POINTS TO REMEMBER FOR CIVIL SITE ENGINEERS

Following are few general points to remember for civil site engineers to make the construction work easier while maintaining quality of construction.

• Lapping is not allowed for the bars having diameters more than 36 mm.
• Chair spacing maximum spacing is 1.00 m (or) 1 No per 1m2.
• For dowels rod minimum of 12 mm diameter should be used.
• Chairs minimum of 12 mm diameter bars to be used.
• Longitudinal reinforcement not less than 0.8% and more than 6% of gross C/S.
• Minimum bars for square column is 4 No’s and 6 No’s for circular column.
• Main bars in the slabs shall not be less than 8 mm (HYSD) or 10 mm (Plain bars) and the distributors not less than 8 mm and not more than 1/8 of slab thickness.
• Minimum thickness of slab is 125 mm.
• Dimension tolerance for cubes + 2 mm.
• Free fall of concrete is allowed maximum to 1.50m.
• Lap slices not be used for bar larger than 36 mm.
• Water absorption of bricks should not be more than 15 %.
• PH value of the water should not be less than 6.
• Compressive strength of Bricks is 3.5 N / mm2.
• In steel reinforcement binding wire required is 8 kg per MT.
• In soil filling as per IS code, 3 samples should be taken for core cutting test for every 100m2.

Density of Materials:

Material	Density
Bricks	1600 – 1920 kg/m3
Concrete block	1920 kg/ m3
Reinforced concrete	2310 – 2700 kg/ m3

Curing time of RCC Members for different types of cement:

Super Sulphate cement: 7 days

Ordinary Portland cement OPC: 10 days

Minerals & Admixture added cement: 14 days

De-Shuttering time of different RCC Members

RCC Member	De-shuttering time
For columns, walls, vertical form works	16-24 hrs.
Soffit formwork to slabs	3 days (props to be refixed after removal)
Soffit to beams props	7 days (props to refixed after removal)
Beams spanning upto 4.5m	7 days
Beams spanning over 4.5m	14 days
Arches spanning up to 6m	14 days
Arches spanning over 6m	21 days

Cube samples required for different quantity of concrete:

Quantity of Concrete	No. of cubes required
1 – 5 m3	1 No’s
6 0 15 m3	2 No’s
16 – 30 m3	3 No’s
31 – 50 m3	4 No’s
Above 50 m3	4 + 1 No’s of addition of each 50 m3

TYPICAL DETAILING OF RCC BEAMS AND SLABS

Simple Beams and Slabs

Simply supported slab:

Curtailment of tension steel in simply supported slab construction.

REINFORCEMENT DETAILING IN BEAMS

Reinforcement are provided to resist tensile stresses due to bending and shear in beams for singly reinforced sections. But when depth of section is restricted due to reasons such as architectural reasons, or basement floors, the beam section is designed as doubly reinforced to resist tensile as well as compressive stresses and are provided with reinforcement in compression face.

Beams when subjected to torsion are provided additional longitudinal and shear reinforcement to resist bending and shear stresses developed due to torsion.

In practice, for singly reinforced beams, two additional bars are provided in compression face of the beam so that stirrups can be tied with bars. These additional reinforcement are of nominal diameter of 8mm or 10mm.

When the width of beam is insufficient for number of bars to be provided, keeping minimum clear distance bars based on maximum size of aggregate as per standard code guidelines, the reinforcement can be provided in layers in both tension and compression face of the beam. These reinforcement are provided as straight bars in both zones.

When a beam is designed with slab, called slab beam or Tee-beam, reinforcements are provided as shown in figure below. The beam is generally designed as simple beam but additional reinforcement are provided on top with slab to make it behave like a Tee-beam.

Fig: Mid-span Details of Tee-Beam.

Fig: Slab-Beam Details

Following are the types of stirrups provided in beams­:

Fig: Types of stirrups in beams

Standard bends and hooks required for reinforcing bars are shown in figure below. The specification for detailing of reinforcement in beams are given in cl.26.5.1 of IS 456 – 2000.

