Inclined beams (often called raker beams) are often found in structures like pedestrian bridges, ramps, staircases, stadiums, etc. Due to their geometry, these beams are often subjected to bending moment, shear force, and axial force. For indeterminate raker beams, it is not uncommon to see the axial forces varying from tension to compression.

The analysis and design of such elements have been presented here using Eurocode 2. For the inclined rectangular raker beam loaded as shown below, we are to fully analyse and design the structure at ultimate limit state.

**Design Information**

Dimensions = 600 × 300mm

Concrete cover = 40mm

Yield strength of reinforcement = 500 N/mm^{2}

Grade of concrete = 35 N/mm^{2}

**Geomterical Properties of the structure**

**
**Length of AB = Length of BC = 7/cos 25 = 7.7236m

cos 25 = 0.9063

sin 25 = 0.4226

At ultimate limit state, load on beam

P_{Ed} = 1.35Gk + 1.5Qk = 1.35(25) + 1.5(5) = 41.25 kN/m

**Analysis of the structure by slope deflection method**

In all cases;

L = 7.0m

L’ = 7.7236m

**Fixed End Moments**

F_{BA} = qcosθL’^{2}/8 = (41.25 × 0.9063 × 7.7236^{2}) / 8 = 278.769 kNm

F_{BC} = -qcosθL’^{2}/12 = (41.25 × 0.9063 × 7.7236^{2}) / 12 = -185.846 kNm

F_{BC} = qcosθL’^{2}/12 = (41.25 × 0.9063 × 7.7236^{2}) / 12 = 185.846 kNm

K_{11} = 3EI/L’ + 4EI/L’

K_{11} = (3EI/7.7236) + (4EI/7.7236) = 0.9063EI

K_{1P} = F_{BA} + F_{BC} = 278.769 – 185.846 = 92.923 KNm

For equilibrium and compatibility;

K_{11}Z_{1} + K_{1P} = 0

0.9063Z_{1}+ 92.923 = 0

On solving;

Z_{1} = -102.53/EI (radians)

Therefore;

M_{BA} = 278.769 + (-102.53/EI × 3EI/7.7236) = 238.944 KNm

M_{BC} = -185.846 + (-102.53/EI × 4EI/7.7236) = -238.944 kNm

M_{CB} = 185.846 + (-102.53/EI × 2EI/7.7236) = 159.296 kNm

**Shear Forces and Span Moments**

**Span A-B**

**Support Reactions**

∑M_{B} = 0

7.7236R_{AB} – (41.25 × 0.9063 × 7.7236^{2})/2 = -238.944

R_{AB} = 113.436 KN

A little consideration will show that;

Ay = 131.360 KN

Ax = 13.029 KN

∑M_{A} = 0

7.7236R_{BA} – (41.25 × 0.9063 × 7.7236^{2})/2 – 238.944 = 0

R_{BA} = 175.309 KN

**Maximum span moment;**

Mz = R_{A}.z – (41.25 × 0.9063 × z^{2})/2 = 0

Mz = 113.436z – 18.692z^{2}

∂Mz/∂z = Qz = 113.436 – 37.385z

The maximum moment occurs at the point of zero shear;

Therefore, let ∂Mz/∂z = Qz = 113.436 – 37.385z = 0

On solving; z = 3.034m

M*max* = 113.436(3.034) – 18.692(3.034)^{2 }= 172.102 KNm

**Span B – C**

∑M_{C} = 0

7.7236R_{BC} – (41.25 × 0.9063 × 7.7236^{2})/2 – 238.944 = -159.296

R_{BC} = 154.685 KN

A little consideration will show that;

Ay = 131.360 KN

Ax = 13.029 KN

∑M_{B} = 0

7.7236R_{CB} – (41.25 × 0.9063 × 7.7236^{2})/2 – 159.296 + 238.944 = 0

R_{CB} = 134.060 KN

**Maximum span moment;**

Mz = R_{BC}.z – (41.25 × 0.9063 × z^{2})/2 – 238.944 = 0

Mz = 154.685z – 18.692z^{2}– 238.944 = 0

∂Mz/∂z = Qz = 154.685 – 37.385z

The maximum moment occurs at the point of zero shear;

