Beams are Basically Horizontal Structural members which support structural members like slabs, staircase etc. and transfer the loads to vertical members. They resist the loads through bending action, due to which they fall in category of flexure members.

Flexure Member

A flexure member is a structural body that resist loads and Moments through bending action which forms zones of both compression and tension within its body.

Examples

• Beams
• Spandrel
• Girder
• Purlins
• Joist
• Lintels

Beam as a flexure member:

Under the action of loads and concentrated moments, Beams undergoes bending which creates zone of tension at top fibers near ends and at bottom fiber near the middle portion. Similarly, Compression zones are formed opposite of tension zones.

The moment generated at end portions is called Hogging Moment and Moment at Mid Portion is called Sagging Moment.

Hence, we design a beam at these three sections i.e. at Middle and at Ends.

The distribution of tensile and compressive zones so created in beams calls for design of singly reinforced sections. The section at ends where hogging moment causes tension at top, can be visualized as inverted section (Tension Bars above the Neutral Axis).

Theoretically, In compression zones, compression steel is provided to counter extra moments.

But In Practice, we have to provide at least 2 bars at top as well as bottom. Also in practical scenario, due to several factors, steel is widely required everywhere, as moments are generated due to torsion and other factors. We design beams for three section namely – left, mid and right for both top and bottom portions.

Types of Beams

• Simply Supported or End Bearing
• Continuous
• Cantilever

Manuel Design of Beams

We Consider a Beam of width B and overall depth D.

Grade of Concrete = f_ck

Grade of Steel = f_y

We assume a cover of 25 mm for the Beam for general case.

Now Effective depth = d = Overall depth (D) – Net Clear Cover – Diameter of Bar (assumed initially) / 2

Effective depth is calculated upto the centre of gravity of bottom bar arrangement.

Effective Depth calculation for 2 layers of reinforcement (Industry level Design) :

We have, Effective depth = d = Overall depth (D) – Net Clear Cover – Diameter of assumed bar in bottom most layer(16 mm commonly) – Spacer Bar Diameter / 2

Effective Depth calculation for 3 layers of reinforcement (Industry level Design) :

Effective depth = d = Overall depth (D) – Net Clear Cover – Diameter of assumed bar in bottom most layer(16 mm commonly) – Spacer Bar Diameter – Assumed bar diameter/ 2

Here, clear cover also includes stirrups diameter.

Spacer bars are placed in order to maintain gap between the layers of main longitudinal reinforcements. They are placed in transverse direction. The Main bars of the layers can be easily anchored to spacer bars using binding wires. We consider 25 mm Diameter spacer bar most commonly.

Assumption of number of layers of Reinforcement (As per RCDC Software)

For Beam depth greater than 400 mm and less than 1200 mm, we assume 2 layers.

For Beam depth greater than 1200 mm and less than 1350 mm, we assume 3 layers. And So on.

Nominal Cover for Different Exposure conditions

Clear cover = Nominal cover + stirrup diameter (8 mm)

Given Moment at Section = M_u

Now, Limiting Moment of Resistance of the Section = M_{u limit}

As per IS 456

M_{u limit} = 0.138f_ck bd² (For Fe 415)

M_{u limit} = 0.149f_ck bd² (For Fe 250)

Likewise for Fe 500 and Fe 550

M_{u limit} = 0.133f_ck bd² (For Fe 500)

M_{u limit} = 0.129f_ck bd² (For Fe 550)

Now, If the Moment at the section is less than the Limiting Moment of Resistance, then Section is considered to be Singly Reinforced. But if the case is otherwise, the section becomes Double Reinforced, since there is additional leftover moment.

Singly Reinforced Beams Sections

Next, We refer IS Code SP 16 table for singly reinforced section percentage of steel. The tables gives value of percentage of steel based on the input values of grade of steel, Concrete grade, and Value of M_u/bd²

Now, We can find out the steel required by multiplying the percentage value by Area of Section.

A_st = P_tbd/100

Alternatively, We can use a general formula given below to calculate steel percentage.

As We proceed further, we select a bar diameter or combination of bar diameters (as per our judgement), then we provide a bar arrangement such that the Provided Steel area is greater than the Area Required.

The Code specifically restrict maximum steel percentage to be 6 % , practically taken as 4 % (to avoid congestion).

Double Reinforced Beams

The additional moment of resistance needed is obtained by compression reinforcement and additional tensile reinforcement.

Additional Moment Required M_u2 = M_u – M_{u lim}

It is needed as beam size is pre-determined due to design, economy and site constraints.

