Cantilever beams are beams that are free at one end and rigidly fixed at the other end. Beams that are free at one end and continuous through the other support (beams with overhangs) are also treated as cantilever beams. The primary design of cantilever beams involves the selection of an adequate cross-section and reinforcements to resist the internal stresses due to the applied loads and to limit the deflection to an acceptable minimum.

Due to the structural system of cantilever beams, they are very sensitive to deflection and vibration. When loaded, the maximum shear force and bending moment occur at the fixed support, while the maximum deflection occurs at the free end. As a result, under normal circumstances in reinforced concrete design, the length of cantilevers is usually kept to a minimum in order not to have bulky sections and heavy reinforcements. In order to save material and to reduce the load due to self-weight, cantilever beams can be tapered, increasing linearly from the free end to the fixed support.

For cantilever beams, the tensile moment occurs at the top, therefore the main reinforcements are provided at the top. At the bottom, standard beam detailing requirements recommend that at least 50% of the reinforcement provided at the top be provided at the bottom. The anchorage length of the top reinforcement is expected to enter at least 0.25 times the effective span of the backspan or 1.25 times the effective length of the cantilever (whichever is greater).

A cantilever beam relies on the backspan or an alternative counterweight for equilibrium or structural stability.

**Design Example of a Cantilever Beam**

Design a two-span cantilever beam (beam with overhang with the following information provided). The beam is to support a rendered 230 mm hollow block wall up to a height of 2.7m, in addition to the load transferred by the floor slab. The ultimate design load on the slab is 12 kN/m^{2}. f_{ck} = 25 N/mm^{2}; f_{yk} = 500 N/mm^{2}; Concrete cover = 35mm; Unit weight of concrete = 25 kN/m^{3}; Unit of block wall = 3.5 kN/m^{2 }

**Load Analysis**

Aspect ratio of the slab k = L_{y}/L_{x} = 6/5 = 1.2

Factored load transferred from slab to beam B1 = 0.5(12 × 5) × [1 – 0.333(1.2)^{2}] = 15.61 kN/m

Factored load transferred from slab to beam B2 (factored) = γ_{G}nl_{x}/4 = 1.35 × (12 × 2.5)/4 = 10.125 kN/m

Factored self-weight of the beam (considering the 300 mm drop) = 1.35(25 × 0.3 × 0.23) = 2.33 kN/m

Factored weight of block wall = 1.35(3.5 × 2.7) = 12.76 kN/m

Load on Beam B1 = 15.61 + 2.33 + 12.76 = 30.7 kN/m

Load on neam B2 = 10.125 + 2.33 + 12.76 = 25.22 kN

**Concrete details – Strength and deformation characteristics for concrete**

Concrete strength class; C25/30

Aggregate type; Quartzite

Aggregate adjustment factor – cl.3.1.3(2); AAF = 1.0

Characteristic compressive cylinder strength; f_{ck} = 25 N/mm^{2}

Mean value of compressive cylinder strength; f_{cm} = f_{ck} + 8 N/mm^{2} = 33 N/mm^{2}

Mean value of axial tensile strength; f_{ctm} = 0.3 N/mm^{2} × (f_{ck}/ 1 N/mm^{2})^{2/3} = 2.6 N/mm^{2}

Secant modulus of elasticity of concrete; E_{cm} = 22 kN/mm^{2} × [f_{cm}/10 N/mm^{2}]^{0.3} × AAF = 31476 N/mm^{2}

Ultimate strain – Table 3.1; ε_{cu2} = 0.0035

Shortening strain – Table 3.1; ε_{cu3} = 0.0035

Effective compression zone height factor; λ = 0.80

Effective strength factor; η = 1.00

Coefficient k_{1}; k_{1} = 0.40

Coefficient k_{2}; k_{2} = 1.0 × (0.6 + 0.0014 / ε_{cu2}) = 1.00

Coefficient k_{3}; k_{3} = 0.40

Coefficient k_{4}; k_{4} = 1.0 × (0.6 + 0.0014 / ε_{cu2}) = 1.00

Partial factor for concrete -Table 2.1N; γ_{C} = 1.50

Compressive strength coefficient – cl.3.1.6(1); α_{cc} = 0.85

Design compressive concrete strength – exp.3.15; f_{cd} = α_{cc} × f_{ck} / γ_{C} = 14.2 N/mm^{2}

Compressive strength coefficient – cl.3.1.6(1); α_{ccw} = 1.00

Design compressive concrete strength – exp.3.15; f_{cwd} = α_{ccw} × f_{ck} / γ_{C} = 16.7 N/mm^{2}

Maximum aggregate size; h_{agg} = 20 mm

Monolithic simple support moment factor; β_{1} = 0.25

**Reinforcement details**

Characteristic yield strength of reinforcement; f_{yk} = 500 N/mm^{2}

Partial factor for reinforcing steel – Table 2.1N; γ_{S} = 1.15

Design yield strength of reinforcement; f_{yd} = f_{yk} / γ_{S} = 435 N/mm^{2}

**Nominal cover to reinforcement**

Nominal cover to top reinforcement; c_{nom_t} = 35 mm

Nominal cover to bottom reinforcement; c_{nom_b} = 35 mm

Nominal cover to side reinforcement; c_{nom_s} = 35 mm

**Fire resistance**

Standard fire resistance period; R = 60 min

Number of sides exposed to fire; 3

Minimum width of beam – EN1992-1-2 Table 5.5; b_{min} = 120 mm

**Flexural** **Design of the Cantilever Section**

Design bending moment; M_{Ed} = 78.8 kNm

Distance between points of zero moment; L_{0} = (0.15 × L_{m1_s1}) + L_{m1_s2} = (0.15 × 6000) + 2500 = 3400 mm