Fig: Standard bends and hooks in beams

Curtailment of Reinforcement in Beams:

Reinforcements are curtailed along its length in beams depending on the bending moment at the section. Anchorage or development length required at support is provided during curtailment of reinforcement. The anchorage length required for main reinforcement in tension and compression is given in cl.26.2 of IS 456 – 2000.

Typical details of curtailment of reinforcement in cantilever and continuous beams are shown in figure below:

Fig: Typical Details of Reinforcement curtailment in beams

Details of anchorage length required for main reinforcement in tension and compression is shown below:

 Fig: Typical Details of anchorage length of reinforcement in beams

Curtailment of steel in beam and slab construction.

Typical steel detail for concrete beam.

Reinforcement detailing in RCC Footings

2. The placement of reinforcement in reinforced concrete footing depends on the type of footing, i.e. whether it is one-way footing, two way square footing or two way rectangular footing.

   In case of one-way RCC footing, such as footing under a wall, the main reinforcement (short) is distributed uniformly across the full width of footing, i.e. the short reinforcements are placement at the bottom and secondary reinforcement is placed above it.

   In case of two-way square footings, the shape of footing is square because the moments in both directions are same. Thus, reinforcement in both direction will also be same. So, reinforcement in two-way square footing are distributed uniformly across the full width and length of the footing. Reinforcement in bottom can be provided in any direction.

   In case of two-way rectangular footings, the moment in longitudinal direction is more than the other direction, thus rectangular shape is selected.

   Thus, in this case, the main reinforcement is the one which resists the larger moment. Thus reinforcement is distributed across the full width of footing in long direction. Thus, long reinforcement is provided at the bottom. For short direction, the reinforcement is distributed in the central band. The rest reinforcement in short direction is distributed equally on both sides of the central band. Thus, short reinforcement is placed above the long reinforcement.
   
3. Design of reinforced concrete columns

   Type of columns

   

   Failure of reinforced concrete columns

   Short column Column fails in concrete crushed and bursting.  Outward pressure break horizontal ties and bend vertical reinforcements	Long column Column fails in lateral buckling.
   See test picture from web-site below	See picture from web-site below

   Short column or Long column?

   ACI definition

   For frame braced against side sway:	For Frame not braced against side sway:
   Long column if klu/r > 34-12(M1/M2) or 40	Long column if  klu/r > 22

   Where k is slenderness factor, lu is unsupported length,  and r is radius of gyration.  M1 and M2 are the smaller and larger end moments. The value, (M1/M2) is positive  if the member is bent in single curve, negative if the member is bent in double curve.

   Determine the slenderness factor, k

   The slender factor, k should be determined graphically from the Jackson and Moreland Alignment charts.

   
   

   (Charts will be added later)

   
   

   where y E( å =cIc/lc) of column /å E(bIb/lb) of beam, is the ratio of effective length factors.  

   Ec and Ec are younger modulus of column and beams.

   lc and lc are unbraced length of column and beams.

   The cracked  moment of inertia, Ic is taken as 0.7 times gross moment of column and Ib is taken as 0.35 times gross moment of inertia of beam.

   Alternatively, k can be calculated as follows:

   1. For braced frame with no sway, 

   k can be taken as the smaller value of the two equations below.

   k = 0.7 + 0.05 (yA+yB) £ ,1

   k = 0.8 + 0.05 (ymin) £ 1

   yA and yB are the y at both ends, ymin is the smaller of the two y values.

   2. For unbraced frame with restrains at both ends, 

   For ym  < 2

   k =  [(20- ym)/20] Ö(1+ym)

   For ym  ³ 2

   k = 0.9 Ö(1+ymin)

   ym is the average of the two y values.

   2. For unbraced frame with restrain at one end, hinge at the other.

   k = 2.0 + 0.3 y

   y is the effective length factor at the restrained end.

   Example:

   Beam information:

   Beam size: b = 18 in, h = 24 in

   Beam unsupported length: lb = 30 ft

   Concrete strength: 4000 psi

   Young's modulus, Eb = 57 0004Ö 5063 = ksi

   Moment of inertia of beam: Ib = 0.35bh3/12 = 7258 in4.