\blue {M_{u2}=A_{sc}(f_{sc}-f_{cc})(d-d^{'})}

\blue {M_{u2}=A_{st2}(0.87f_{y})(d-d^{'})}

A_st2 – Area of additional tensile reinforcement.

A_sc – Area of Compression reinforcement.

f_sc – Stress in Compression reinforcement.

f_cc – Compressive stress in concrete at level of centroid of compression reinforcement.

Stress at level of compression reinforcement = 0.35(1-d’/X_{u max})

Value of f_cc for values of d’/d upto 0.2, is 0.446f_ck

Value of f_sc is taken from table F using values of d’/d and grade of steel.

For Mild Steel : f_sc = 0.87f_y

Additional tensile bar A_st2

\blue {\large A_{st2}= \frac{A_{sc(f_{sc}-f_{cc})}}{0.87f_{y}}}

Maximum percentage of tensile reinforcement for singly reinforced section (p_{t lim} )

A_st1 is taken up from table E using values of grade of concrete and steel.

A_st1 = p_{t lim} b d / 100

Total Tensile reinforcement A_st = A_st1 + A_st2

Now we detail top bars as per A_sc and bottom bars as per A_st

Total steel at section should be less than 4% and should be greater than 0.87 bd/f_y i.e. ( 0.17 % for Fe 500 and 0.15% for Fe 550).

Control for Beams Deflection

Span to depth ratio should not be greater than vertical deflection limit values.

Limit Values :

Cantilever – 7

Simply Supported – 20

Continuous – 26

Numerically We can say that the given section is safe if it satisfy the following condition :

\blue{\frac{L}{D}< K_{1}K_{2}K_{3}K_{4}}

Here, K₁ is the codal limit given above.

K₂ is modification factor for tensile steel.

K₃ is modification factor for compressive steel.

And, K₄ is a factor for flanged beam.

For spans above 10 m in length, we need to multiply this value by 10/span (m).

Modification factor for tensile reaction K₂ can be found in Figure 4 (IS 456).

Percentage tensile reaction, f_s yield value of K

\blue {f_{s}=0.58f_{y}\frac{Area\; of \; steel \;required}{Area\; of \; steel \;provided}}

Modification factor for compressive reaction K₃ is determined from Fig 5 using value of percentage compression reaction.

For flanged beam, section area = b_fd

Ration of web width to flange width gives K₄ (Fig 6)

Extra modification factor for cantilever beams = 0.35

Extra modification factor for Continuous beams = 1.3

For 2 way slab, shorter span shall be considered for deflection check.

Beams Shear Reinforcement

Given area of tensile (longitudinal) steel :

\blue {p_{t}=100\frac{A_{st}}{bd}}

Critical shear stress is determined using Table 19 of IS 456 using concrete grade value Versus p_t value.

Maximum critical stress for grade (Table 20) :

Concrete Shear Resistance

\blue {V_{uc}=\frac{\tau_{c}bd}{100}}

No. of legs of stirrups (n) and Diameter of stirrups (d) are specified accordingly as per requirements. Then, we calculate area of stirrup A_sv

Net Given Shear = V_u

τ_v = V_u/bd

If τ_v > τ_c – Provide Shear Reinforcement.

Spacing of Stirrups :

Additional Shear = V_us = V_u – V_uc

Area of Provided Stirrups is given by :

\blue{A_{sv}=n\times\frac{\pi}{4}\times d^2}

Spacing of stirrups shall be minimum of following :

• 0.75d
• 300 mm
• (0.87f_yA_svd)/(V_us)
• (0.87f_yA_sv)/(0.4b)

Now, Shear for taken by minimum reinforcement V_s = (0.87f_yA_svd)/(S)

where ‘S’ is a specified spacing.

Total Shear Resistance = Concrete Shear Resistance (V_uc) + Shear Resistance by minimum Reinforcement (V_s)

Beams Torsional Reinforcement

We consider an equivalent shear force and an equivalent bending moment for the torsion.

Torsion – T

Factored Shear Force = V

Factored Moment = M

Width of section = B

Overall Depth of section = D

Now, Equivalent Shear force

\blue {V_{e}=V+1.6\frac{T}{B}}

Equivalent Bending Moment

\blue {M_{e}=M+\frac{T(1+\frac{D}{B})}{1.7}}

Now, we design section using these total equivalent moments and shear forces in similar ways.

