How to Calculate shear force in Simply Supported Beam

 Shear Force is the algebraic sum of all vertical forces acting on either side of the point on the beam. Shear force is to shear off the beam along with the point where it is acting.

                                                            Shear Off Diagram

Here submitting an example of shear force acting on the simply supported beam.

Loading Diagram 

1- Calculate Reactions

we will use Minus (-) Sign for downward forces (Loads) and Positive (+) Signs for Upward Forces (Reactions or Supports)

Taking Moment at Reaction "A"

+RAx0 - (1200x60)-(3000x20)-(4000x47)+(RBx60)

-(72000)-(60000)-(188000)+RBx60

RB =  320000/60 =  5333.333 Lbs

We know that (RA+RB) = Total Load

RA+RB =  (3000+4000)+(1200x60)

RA+RB =  79000 Lbs_________Eq.1

Putting value of RB in Eq.1

RA+(5333.333) =  79000 Lbs_________Eq.1

RA = 73666.667  Lbs

Now Calculating Shear forces at different points

Shear force at point "a" due to reaction A = +73666.667 Lbs

Shear force at point "b" = 73666.667-(1200x20)-(3000) 

SF due to UDL              =73666.667-(24000)= +49666.667 Lbs

SF due to Point Load   =49666.667-(3000)= +46666.667 Lbs     

Shear force at point "c" = +46666.667-(1200x27)-(4000)

SF due to UDL               =+46666.667-(1200x27)=  +14266.667 Lbs

SF due to Point Load     =+14266.667-(4000)       = +10266.667 Lbs

Shear force at point "d" = +10266.667-(1200x13)+(5333.333)

SF due to UDL               = +10266.667-(1200x13) = - 5333.333

SF due to Reaction RB  =  -5333.333+5333.333  =0

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Function of Shear Wall

 This is a structural member of RCC structures to provide lateral stability against horizontal forces caused by Wind pressures and Earthquakes.

Shapes of Shear Wall.

Shear walls are designed in different shapes using Reinforced cement concrete materials which produce resistance against the horizontal forces that act to over-turn the structures.

Placement of Shear Walls.

Shear walls are placed at three locations

1- On the edges of the building.

2-Inner side of Building on various locations.

3-On the Center of High rise buildings.

The function of Shear Wall.

There are various useful functions of shear walls.

1-Shear Wall provides resistance against self-load of structure, movement loading, and resultant loading such as uplift loads acting on shear walls.

2-Shear Wall holds gravity lines of all areas of the building that can create extra stresses during movements and may damage nonstructural elements in the building.

3-Shear wall keeps stiffness of junctions (on column and beams)  within the design limit.

4-Shear wall provides protection in lateral displacement and sway of the building.

5-Shear wall provides safe structural movement under earthquake and wind pressures. 

The Calculation for Loading Function.

Let's have a calculation session on design of shear walls as shown in this building plan.

Suppose the each span is 20 ft and slab design load is 300 Pound/Sq.Ft.(Including weight of walls ).

Total Load of one floor = (80'x40'x300)=  960000 Pounds

Factored load                 =(960000x1.5) =1440000 Pounds

Now whole Plan will be divided 5 Parts as showing  (Area 1=800 Sft), (Area 2, =600 Sft + 600 Sft ),(Area 3,= 300 Sft+300 Sft).

Now load can be calculated for each shear force

Load on Area 1 Shear Wall =(800x1440000)/(40x80) =360000 Pounds

Load on Area 2 Shear wall = (600x1440000)/(40x80)=270000 Pounds

Load on Area 3 Shear Wall =(300x1440000)/(40x80) =135000 Pounds

 

Uniformly Distributed Load Calculation

Considering shear wall as a beam between two junctions of the slab and beams vertical height is  12 ft.