   Column information:

   Square Column: D = 18 in, unsupported length lc =10 ft

   Concrete strength: 5000 psi

   Young's modulus: Ec = 57 0005Ö 0304 = ksi

   moment of inertia of column: Ic = 0.7D4/12 = 6124 in4.

   Column top condition:

   There are beams at both sides of column at top of column, no column stop above the beams

   The effective length factor: yA E( =cIc/lc) /[2 E(bIb/lb)] = 1.4

   Column bottom condition:

   There are beams at both sides of column at bottom of column and a column at bottom level

   The effective length factor: yA E( 2[ =cIc/lc)] / [2 E(bIb/lb)] = 2.8

   From chart:

   If the column is braced: k » 0.84

   If the column is unbraced: k » 1.61

   From equation

   If the column is braced: 

   k = 0.7 + 0.05 (yA+yB) =  19.0

   k = 0.8 + 0.05 (ymin) = 29.0

   If the column is unbraced: ym = (yA+yB)/2 = 2.12

   k = 0.9 Ö(1+ymin) = 1.6

   Design of reinforced concrete columns

   Short column	Long non-sway column & Long sway columns
   1. Column shall be designed to resist factored axial compressive load and factored moments. 2. Column strength shall be determined based on strain compatibility analysis.	1. Column shall be designed to resist factored axial compressive load.  Factored moment shall be magnified with magnification factors. 2. Column strength shall be determined based on strain compatibility analysis.

   Column ties and spiral

   ACI code requirements for column ties

   1. No. 3 ties for longitudinal reinforcement no. 10 bars or less, no. 4 ties for no. 11 bars or larger and bundled bars.
   2. Tie spacing shall not exceed 16 diameter of longitudinal bars, 48 diameters of tie bars, nor the least dimension of column.
   3. Every corner bar and alternate bars shall have lateral tie provide the angle shall not exceed 135 degree.
   4. No longitudinal bar shall be spacing more than 6 inches without a lateral tie.

   

   ACI code requirements for spiral

   1. Sprial shall be evenly space continuous bar or wire, no. 3 or larger.
   2. Sprial spacing shall not exceeds 3 in, nor be less than 1 in.
   3. Anchorage of spiral shall be provided by 1-1/2 extra turn.

   Design of short columns

   Design of long non-sway columns

   Design of long column with sway

   Design of short concrete columns

   Strength of column subjected to axial load only

   Ideally, if a column is subjected the pure axial load, concrete and reinforcing steel will have the same amount of shortening.  Concrete reaches its maximum strength at 0.85fc' first.  Then, concrete continues to yield until steel reaches its yield strength, fy, when the column fails. The strength contributed by concrete is 0.85f’c(Ag-Ast), where fc' is compressive strength of concreter, Ag is gross area of column, Ast is areas of reinforcing steel.  The strength provided by reinforcing steel is Astfy.  Therefore, the nominal strength of a reinforced concrete column,  is

   Pn = 0.85f’c(Ag-Ast)+Astfy                                                                                            [1]

   For design purpose, ACI specify column strength as follows

   For a spiral column, the design strength is

   fPn = 0.85f[0.85f’c(Ag-Ast)+Astfy] [2]

   For a regular tie column, the design strength is

   fPn = 0.80f[0.85f’c(Ag-Ast)+Astfy] [3]

   where f is strength reduction factor.  For spiral column f 57.0 = (ACI 318-99), f 7.0 = (ACI 318-02, 05). For spiral column f 7.0 = (ACI 318-99), f 56.0 = (ACI 318-02, 05)

   The factors 0.85f and 0.8f are considering the effect of confinement of column ties and strength reduction due to failure mode.  Nevertheless, column loads are never purely axial.  There is always bending along with axial load.

   Strength of column subjected to axial load and bending

   Consider a column subjected to axial load, P and bending moment, M.  Axial load, P produces an uniform stress distribution across the section while bending moment produces tensile stress on one side and compressive stress on the other.

