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.
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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)2 = 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)2 – 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) + k1.σcp]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)/(cotθ + 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) + k1.σcp]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.
VRd,max = (bw.z.v1.fcd)/(cotθ + 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
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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).
Thank you.Such a useful content. Could you please explain what the purpose of finding K11 and K1P?
After getting fixed ends moments, without considering Z1, can it be used directly to find support reactions?? But, here, there is one more step than I have known- modifying the inital fixed end moments by considering Z1.
Good job.