UDL  For Area 1, Shear Wall = 360000/12= 30,000 Pound/Ft

UDL  For Area 2, Shear Wall = 270000/12= 22,500  Pound/Ft

UDL  For Area 1, Shear Wall = 135000/12=  11,250 Pound/Ft

Moment of Area 1, Shear Wall=( w.L.L.x12) /8=

                                               = 30000x12x12x12/8=6,480,000 (Lbs-Inch)

Moment of Area 2, Shear Wall=( w.L.L.x12) /8=

                                               = 22500x12x12x12/8=4,860,000 (Lbs-Inch)

Moment of Area 3, Shear Wall=( w.L.L.x12) /8=

                                               = 11250x12x12x12/8=2,430,000 (Lbs-Inch)

Moment of Resistance = 250xbxd1^2   ................    EQ.1

we knows that M.r = Moment

Putting values into Eq.1 for Area 1 Shear Wall

 

Longer side in Inches =10x12= 120 Inch

Shorter side in Inches =8x12=  96  Inch.

(6,480,000/2) (Lbs-Inch) = 250x 120xd1^2

d1 = 7.34 say 7.5" Adding cover 1" on both faces = 9.5"

Repeating for Shorter Side

(6,480,000/2) (Lbs-Inch) = 250x 96 xd1^2

d1= 11.61 Say 12"  Adding cover of 1" at both faces = 14"

Maximum size of wall thickness is 14" so adopting this

Area of Steel = M/Fst.Lever Arm

6,480,000/2x20000x.75x12 =18 Sq.Inch

For Spacing Using 1 " Dia Bars = Area of Bar x width /Area of Steel

                            Spacing  =(0.785x96)/18 =4.18 say 4" C/C (Vertical Bars)

Now Calculating the stirrups spacing

Spacing of 1/2" Dia bar =Area of 6-Leg Stirrups x Fst x Lever Arm/Shear Force

                                          6x0.302x20000x0.75x12/(30000x12x.5x.5) =3.5" C/c

Shear Wall Area 2

Longer side in Inches =15x12= 180 Inch

 (4,860,000 (Lbs-Inch)= 250x 180xd1^2

d1 = 10.39 say 10.5" Adding cover 1" on both faces = 12.5" (Shear wall Thickness)

Area of Steel = M/Fst.Lever Arm

4,860,000 /20000x.75x10.5= 30.85 Sq.Inch

For Spacing Using 1 " Dia Bars = Area of Bar x width /Area of Steel

                            Spacing  =(.785x180)/30.85 =4.5 "c/c  (Vertical Bars)

Now Calculating the stirrups spacing

Spacing of 1/2 " Dia bar =Area of 12-Leg Stirrups x Fst x Lever Arm/Shear Force

                                      =  12x.302 x20000x0.75x10.5/(22500x12x.5) =4.25" C/c

 

 

Shear Wall Area 3

Longer side in Inches =8x12"= 96 Inch

 2,430,000 (Lbs-Inch)= 250x 96xd1^2

d1 = 10  Adding cover 1" on both faces = 12" (Shear Wall Thickness)

Area of Steel = M/Fst.Lever Arm

2,430,000 /20000x.75x10= 16.20 Sq.Inch

For Spacing Using 1 " Dia Bars = Area of Bar x width /Area of Steel

                            Spacing  =(.785x96)/16.2=4.5 "c/c  (Vertical Bars)

Now Calculating the stirrups spacing

Spacing of 1/2 " Dia bar =Area of 6-Leg Stirrups x Fst x Lever Arm/Shear Force

                                      =  6x.302 x20000x0.75x10/(11,250 x12x.5) =4" C/c

 

Why Curing of concrete is Important.

 

The curing process is keeping the concrete moist awaiting the safe  hydration of concrete is complete and strength is accomplished. Curing of concrete should start on soon after initial setting time of concrete and should be continued up to 28 days to attain maximum strength of concrete.

Good Curing of concrete results in the following strength.

Comparative Strength of Ordinary (Portable) Cement concrete at various ages
Curing Time	Compressive Strength
3-Days	40%
7-Days	65%
14-Days	90%
28-Days	99%
3-Months	115%
6-Months	120%
1-Year	130%

Why Curing is Important.