   Strain and stress distributions of short concrete column at failure and interactive diagram

   Assumption:  

   1. A plan section remains a plan at failure.  Strain distributes linearly across section

   2. Concrete fails at a strain of 0.003.

   3. Reinforcing steel fails at a strain of 0.005.

   
   

   Axial load only (moment = 0) Failure occurs when concrete strain reaches 0.003	Moment only (Axial load = 0) Failure occurs when steel strain reaches 0.005 first.
   Large axial load with small moment Failure occurs when concrete strain reaches 0.003	Small axial load with large moment Failure occurs when steel strain reaches 0.005
   Balanced condition Failure occurs when concrete strain reaches 0.003 and steel strain reaches 0.005 at the same time.	Interaction diagram for Pn and Mn

   
   

   Design aid:  

   The interaction diagrams of concrete column with strength reduction factor is available on ACI design handbook.  The vertical axis is fPn /Ag and the horizontal axis is fMn /Agh, where h is the dimension of column in the direction of moment.  The chart is arranged based on the ratio, g which is the ratio of the distance between center of longitudinal reinforcements to h.

   

   Column strength interaction diagram for rectangular column with g =0.6 (Coursey of American Concrete Institute)

   
   

   Design of short concrete column

   Design requirements:

   1. Design strength:fPn ³ Pu   and fMn ³ Mu

   2. Minimum eccentricity, e = Mu/Pu ³ .1.0

   Design procedure:

   1. Calculate factored axial load, Pu and factored moment, Mu.

   2. Select a trial column column with b and column depth, h in the direction of moment.

   3. Calculate gross area, Ag and ratio, g = distance between rebar/h.

   4. Calculate ratio, Pu/Ag and Mu/Agh.

   5. Select reinforcement ratio,,rfrom PCA design chart based on concrete strength, fc', steel yield strength, fy, and the ratio, g.

   6. Calculate area of column reinforcement, As, and select rebar number and size.

   7. Design column ties.

   
   

   Design example:

   
   

   Example: A 12"x12" interior reinforced concreter short column is supporting a factored axial load of 350 kips and a factored moment of 35 kip-ft.

   Desogn data:

   Factored axial: Pu = 350 kips

   Factored moment: Mu = 35 ft-kips

   Compressive strength of concrete: fc'= 4000 psi

   Yield strength of steel: fy = 60 ksi

   Requirement: design column reinforcements.

   Column size: b = 12 in, h = 12 in

   Gross area, Ag = 144 in2.

   Concrete cover: dc = 1.5 in

   Assume #4 ties, dt = 0.5 in and #6 bars, ds = 0.75 in

   Calculate  g ( =h - 2 dc-2 dt - ds)/h = 0.6

   Calculate, 

   Pu/Ag = 300/144 = 2.43 ksi, 

   Mu/Agh = 35/[(144)(12)] = 0.243

   From ACI design handbook, reinforcement ratio, r 810.0 =

   Area of reinforcement, As = (0.0018)(144) = 2.6 in2.

   Use 6#6, area of reinforcement,  As = (6)(0.44) = 2.64 in2.

   Check Bar spacing, s = (h - 2 dc - 2 dt - ds)/2 = 3.625 in  (O.K.)

   Calculate minimum spacing of column ties:

   48 times of tie bar diameter = (48)(0.5) = 24 in

   16 times of longitudinal bar diameter = (16)(0.75) = 12 in

   Minimum diameter of column = 12 in

   Use #4 ties at 12 inches on center.

   Design of long column in non-sway frame (ACI 318-02,05)

   Moment magnification for columns in braced frames (non-sway)

   For a slender column in a braced frame that is subjected to axial compression and moments.  If M1 is the smaller and M2 is the larger moment, the moment need to be design for magnified moment if the ratio

   klu/r > M(21 - 431M/2)

   where k is slenderness ratio, lu is unsupported length, r is radius of gyration. k shall not be taken as 1 unless analysis shows that a lower value is justified.

   M1M/2 is positive if the column is bent in single curve and M1M/2 is not to be taken less than -0.5 (klu/r = 40).