Beams Side Face Reinforcement

If the depth of beam is more than 750 mm, then we need to provide side face reinforcement equal to at least 0.1 % of cross section area (web area). The spacing of side face bars should not exceed minimum of 300 mm or width of section. For each side the percentage becomes half that is 0.05 %.

We Provide Side Face Reinforcement in order to control the crack width in tensile zone of concrete. It also prevents cracking in side exposed faces of beam due to temperature variations and shrinkage.

For torsion moment case, The limit of depth is 450 mm.

\purple {A_{req}=\frac{0.05bd_{s}}{100}}

Development Length

It is the minimum length required by bars to get proper anchorage.

It includes anchorage values of hooks and bends in tensile reinforcement. Also, It increases by 60 % for deformed bars and for Compression bars we multiply length by 1.25 factor.

\blue {L_{d}=\frac{\phi \sigma_{s}}{4\tau_{bd}}}

L_d : Development Length

τ_bd : Design Bond Stress

σ_s : Stress in Bar

ϕ : Diameter of bar

There should be no lapping in bars of diameter greater than 36 mm. They are either welded or spliced.

Illustration with An Example : 

Give Data 
Beam Type : Cantilever
Section to be considered : Top left
fck = 30 N/mm2
fy  = 550 N/mm2
Clear Cover = 38 mm    (30 mm nominal cover + 8 mm diameter of stirrup)
Length of beam = 2.9 m = 2900 mm
Width of beam = 200 mm
Depth of beam = 700 mm
Applied Moment Mu (top left) = 225 kNm (Factored Bending moment in limit state) 
Mu (top right) = 852.92 kNm (Additional data for developing judgment for design)
Torsion T = 0 kNm 
MT = 0 kNm
Now, Design Moment Mud = Mu + MT = 225 kNm
Width is 700 mm, hence we can consider 2 layers of reinforcement in tensile zone. But let us design with three layers as Moment is high in value in top right section, which fails for 2 layers of bars even for largest diameter (32 mm) and also we need to give uniform reinforcement all over. We take trial bar diameter to be 16 mm.
Spacer bar diameter = 25 mm
Effective Depth = d = 700 (D) - 38 (Clear cover) - 16 (Assumed bar)- 25 (Spacer bar) - 16/2 (Half of assumed bar present in central band of 3 layers) = 613 mm ≈ 612 mm
Now, Mu limit = 0.129fck bd2 (For Fe 550) = (0.129 x 30 x 200 x 6122 )/106 = 289.89 kNm
Here, Mud < Mu limit, hence, the section will be designed as singly reinforced section.

Now, Moment ratio R = Mu / bd2 = 2.997
and grade of concrete is M30, 
From SP 16 table reference, we get percentage of steel required pt = 0.723
So, Area of steel required = (0.723 x 200 x 612) / 100 = 884.95 mm2
Since, we came up with idea of three layer of reinforcement, let us provide 3 layers of 2 no. of 16 mm bars (Ast = 2 x 200 = 400 mm2 ), making a total of 1200 mm2 area.

Note:
Here, we have given 3 layers of 2 bars of 16mm diameter, but we need to design top right section (higher moment value), and finally provide maximum of two.  

Data for Shear Design : (left part)
Given Shear Vu = 233.2 kN
Shear due to torsion Vtu = 0 (since Torsion T = 0 kNm)
So, Design Shear V = Vu = 233.3 kN
Now, τv = Vu/bd = 233.2 X 1000/(200 x 612) = 1.91 N/mm2
τc = 0.92 N/mm2(based on pt and grade of concrete M30)
Shear taken by concrete Vc = τc bd = 0.92x200x612/1000 = 112.6 kN
So, Shear taken by steel Vus= Vu - Vc = 233.3 - 112.6 = 120.7 kN
For width 200 mm beam with 2 bars in one layer, we will provide 2 legged stirrup of diameter 10 mm.
Asv = 2 x 3.14 x 10 x 10 / 4 = 157 mm2
Next, let us check the conditions to determine  minimum spacing of stirrups
1. 0.75 d = 0.75 x 612 = 459 mm
2. 300 mm
3. 0.87fyAsvd/Vus = 0.87 x 550 x 157 x 612 / (120.7 x 1000) = 380.91 mm
4. 0.87fyAsv/0.4b = 0.87 x 550 x 157 / (0.4 x 200) = 939 mm

We will take spacing of 100 mm as the beam is cantilever (more strength needed). 

The depth of beam is 700 mm which is less than 750 mm, so side face reinforcement is  not required.