1-Curing is required to avoid dryness of concrete during setting and hardening of cement that is only possible with effective hydration of cement, Now how hydration of cement will be effective, it's only possible when mixing water will remain in concrete during setting time period of concrete which is 28-days.

If curing is not started after placing the concrete it will produce only 50% of the total designed compressive strength in concrete.

2-Curing ensures the safe processing for the attainment of strength in concrete structures and also helps in speedy construction practices.

For Example, Concrete Structure avails 90 % strength in 14-Days and mostly engineers remove formworks after 14-days because structures can afford to hold their self loads after 14-days. The self-load of structures remains around 50% of the total designed load so when concrete avails its 90 % strength it provides easy and safe decision to remove the formwork without any hesitation.

3-Curing can raise the strength of concrete up to 130% which is the guarantee of long life and durability of concrete in less budgeting, so it is easy to design the structures for 100 years with effective curing of concretes.

4- Curing helps to attain high and required strength in concrete which helps to avoid faulty constructions and also finally  provides a safe and reliable structures. 

Curing Methods

1- Shading.

In this method, concrete is covered to avoid evaporation from sun rays. 

2-Ponding.

This method is used only for horizontal surfaces of slabs and concrete roads but require a lot of water.

3-Spraying of Water.

This method involves the sprinkling of water on concrete surfaces to keep moist surfaces but this is difficult as requires periodic showering of water.

4-Membrane Curing.

In this method of curing a chemical coating /Using some membrane is provided on the surface of the concrete to avoid evaporation of concrete water.

5-Steam Curing/ Hot water Curing.

Curing with Steam and hot water is sometimes adopted to gain rapid development in the strength of concrete. This process of curing is best for the construction of Pre-Cast Structures.

6-Covering.

This is a broadly used technique of curing, mainly for structural concrete. The uncovered surface of the concrete is prohibited from drying out by casing it with a hessian cloth or canvas cloth.

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How to design a simply supported beam manually

 Suppose we have a beam whose span is 30 Ft , and a Point Load "F"=10000 Pounds, which includes weight of beam and imposed load.

In this case we are going to design a RCC Beam using Load Factor Theory in following steps.

1-Find out Support or Reactions

Left and Right Support = F/2= 5000  Pounds

In given symmetrical loading case both reactions are equal

2-Find out Max shear force = Load /2 = 10000/2 =5000 Pounds

Shear force at support = +5000 Pounds

Shear force at Mid      = +5000 + (-10000) =   -5000 Pounds

3-Max Bending Moment at Center of  Beam =    Reaction x L/2

Max MB=            5000x (30/2)x12= 900000 Pound-Inch--------1

4-Moment of Resistance =  K.b.d1^2----------------EQ.2

5-b (width of beam ) = (Span x12)/30 =12"

Value of  "K" = 250  in Load Factor Method using (1000-20000) Concrete

Where fc =1000 PSI  and Fs = 20000 PSI

We know that Moment of Resistance =Bending Moment  so comparing both equations EQ----1 AND EQ-----2

6-       900000 = 250x12xd1^2

d1 (Effective Depth) = 17.33 inch Say 17.5"

7-   Area of Steel in Tension zone = BM /(FstxLever Arm)

                        Ast  =900000/(20000x0.75/17.5)= 3.42 Sq.in

 No of bars according to 3/4" dia bars =  Ast/area of one bar =8 nos.

 

8-Area of  steel in compression zone =(B.M - M.r)/fst.Lever Arm

               =((900000-(250x12x17.5^2))/20000x 0.75x17.5

               =(900000-918750)/20000x0.75x17.5= - (0.07)

Use (4+4) nos of 3/4" dia bars and 2 no of 1/2" dia bar at top sides.

Use  3/8" dia bar stirrups 2-Leg @ 10 " c/c

Minus sign shows that the M.r value is greater than the applied value of B.m so no need of Asc (Area of steel in compression zone ), Only two steel bars of 1/2" dia bars are enough to hold stirrups.