   The design moment shall be amplified as 

   Mc = dnsM2

   where

   dns = Cm/(1-Pu/0.75Pc)  ³ 1 is moment magnification factor  for non-sway frame,

   Cm = 0.6+0.4(M1/M2) ³ 4.0

   Pu is factored column load, and 

   Pc = p2(/IEklu)2  is Euler's critical buckling axial load,

   EI shall be taken as

   EI = (0.2EcIg+Es/Ise)/(1+bd) or EI = 0.4EcIg/(1+bd)

   Where  is the ratio of maximum factored axial dead load to total factored load

   Example: 

   A 12"x12" interior reinforced concrete column is supporting a factored axail dead load of 200 kips and a factored axial live load of 150 kip. Factored column end moments are -35 kip-ft and 45 kip-ft. The column is a long column and has no sway.

   Design data:

   Total Factored axial: Pu = 350 kips

   Factored moment: M1 = -35 kips, M2 = 45 kips (bent in single curve)

   Compressive strength of concrete: 4000 ksi

   Yield strength of steel: 60 ksi

   Unsupported length of column: 10 ft

   Requirement: Determine the magnified design moment

   Column size: b = 12 in, h = 12 in

   Gross area: Ag = 144 in2.

   Concrete cover: 1.5 in

   Assume #4 ties and #8 bars, dt = 0.5 in, ds = 1 in

   Gross moment of inertia: Ig = (12 in)4/12 = 1728 in4.

   Radius of gyration, r = Ö(1728/144) = 3.5 in or r = 0.3(12 in) = 3.6 in

   Assume slenderness factor, k = 1 without detail analysis

   Slenderness factor, klu/r = (1)(120 in)/3.6 in = 35 > 34-12(35/45) = 25, long column

   Young's modulus of concrete, Ec = 570004Ö 5063 = ksi

   Elastic modulus of steel: Es = 29000 ksi

   Assume 1% area of reinforcement: As = (0.01)(144 in2) =1.44 in2.  

   Assume half of the reinforcement at each side of column, distance between rebas = 12-1.5*2-0.5*2-1 = 7 in

   Moment of inertia of steel reinforcement: Ise = (0.72 in2)(7/2)2*2 = 17.6 in4.

   The ratio of factored load, bd = 200/350 = 0.57

   The flexural stiffness, EI = 0.4EcIg/(1+bd)= 0.4(3605)(144)/(1+0.57) = 1.58x106 kip-in2. 

   or EI = (0.2EcIg+Es/Ise)/(1+bd) 11.1 = )75.0+1(/])6.71()00092(+)441()5063( 2.0[ =x106 kip-in2. 

   The critical load, Pc = p2(/IEklu)2  = p2(1.58x106)/(35)2 = 1087 kip

   Factor Cm = 0.6+0.4(35/45) = 0.911

   Moment magnification factor, dns = Cm/(1-Pu/0.75Pc)  = (0.911)/[1-350/1087] = 1.6

   The magnified design moment, Mc = 1.6 (45) = 71.8 ft-kip

   Design of long column in sway frame

   Moment magnification for columns in unbraced frames (sway)

   Determine if the frame is a sway frame

   1. The frame can be assumed as non-sway if the end moment from second-order analysis not exceeds 5% of the first-order end moment.

   2. The frame can be assumed as non-sway if stability index, 

   Q = åPu Do/ Vulc ³ 50.0

   where åPu and Vu are the total vertical load and the story shear,  Do is the first-order relative deflection between the top and bottom of that story due to Vu, and  lc is the length of the column.

   3. The moment need not to be design for magnified moment if the ratio

   klu/r £ 22

   where k is slenderness ratio, lu is unsupported length, r is radius of gyration. 

   Calculating magnified moments

   If M1 is the smaller and M2 is the larger moment, the design moment shall be amplified as 

   M1 = M1ns+dsM1s

   M2 = M2ns+dsM2s

   where M1ns and M2ns are moments from loadings that are not contribute to sway (i.e. gravity load), and M2s and M2s are moments from loading that contribute to sway (i.e. wind and seismic)

   1. The magnified sway moment dsMs can be determined by a second-order analysis based on the member stiffness reduced for crack section.

