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Thursday, November 30, 2017

Design of Reinforced Concrete Slanted / Raker Beams


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/mm2
Grade of concrete = 35 N/mm2

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
PEd = 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
FBA = qcosθL'2/8 = (41.25 × 0.9063 × 7.72362) / 8 = 278.769 kNm
FBC = -qcosθL'2/12 = (41.25 × 0.9063 × 7.72362) / 12 = -185.846 kNm
FBC = qcosθL'2/12 = (41.25 × 0.9063 × 7.72362) / 12 = 185.846 kNm



K11 = 3EI/L' + 4EI/L'
K11 = (3EI/7.7236) +  (4EI/7.7236) = 0.9063EI

K1P = FBA + FBC = 278.769 - 185.846 = 92.923 KNm



For equilibrium and compatibility;
K11Z1 + K1P = 0
0.9063Z1+ 92.923 = 0

On solving;
Z1 = -102.53/EI (radians)

Therefore;
MBA = 278.769  + (-102.53/EI × 3EI/7.7236) = 238.944 KNm
MBC = -185.846 + (-102.53/EI × 4EI/7.7236) = -238.944 kNm
MCB = 185.846 +  (-102.53/EI × 2EI/7.7236) = 159.296 kNm

Shear Forces and Span Moments

Span A-B


Support Reactions
∑MB = 0
7.7236RAB - (41.25 × 0.9063 × 7.72362)/2 = -238.944
RAB = 113.436 KN

A little consideration will show that;
Ay = 131.360 KN
Ax = 13.029 KN

∑MA = 0
7.7236RBA - (41.25 × 0.9063 × 7.72362)/2  - 238.944 = 0
RBA = 175.309 KN

Maximum span moment;
Mz = RA.z - (41.25 × 0.9063 × z2)/2 = 0
Mz = 113.436z - 18.692z2

∂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

Mmax  = 113.436(3.034) - 18.692(3.034)= 172.102 KNm

Span B - C



∑MC = 0
7.7236RBC - (41.25 × 0.9063 × 7.72362)/2 - 238.944 = -159.296

RBC = 154.685 KN

A little consideration will show that;
Ay = 131.360 KN
Ax = 13.029 KN

∑MB = 0
7.7236RCB - (41.25 × 0.9063 × 7.72362)/2  - 159.296 + 238.944 = 0
RCB = 134.060 KN

Maximum span moment;
Mz = RBC.z - (41.25 × 0.9063 × z2)/2 - 238.944 = 0
Mz = 154.685z - 18.692z2- 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)- 238.944 =  81.078 KNm




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




Structural Design
Span
MEd = 172.102 KN.m

Effective depth (d) = h – Cnom – 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 = MEd/(fckbd2) = (172.102 × 106)/(35 × 300 × 5422) = 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

As1 = MEd/(0.87yk z) = (172.102 × 106)/(0.87 × 500 × 0.948× 542) = 770 mm2
Provide 4H16mm BOT (ASprov = 804 mm2)



Check for deflection

ρ = As,prov /bd = 804 / (300 × 542) = 0.004944
ρ0 = reference reinforcement ratio = 10-3√(fck) = 10-3√(35) = 0.005916
Since if ρ ≤ ρ0;

L/d = K [11 + 1.5√(fck) ρ0/ρ + 3.2√(fck) (ρ0 / 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

βs = (500 Asprov)/(Fyk Asreq) = (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
MEd = 238.944 KN.m

Effective depth (d) = h – Cnom – 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 = MEd/(fckbd2) = (238.944 × 106)/(35 × 300 × 5422) = 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

As1 = MEd/(0.87fyk z) = (238.944 × 106)/(0.87 × 500 × 0.926 × 542) = 1094 mm2
Provide 6H16 mm TOP (ASprov = 1206 mm2)

Shear Design (Support A)
VEd = 113.436 kN
NEd = 67.323 KN (compression)

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

VRd,c = [CRd,c.k.(100ρ1 fck)(1/3) + k1cp]bw.d

Where;
CRd,c = 0.18/γc = 0.18/1.5 = 0.12
σcp = NEd / bd = (67.323 × 1000) / (542 × 300) = 0.414 N/mm2
k = 1 + √(200/d) = 1 + √(200/542) = 1.607 > 2.0, therefore, k = 1.702
ρ1 = As/bd = 804/(300 × 542) = 0.004944 < 0.02; K1 = 0.15

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

Since VRd,c (91.999 KN) < VEd (113.436 KN), shear reinforcement is required.



The compression capacity of the compression strut (VRd,max) assuming θ = 21.8° (cot θ = 2.5)

VRd,max = (bw.z.v1.fcd)/(cot⁡θ + tanθ)
V1 = 0.6(1 - fck/250) = 0.6(1 - 35/250) = 0.516
fcd = (αcc ) fck)/γc = (1 × 35)/1.5 = 23.33 N/mm2
Let z = 0.9d

VRd,max = [(300 × 0.9 × 542 × 0.516 × 23.333) / (2.5 + 0.4)] × 10-3 = 607.554 KN

Since VRd,c < VEd < VRd,max
Hence, Asw/S = VEd/(0.87Fykzcot θ) = 113436/(0.87 × 500 × 0.9 × 542 × 2.5 ) = 0.21383

Minimum shear reinforcement;
Asw/S = ρw,min × bw × sinα (α = 90° for vertical links)
ρw,min = (0.08 × √(fck))/fyk = (0.08 × √35)/500 = 0.0009465
Asw/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 (Asw/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)
VEd = 175.309 KN
NEd = 67.323 KN (tension)

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

VRd,c = [CRd,c.k.(100ρ1 fck)(1/3) + k1cp]bw.d

Where;
CRd,c = 0.18/γc = 0.18/1.5 = 0.12
σcp = NEd / bd = (67.323 × 1000) / (542 × 300) = - 0.414 N/mm2
k = 1 + √(200/d) = 1 + √(200/542) = 1.607 > 2.0, therefore, k = 1.702
ρ1 = As/bd = 1206/(300 × 542) = 0.00741 < 0.02; K1 = 0.15

VRd,c = [0.12 × 1.607(100 × 0.00741 × 35 )(1/3) - (0.15 × 0.414)]  × 300 × 542 =  82716.452 N = 82.716 KN

Since VRd,c (82.716 KN) < VEd (175.309 KN), shear reinforcement is required.

The compression capacity of the compression strut (VRd,max) assuming θ = 21.8° (cot θ = 2.5)

VRd,max = (bw.z.v1.fcd)/(cot⁡θ + tanθ)
V1 = 0.6(1 - fck/250) = 0.6(1 - 35/250) = 0.516
fcd = (αcc ) fck)/γc = (1 × 35)/1.5 = 23.33 N/mm2
Let z = 0.9d

VRd,max = [(300 × 0.9 × 542 × 0.516 × 23.333) / (2.5 + 0.4)] × 10-3 = 607.554 KN

Since VRd,c < VEd < VRd,max
Hence, Asw/S = VEd/(0.87Fykzcot θ) = 175309/(0.87 × 500 × 0.9 × 542 × 2.5 ) = 0.33047
Provide 2H8mm @  250mm c/c (Asw/S = 0.0.402) Ok

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