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**

_{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;

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.

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) = 175309/(0.87 × 500 × 0.9 × 542 × 2.5 ) = 0.33047

Provide 2H8mm @ 250mm c/c (A

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_{sw}/S = 0.0.402) OkThank you for visiting Structville today.

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