظهر يحتل يزيد شقوق يمتد يغرز معا اجهاد انفعال فشل خضوع سعة ضغط مطيلي مسموح يستطيل يحافظ على تشوه ثابت وافر معتدل مصحوبة قصف مادة اسلوب اقصى مقاومة تجاهل تتالف موضح محصلة
Word Diagram Plane Simplified Rearranging Multiplied Equilibrium Desirable Achieve Preceded Utilises Distribution Series Intermediate Blind layer Sustain Assume Distance Possibilities Extend Sufficient Treat Lateral stability Shear wall Soil investigation Grade Exposure Aggregate Braced Unbraced Density Requirements Estimation According Deflection Obtained Experience Partition Connection coefficient Classification Bearing capacity Generalized
Translation مخطط مستوى مبسط اعادة ترتيب مضروب اتزان مرغوب تحقيق يتقدم يستخدم توزيع مجموعه متوسط طبقة مصمته تحمل يفرض مسافه احتماالت يمتد كاف يعالج استقرار جانبي حائط قص فحص التربة درجه تعرض حصي مقيد غير مقيد كثافه متطلبات تقدير تبعا ل ترخيم تم الحصول عليها خبرة فواصل اتصال معامالت تصنيف سعة التحميل يعمم
Page iv
Word Tiles Plaster Sand Finishing Condition Clause Support Reduced Internal External Redistribution Bending moment Shear force Analysis Design Provision Arrangements Single load case Exceed Satisfactory Characteristic Envelope Relative stiffness Allowance Fixed end moment Cantilever Balance Carry over Strip Economic Isolated Transferable breadth Equivalent Zone Approximately Respectively Positive Negative Critical Punching shear Conservative Percentage deep Division Curtailment Apportionment
Translation بالط بياض رمل تشطيبات شروط فقرة دعامه يخفض وسطي خارجي اعادة توزيع عزم انحناء قوة قص تحليل تصميم اشتراط ترتيبات حالة الحمل المفرد يتخطى محقق مميز مجمع جساءات نسبيه زيادة او سماحيه عزوم التثبيت كابولي اتزان انتقال شريحه اقتصادي معزول منتقل عرض معادل منطقه تقريبا علي التوالي موجب سالب حرج قص ثاقب تحفظ % نسبة عميق تقسيم تقطيع تخصيص
Mujtaba Haddad [0926206662]
Word Reference Accurate Ratio Average Weight Details cover Including Links Content Adjustments Specified Limitation Slenderness Appropriate Uniformly Variations Substantially Effect Symbols Formulae Applicable Neutral axis Expression Buckling homogenous Traversed Recommendation Combined Cast statement Enhancement Particularly Intersect Torsion Adverse Beneficial Adequately Controlled Modify Interpolation Shrinkage Creep Shallow Pressure In-plane moment Out of plane moment Eccentricity
Translation مرجع دقيق نسبة متوسط وزن تفاصيل غطاء يتضمن كانات محتوى تعديالت محدد حدود نحافه مناسب بانتظام اختالفات جوهريا تاثير رموز صيغه قابل للتطبيق المحور المحايد تعبير انبعاج متجانس اجتاز توصيات مجمع حالة الصب تعزيز خاصة تقاطع لي معكوس مفيد كاف تحكم يعدل استكمال انكماش زحف سطحي ضغط عزم باتجاه المحور الرئيسي عزم باتجاه المحور الثانوي المركزية Page v
Word Applied Abnormally Service stress Alternative Significant Unfavorable Three-quarters Lifting Prevent Observe Combination Ribbed slab Measured Flight Permanent Permitted Prohibit Determine Narrow Proportion Dictate Beyond Durability Definition Adjacent Purpose Finite Element Approaches Drop Within Depend Aspect Extremities Describe
Translation مسلط غير طبيعي اجهاد الخدمه بديل هام غير مرغوب ثالثة ارباع رفع يمنع رصد مجموعه بالطة معصبه مقاس الجزء من السلم البه درجات دائم مسموح ممنوع حدد ضيق تناسب تفرض او تملي وراء ديمومة تعريف مجاور غرض عنصر محدد نهج ساقط ضمن يعتمد جانب اطراف يوصف
Mujtaba Haddad [0926206662]
Word Bounded Subclauses Intended Clarify Perpendicular Assessed Empirical Anchored Justification Demonstrate Validity Subsequent Perimeter Further Modification Adopted Major axis Minor axis Restraints Additional Symmetrically Commenced Buttressing Function Rigorous Corresponds Decrease Transmitting Influence Attached Analytical Guidance Precast concrete Prestressed concrete
Translation محدود فقرات فرعية معد توضيح عمودي تقييم تجريبي تقوية تبرير تظاهر صالحية الحقا محيط اضافه تعديل اعتمد المحور الرئيسي المحور الثانوي تقييد اضافي متناظر بدات او شرعت تدعيم دالة دقيق او صارم يطابق تصغير او تقليل ارسال تاثير ارفاق تحليلي توجيه خرسانة مسبقة الصب خرسانة مسبقة الشد
Page vi
Appendix-c: Tables:
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Figures:
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