   2. The magnified sway moment can be calculated as

   dsMs =Ms/(1-Q)  ³ Ms when 1 £ ds £ 5.1

   3. The magnified sway moment can be calculated as

   dsMs =Ms/[1-åPu/(0.75åPc )]³ Ms 

   where åPu is the summation for all the factored vertical loads in a story and åPc = på2(/IEklu)2  is the summation of critical bucking loads for all the columns  in a story from non-sway frame.

   Limitations

   1. The ratio of second-order lateral deflection to the first-order lateral deflection based on factored dead and live loads plus lateral load shall not exceed 2.5.

   2. The value of Q shall not exceed 0.6.

   3. The magnification factor ds shall not exceed 2.5.

   Example: 

   A reinforced concrete moment frame has four 18"x18" reinforced concrete columns.

   Design data:

   Factored column axial loads:

   Column 1 & 4 (Exterior columns)

   Live load: PL1 = 150 kips

   Dead load: PD1 = 200 kips

   Lateral load: PW1 = 40 kips

   Column 2 & 3 (interior columns)

   Live load:  PL2 = 300 kips

   Dead load: PD2 = 400 kips

   Lateral load: PW2 = 20 kips

   Factored column moments:

   Column 1 & 4 (Exterior columns)

   Live load moment: ML11 = -25 ft-kips, ML12 = 40 ft-kips

   Dead load moment: MD11 = -35 ft-kips, MD12 = 45 ft-kips

   Lateral load moment: MW11 = -75 ft-kips, MW12 = 80 ft-kips

   Column 2 & 3 (interior columns)

   Live load moment: ML21 = -15 ft-kips, ML22 = 25 ft-kips

   Dead load moment: MD21 = -30 ft-kips, ML22 = 40 ft-kips

   Lateral load moment: MW21 = -65 ft-kips, MW22 = 75 ft-kips

   Compressive strength of concrete: fc' = 4000 psi

   Yield strength of steel: fy = 60 ksi

   Unsupported length of column: lu = 10 ft

   Requirement: Calculate magnification design moments for column 2 and 3.

   1. Calculate total factored loads

   Column 2 & 3: Pu = (PL2+PD2+PW2) = 720 kips

   All column:

   åPu = (PL1+PD1+PW1+PL2+PD2+PW2) = 2220 kips

   Column size: b = 18 in. h = 18 in.

   Gross area of column: Ag = bh = 324 in2.

   Assume concrete cover: 1.5 in for interior column

   Assume #4 ties, dt = 0.5 in and #8 bars, ds = 1 in, for vertical reinforcement

   Calculate gross moment of inertia of column: Ig = bh3/12 = 8748 in4.

   Radius of gyration, r = ÖIg/Ag = 5.2 in

   Assume slenderness factor, k =2 for moment frame with side-sway, 

   the slenderness ratio, k lu/r = 46. > 22 (long column)

   Young's modulus of concrete, Ec = 57 ()0004Ö 5063 = ksi

   Elastic modulus of steel: Es = 29000 ksi

   Assume 1% of reinforcement area, As = 0.01 Ag = 3.24 in2.

   Assume that reinforcement are placed equally at each side of the column, the distance between vertical reinforcement is

   18-2(1.5)-2(0.5)-1 = 12 in

   the moment of inertia of reinforcement, Ise = As (13/2)2 = 137 in4.

   Factor, bd = PD2/(PD2+PL2+PW2) = 0.556

   The stiffness factor, EI = (0.2EcIg+Es/Ise)/(1+bd) 16.6 =x106 kip-in2.

   or EI = 0.4EcIg/(1+bd) 11.8 =x106 kip-in2.

   Calculate magnification factor for non-sway moment:

   Pc = p2(/IEklu)2  = 1390 kips.

   Assume Cm = 1 (Transverse load at top and bottom of column)

   dns = Cm/[1-Pu/(0.75Pc )] = 3.235

   Calculate magnification factor fo sway moment:

   Euler's critical buckling axial load of all columns, åPc = p2(/IEklu)2  x 4 = 5558 kips.

   The moment magnification factor, ds = 1/[1-åPu/(0.75åPc )] = 2.14

   Magnified design for column 2 and 3, Mu2 = dns (ML22+MD22)+ds (MW2) = 370.7 ft-kip