Therefore, let ∂Mz/∂z = Qz = 154.685 – 37.385z = 0

On solving; z = 4.1376m

Mmax = 154.685(4.1376) – 18.692(4.1376)^{2 }– 238.944 = 81.078 KNm

Kindly verify that the axial force diagram for this structure is given by;

**Structural Design**

**Span**

M_{Ed} = 172.102 KN.m

Effective depth (d) = h – C_{nom} – ϕ/2 – ϕlinks

Assuming ϕ16 mm bars will be employed for the main bars, and ϕ10mm bars for the stirrups (links)

d = 600 – 40 – 8 – 10 = 542 mm

k = M_{Ed}/(f_{ck}bd^{2}) = (172.102 × 10^{6})/(35 × 300 × 542^{2}) = 0.0557

Since k < 0.167, no compression reinforcement required

z = d[0.5+ √(0.25 – 0.882K) ]

z = d[0.5+ √(0.25 – 0.882(0.0557))] = 0.948d

A_{s1} = M_{Ed}/(0.87_{yk} z) = (172.102 × 10^{6})/(0.87 × 500 × 0.948× 542) = 770 mm^{2}

Provide 4H16mm BOT (AS_{prov} = 804 mm^{2})

**Check for deflection**

**
**ρ = A

_{s,prov}/bd = 804 / (300 × 542) = 0.004944

ρ

_{0}= reference reinforcement ratio = 10

^{-3}√(f

_{ck}) = 10

^{-3}√(35) = 0.005916

Since if ρ ≤ ρ

_{0};

L/d = K [11 + 1.5√(f_{ck}) ρ_{0}/ρ + 3.2√(f_{ck}) (ρ_{0} / ρ – 1)^{(3⁄2)}

^{
}k = 1.3 (One end continuous)

L/d = 1.3 [11 + 1.5√(35) × (0.005916/0.004944) + 3.2√(35) × [(0.005916 / 0.004944) – 1]^{(3⁄2)}

L/d = 1.3[11 + 10.619 + 1.650] = 30.250

β_{s} = (500 As_{prov})/(F_{yk} As_{req}) = (500 × 804) / (500 × 770) = 1.0441

Therefore limiting L/d = 1.0441 × (7/7.7236) × 30.250 = 28.625

Actual L/d = 7723.6/542 = 14.25

Since Actual L/d < Limiting L/d, deflection is satisfactory.

**Support B**

M_{Ed} = 238.944 KN.m

Effective depth (d) = h – C_{nom} – ϕ/2 – ϕlinks

Assuming ϕ16 mm bars will be employed for the main bars, and ϕ8mm bars for the stirrups (links)

d = 600 – 40 – 8 – 10 = 542 mm

k = M_{Ed}/(f_{ck}bd^{2}) = (238.944 × 10^{6})/(35 × 300 × 542^{2}) = 0.0775

Since k < 0.167, no compression reinforcement required

z = d[0.5+ √(0.25 – 0.882K) ]

z = d[0.5+ √(0.25 – 0.882(0.0775))] = 0.926d

A_{s1} = M_{Ed}/(0.87f_{yk} z) = (238.944 × 10^{6})/(0.87 × 500 × 0.926 × 542) = 1094 mm^{2}

Provide 6H16 mm TOP (AS_{prov} = 1206 mm^{2})

**Shear Design (Support A)**

V_{Ed} = 113.436 kN

N_{Ed} = 67.323 KN (compression)

We are going to anchor the 4No of H16mm reinforcement provided fully into the supports.

V_{Rd,c} = [C_{Rd,c}.k.(100ρ_{1} f_{ck})^{(1/3)} + k_{1}.σ_{cp}]b_{w}.d

Where;

C_{Rd,c} = 0.18/γ_{c} = 0.18/1.5 = 0.12

σ_{cp} = N_{Ed} / bd = (67.323 × 1000) / (542 × 300) = 0.414 N/mm^{2}

k = 1 + √(200/d) = 1 + √(200/542) = 1.607 > 2.0, therefore, k = 1.702

ρ_{1} = As/bd = 804/(300 × 542) = 0.004944 < 0.02; K_{1} = 0.15

V_{Rd,c} = [0.12 × 1.607(100 × 0.004944 × 35 )^{(1/3)} + (0.15 × 0.414)] × 300 × 542 = 91199.759 N = 91.199 KN

Since V_{Rd,c} (91.999 KN) < V_{Ed} (113.436 KN), shear reinforcement is required.