Maximum flange outstand; b_{1} = b_{f} – b = 720 mm

Effective flange outstand; b_{eff,1} = min(0.2 × b_{1} + 0.1 × L_{0}; 0.2 × L_{0}; b_{1}) = 484 mm

Effective flange width; b_{eff} = b_{eff,1} + b = 714 mm

Effective depth of tension reinforcement; d = 399 mm

K = M / (b_{eff} × d^{2} × f_{ck}) = 0.028

K’ = (2 × η × α_{cc} / γ_{C}) × (1 – λ × (δ – k_{1}) / (2 × k_{2})) × (λ × (δ – k_{1}) / (2 × k_{2})) = 0.207

Lever arm; z = min(0.5 × d × [1 + (1 – 2 × K / (η × α_{cc} / γ_{C})^{0.5}], 0.95 × d) = 379 mm

Depth of neutral axis; x = 2 × (d – z) / λ = 50 mm

λx < h_{f} – Compression block wholly within the depth of flange

K’ > K – No compression reinforcement is required

Area of tension reinforcement required; A_{s,req} = max(M / (f_{yd} × z), A_{s,min}) = 478 mm^{2}

Tension reinforcement provided;3H16

Area of tension reinforcement provided; A_{s,prov} = 603 mm^{2}

Minimum area of reinforcement – exp.9.1N; A_{s,min} = max(0.26 × f_{ctm} / f_{yk}, 0.0013) × b × d = 122 mm^{2}

Maximum area of reinforcement – cl.9.2.1.1(3); A_{s,max} = 0.04 × b × h = 4140 mm^{2}** PASS **– Area of reinforcement provided is greater than area of reinforcement required

**Deflection control **

Reference reinforcement ratio; ρ_{m0} = (f_{ck} )^{0.5} / 1000 = 0.00500

Required tension reinforcement ratio; ρ_{m} = A_{s,req} / (b_{eff} × d) = 0.00168

Required compression reinforcement ratio; ρ’_{m} = A_{s2,req} / (b_{eff} × d) = 0.00000

Structural system factor – Table 7.4N; K_{b} = 0.4

Basic allowable span to depth ratio ; span_to_depth_{basic} = K_{b} × [11 + 1.5 × (f_{ck})^{0.5} × ρ_{m0} / ρ_{m} + 3.2 × (f_{ck})^{0.5} × (ρ_{m0}/ρ_{m} – 1)^{1.5}] = 31.160

Reinforcement factor – exp.7.17;K_{s} = min(A_{s,prov} / A_{s,req} × 500 N/mm^{2} / f_{yk}, 1.5) = 1.262

Flange width factor; F1 = if(b_{eff} / b > 3, 0.8, 1) = 0.800

Long span supporting brittle partition factor; F2 = 1 = 1.000

Allowable span to depth ratio; span_to_depth_{allow} = min(span_to_depth_{basic} × K_{s} × F1 × F2, 40 × K_{b}) = 16.000

Actual span to depth ratio; span_to_depth_{actual} = L_{m1_s2} / d = 6.266

** PASS** – Actual span to depth ratio is within the allowable limit

**Shear Design**

Angle of comp. shear strut for maximum shear; θ_{max} = 45 deg

Strength reduction factor – cl.6.2.3(3); v_{1} = 0.6 × (1 – f_{ck} / 250) = 0.540

Compression chord coefficient – cl.6.2.3(3); α_{cw} = 1.00

Minimum area of shear reinforcement – exp.9.5N; A_{sv,min} = 0.08 N/mm^{2} × b × (f_{ck} )^{0.5} / f_{yk} = 184 mm^{2}/m

Design shear force at support ; V_{Ed,max} = 63 kN

Min lever arm in shear zone; z = 379 mm

Maximum design shear resistance – exp.6.9; V_{Rd,max} = α_{cw} × b × z × v_{1} × f_{cwd} / (cot(θ_{max}) + tan(θ_{max})) = 392 kN*PASS* – Design shear force at support is less than maximum design shear resistance

Design shear force at 399mm from support; V_{Ed} = 53 kN

Design shear stress; v_{Ed} = V_{Ed} / (b × z) = 0.608 N/mm^{2}

Angle of concrete compression strut – cl.6.2.3; θ = min(max(0.5 × Asin(min(2 × v_{Ed} / (α_{cw} × f_{cwd} × v_{1}),1)), 21.8 deg), 45deg) = 21.8 deg

Area of shear reinforcement required – exp.6.8; A_{sv,des} = v_{Ed} × b / (f_{yd} × cot(θ)) = 129 mm^{2}/m

Area of shear reinforcement required; A_{sv,req} = max(A_{sv,min}, A_{sv,des}) = 184 mm^{2}/m

Shear reinforcement provided; 2 × 8 legs @ 200 c/c

Area of shear reinforcement provided; A_{sv,prov} = 503 mm^{2}/m** PASS** – Area of shear reinforcement provided exceeds minimum required

Maximum longitudinal spacing – exp.9.6N; s_{vl,max} = 0.75 × d = 299 mm** PASS **– Longitudinal spacing of shear reinforcement provided is less than maximum

I get some confusing as

1- transferred load to B1 should be equal (23.05 kN/m) from equation 0.5nlx (1-1/(3k2))

2- transferred load to B2 why you multiply the load with 1.35 while the load is given ultimate which mean factored as B1