 

Use (4+4) nos of 3/4" dia bars and 2 no of 1/2" dia bar at top sides.

9- Spacing of stirrups = Aw.Fw.Lever arm/(Shear Force)

                                    =(0.22x18000x0.75x17.5) /5000 = 10" C/c

Here Aw = Area of 2 leg stirrup bars

      Fw    =   Permissible value of Tensile stress of stirrups

Overall depth of beam = Effective depth+ 2 concrete covers

concrete cover = 1"

Concrete ratio =1:2:4

Factor of safety in Fc = 3000/1000=3 Times (considering compressive strength of concrete)

 The factor of safety in Fst= 60000/20000=3 Times (considering Tensile strength of Steel).

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Concrete Slab Construction Methods

 

What is Concrete Slab.

The concrete slab is a Horizontal structural element to provide a flat surface for arranging floors and ceilings . Concrete Slabs are reinforced with mild steel bars having different diameters and surfaces like deformed steel bars and Twisted/Tor Bars to produce good bonding with concretes.

Steel Types used in Concrete.

According to strength, there are two types of Steel used in construction.

1, Low carbon Steel /Mild steel.

2, High carbon steel.

In construction Mild steel is usually used in all types of Civil engineering structures but High carbon steel is preferred in Precast structures only.

Mild steel or low carbon steel is mostly used in construction because of its more protection against corrosion.

In most structures Grade 60 is preferred because of economic suitability and its minimum Yield strength is used in designing RCC structures. 

STRENGTH in (PSI)	Grade 40	Grade 60	Grade75
Minimum Yield Strength	40000	60000	75000
Maximum Yield Strength	60000	90000	100000

Concrete mix for RCC/PCC slabs.

DETAILS OF CONCRETES
TYPE OF CONCRETE	MIX RATIO	COMPRESSIVE STRENGTH in (PSI) AFTER 28 DAYS
PCC	(1:5:10)	1000
PCC	(1:4:8)	1500
PCC	(1:3:6)	2000
RCC	(1:2:4)	3000
RCC	(1:1.5:3)	4000
RCC	(1:1:2)	4500

Generally, the above mixtures are used in the production of Plain cement concretes and Reinforced cement concretes but special concretes may be produced by special measuring and testing in labs. These special concretes are made for high loading demands using moderated tools and materials and specialists are needed to produce such high-strength concretes.

Some of these special concretes are as follows

M25   (25MPa)= 3625 PSI

M30    (30MPa)= 3625 PSI

M35    (35MPa)= 5075 PSI

M40    (40MPa)= 5800 PSI

M50    (50MPa)= 6525 PSI

M55    (55MPa)= 7250 PSI

M60    (60MPa)= 8700 PSI

M65    (65MPa)= 9425 PSI

M70    (70MPa)= 10150 PSI

Methods of Construction.

There are some methods involved in the Concreting process.

1- Preparation of Flooring for Scaffolding.

Scaffolding can be done on firm ground but there may be chances of settlement so it should be better to lay Plain cement concrete which may be part of the floor in later stages. Arrangement and fixing of scaffolding should be according to the weight of Concrete which may be calculated from the following formula.

Weight of plain cement concrete = Volume of concrete x Density of Concrete

Density of Plain cement concrete=140 Pound/ Cft

Density of Reinforced cement concrete=150 Pound/ Cft

One scaffolding pipe can bear 2240 Pound Safely if braced at vertical interval or 3Ft.

So the number of vertical pipes under PCC /RCC = Weight /2240= No of Pipes

Horizontal pipes should be calculated according to the height of pipes and according to loading of slabs.

2- Fixing of Shuttering

It is a very important stage as shuttering provides a firm platform for placing concrete and steel, so shuttering should be water-tight so the cement paste may not go out from the concrete mix as it will decrease the strength of cement concrete. Shuttering may be made of wooden ply sheets supported by wooden beams, and may also, be made of steel sheets supported by steel girders.