The compression capacity of the compression strut (V_{Rd,max}) assuming θ = 21.8° (cot θ = 2.5)

V_{Rd,max} = (b_{w}.z.v_{1}.f_{cd})/(cotθ + tanθ)

V_{1} = 0.6(1 – f_{ck}/250) = 0.6(1 – 35/250) = 0.516

f_{cd} = (α_{cc} ) f_{ck})/γ_{c} = (1 × 35)/1.5 = 23.33 N/mm^{2}

Let z = 0.9d

V_{Rd,max} = [(300 × 0.9 × 542 × 0.516 × 23.333) / (2.5 + 0.4)] × 10^{-3} = 607.554 KN

Since V_{Rd,c} < V_{Ed} < V_{Rd,max}

Hence, A_{sw}/S = V_{Ed}/(0.87F_{yk}zcot θ) = 113436/(0.87 × 500 × 0.9 × 542 × 2.5 ) = 0.21383

**Minimum shear reinforcement;**

A_{sw}/S = ρ_{w,min} × b_{w} × sinα (α = 90° for vertical links)

ρ_{w,min} = (0.08 × √(f_{ck}))/f_{yk} = (0.08 × √35)/500 = 0.0009465

A_{sw}/S (min) = 0.0009465 × 300 × 1 = 0.2839

Since 0.2839 > 0.21383, adopt 0.2839 (i.e the minimum reinforcement)

Maximum spacing of shear links = 0.75d = 0.75 × 542 = 406.5mm

Provide 2H8mm @ 300mm c/c (A_{sw}/S = 0.335) Ok

Note that this link must be properly closed with adequate anchorage length, because it will assist in resisting torsion.

**Shear Design (Support B just to the left)**

V_{Ed} = 175.309 KN

N_{Ed} = 67.323 KN (tension)

We are going to anchor the 6No of H16mm reinforcement provided fully into the supports.

V_{Rd,c} = [C_{Rd,c}.k.(100ρ_{1} f_{ck})^{(1/3)} + k_{1}.σ_{cp}]b_{w}.d

Where;

C_{Rd,c} = 0.18/γ_{c} = 0.18/1.5 = 0.12

σ_{cp} = N_{Ed} / bd = (67.323 × 1000) / (542 × 300) = – 0.414 N/mm^{2}

k = 1 + √(200/d) = 1 + √(200/542) = 1.607 > 2.0, therefore, k = 1.702

ρ_{1} = As/bd = 1206/(300 × 542) = 0.00741 < 0.02; K_{1} = 0.15

V_{Rd,c} = [0.12 × 1.607(100 × 0.00741 × 35 )^{(1/3)} – (0.15 × 0.414)] × 300 × 542 = 82716.452 N = 82.716 KN

Since V_{Rd,c} (82.716 KN) < V_{Ed} (175.309 KN), shear reinforcement is required.

_{Rd,max}) assuming θ = 21.8° (cot θ = 2.5)

V_{Rd,max} = (b_{w}.z.v_{1}.f_{cd})/(cotθ + tanθ)

V_{1} = 0.6(1 – f_{ck}/250) = 0.6(1 – 35/250) = 0.516

f_{cd} = (α_{cc} ) f_{ck})/γ_{c} = (1 × 35)/1.5 = 23.33 N/mm^{2}

Let z = 0.9d

V_{Rd,max} = [(300 × 0.9 × 542 × 0.516 × 23.333) / (2.5 + 0.4)] × 10^{-3} = 607.554 KN

Since V_{Rd,c} < V_{Ed} < V_{Rd,max}

Hence, A_{sw}/S = V_{Ed}/(0.87F_{yk}zcot θ) = 175309/(0.87 × 500 × 0.9 × 542 × 2.5 ) = 0.33047

_{sw}/S = 0.0.402) Ok

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Good job. Can you please include the rebar detailing diagram. I want to know if the stirrups will be vertical or perpendicular to the axis of the beam.

What is the difference between the raker beam and slant column?

The stirrups should be inclined (perpendicular to the section).