Shuttering should be leveled as per drawing and design and also according to the elevation of the building.

3- Fixing of Steel Reinforcement.

Steel reinforcement should be placed and fixed at shuttering platform according to drawing and design of slab. There are different sizes of steel bars as 1.25" ,1" ,7/8",6/8",5/8",4/8"3/8"2/8" used in building construction.In Oneway slabs Main bars are placed along short spans whereas distribution bars are placed along the longer sides of slabs. In TwoWay slabs both longer and shorter bars contribute equally so up or down does not matter as per design aspects. Negative bars control negative moments that apply on supports and up to the distance of L/4.Supporting bars should be provided beneath the Negative bars usually at intervals of 12 " c/c. Chairs should be provided to control the deflection of steel bars and to keep steel in position during the concreting process.

Concrete Spacers should be provided under the steel reinforcement mesh and also on the sides of slabs, normally concrete cover is equal to the larger dia of steel used in slab but the minimum cover should be up to 1/2".

4- Fixing of Electric and Plumbing Conduits.

After steel electric and plumbing conduits should be placed as per the location of electric fixtures like Fans, Lights, Fan Boxes Etc. Plumbing pipes are kept vertically according to the required layout of lines. The position of lines should not be disturbed during the concrete, Care should be taken to avoid the breaking of pipes from the movement of labor and wheel borrows because broken pipes get choked and hinder the passage of electric cables in later stages which causes problems of rerouting by doing more wall trenching and fixing and inserting new cabling. 

5- Mixing and Placing of Concrete.

Concrete maybe mixed by hand or in the mechanical mixer, it should be methodically mixed and the concrete placed in its final position with the minimum delay. Normal RCC Works are constructed using (1:2:4) mix with concrete incidents as Cement, Sand, Crush, Water, Steel Reinforcement, and Admixtures. Three mixing practices can be adopted for concreting 1, By Hand Mixing, 2, By Machine Mixing 3, By Batching Plant.

The most prominent mixing method is mixing by batching plant as quantities can be measured with controlled mechanism but such mixings are only required to avail high strength concretes which requires more than 4000 PSI compressive strength. Batching plants prepare the concrete using weights of ingredients that are mentioned in the job mix formula. Transportation of concrete Transit mixer trucks is used whose minimum volume may be 5 cum. In Some places, the batching plants are available at the construction site and concrete is transported to the workplace using tower cranes and pumps along with pipelines.

Batching plants are used for massive construction projects and for normal works such a the method is not affordable at all.

Mixing by hand may be adopted for small works in which all ingredients are mixed roughly for two minutes on cleaned and water-tight platforms. An additional quantity of cement up to 10 % should be added in such mixes to ensure the quality of concrete.

In any case, water should be added only according to suitably designed Water Cement Ratios because excessive water reduces the strength of concrete.

On the Construction of slabs for small houses, mixture machines are used along with a tower lift to transporting the concrete on the heights OF slabs, if the quantity of concrete is massive than more than one machine and tower lifts can be fixed to avoid delays in the pouring of slabs.

Compaction of concrete is very important as it provides dense concrete expelling the air voids from the concrete volume and also increases the strength of concretes.

Compacting can be efficiently done using Vibrator machines, the vibration should be 1 minute for 1 cum of concrete, Over vibration should be avoided as it segregates the ingredients of concrete.

Concrete should be smooth and well leveled using wooden and steel floats, Level points may be maintained by the surveyor to ensure the exact thickness of the slab on all portions.

6- Curing of Concrete slabs.

Curing is an important activity as this saves concrete from hydration which causes cracks and strength losses in concrete structures so curing should be executed at least for 14 days for slab structures because in 14 days concrete maintains 90 % of its ultimate strength.

7-Removing of shuttering and scaffolding.

Removing of shuttering may be started after 14 days for medium spans of the building but larger spans should not be suspended before 28 days as concrete avails its 100 % strength in 28-days.