Beams that are curved in plan are often found in buildings, circular reservoirs, bridges, and other structures with curves. Curved beams always develop torsion (twisting) in addition to bending moment and shear forces because the centre of gravity of loads acting normal to the plane of the structure lies outside the lines joining its supports. Therefore to maintain equilibrium in the structure, the supports of a curved beam must be fixed or simply continuous.

In this post, we are going to present in the most simplified manner, how to analyse continuous circular (ring) beams.

For ring beams;

Maximum Negative Moment at any support = K_{1}wr^{2}

Maximum Positive Moment at any span = K_{2}wr^{2}

Maximum Torsional Moment = K_{3}wr^{2}

Total load on each column (support reaction) R = wr(2θ)

Shear force at any support = R/2 = wrθ

The coefficients are given in the table below;

**Solved Example**

A cylindrical reservoir with a diameter of 6m is supported by a ring beam, which is supported on 8 equidistant columns.

It is desired to analyse and design the ring beam to support the load from the superstructure.

The plan view of the structural disposition of the reservoir is shown below;

**Load Analysis**

**(a) Geometry of sections**

Dimension of beams = 450mm x 300mm

Dimension of column = ϕ300mm circular columns

Thickness of reservoir walls = 250mm

Thickness of reservoir slab = 250mm

**(b) Density of materials**

Density of stored material = 10 kN/m^{3}

Density of concrete = 25 kN/m^{3}

**(c) Dead Loads**

Weight of the walls = (25 kN/m^{3} × 0.25m × 4.75m × 18.849m) = 559.579 kN

Weight of bottom slab = (25 kN/m^{3} × 0.25m × 28.274m^{2}) = 176.7125 kN

Weight of water stored = (10 kN/m^{3} × 4.5m × 23.758m^{2}) = 1069.11 KN

**Total = 1805.4015 kN**

Let us now transfer this load to the the ring beam based on the perimeter.

Perimeter of ring beam = πd = π × 6 = 18.849m

w = 1805.4015 kN / Perimeter of ring beam = 1805.4015 kN / 18.849m = 95.782 kN/m

Self weight of the beam = 25 kN/m^{3} × 0.3m × 0.45m = 3.375 kN/m

Total dead load on beam = 95.782 kN/m + 3.375 kN/m = 99.157 kN/m

Factoring the load on the beams at ultimate limit state = 1.35 × 99.157 kN/m = 133.862 kN/m

From the table above;

Number of supports (n) = 8

θ = π/n = 45°

K_{1} = 0.052

K_{2} = 0.026

K_{3} = 0.0040

Radius (r) = 3m

Maximum Negative Moment at the supports = K_{1}wr^{2 }= -0.052 × 133.862 × 3^{2} = -62.647 KN.m

Maximum Positive Moment at the spans = K_{2}wr^{2 }= 0.026 × 133.862 × 3^{2} = 31.323 KN.m

Maximum Torsional Moment = K_{3}wr^{2 }= 0.0040 × 133.862 × 3^{2} = 4.819 KN.m

Shear force at the supports = R/2 = wrθ = 133.862 × 3 × (π/8) = 157.7 KN

**Structural Design**

Design strength of concrete f_{ck }= 35 N/mm^{2}

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

Nominal cover to reinforcement = 30mm

**Span**

M_{Ed} = 31.323 KN.m

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

Assuming ϕ12 mm bars will be employed for the main bars, and ϕ8mm bars for the stirrups (links)

d = 450 – 30 – 6 – 8 = 406 mm

k = M_{Ed}/(f_{ck}bd^{2}) = (31.323 × 10^{6})/(35 × 300 × 406^{2}) = 0.0181

Since k < 0.167, no compression reinforcement required

z = d[0.5+ √(0.25 – 0.882K) ] = z = d[0.5+ √(0.25 – 0.882(0.0181))] = 0.95d

A_{s1} = M_{Ed}/(0.87_{yk} z) = (31.323 × 10^{6})/(0.87 × 500 × 0.95 × 406) = 186.69 mm^{2}

Provide 3H12 mm BOT (AS_{prov} = 339 mm^{2})

**Supports**

M_{Ed} = 62.647 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 = 450 – 30 – 8 – 8 = 404 mm

k = M_{Ed}/(f_{ck}bd^{2}) = (62.647 × 10^{6})/(35 × 300 × 404^{2}) = 0.0365

Since k < 0.167, no compression reinforcement required

z = d[0.5+ √(0.25 – 0.882K) ] = z = d[0.5+ √(0.25 – 0.882(0.0365))] = 0.95d

A_{s1} = M_{Ed}/(0.87f_{yk} z) = (62.647 × 10^{6})/(0.87 × 500 × 0.95 × 404) = 375.23 mm^{2}

Provide 2H16 mm TOP (AS_{prov} = 402 mm^{2})

**Interaction of shear and torsion**

According to clause 6.3.2(2) of Eurocode 2, the effects of torsion and shear for both hollow and solid members may be superimposed, assuming the same value for the strut inclination θ. The limits for θ given in 6.2.3 (2) of EC2 are also fully applicable for the case of combined shear and torsion.

According to clause 6.3.2(4) of EC2, the maximum resistance of a member subjected to torsion and shear is limited by the capacity of the concrete struts. In order not to exceed this resistance the following condition should be satisfied:

T_{Ed} /T_{Rd,max} + V_{Ed} /V_{Rd,max} ≤ 1 ———- Equation (6.29 of EC2)

V_{Ed} = 157.7 KN

T_{Ed} = 4.819 kN.m

T_{Rd,max} = design torsional resistance moment

V_{Rd,max} = maximum shear resistance of the cross-section

**Geometrical Properties for analysis of torsion**

Area (A) = 300mm × 450mm = 135000 mm^{2}

Perimeter (U) = 2(300) + 2(450) = 1500mm

Equivalent thickness = t_{ef,i} = A/U = 135000/1500 = 90mm

The equivalent thin wall section for the rectangular section is given below;

A_{k }= the area enclosed by the centre-lines of the connecting walls, including inner hollow areas = (450 – 90) × (300 – 90) = 75600 mm^{2}

U_{k }= is the perimeter of the area A_{k} = 2(450 – 90) + 2(300 – 90) = 1140 mm

T_{Rd,max} = 2 v α_{cw}f_{cd}A_{k}t_{ef,i} sinθ cosθ

Assuming θ = 21.8° (cot θ = 2.5)

v = 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}

^{
} T_{Rd,max} = 2 × 0.516 × 1.0 × 23.33 × 75600 × 90 × cos 21.8° × sin 21.8° × 10^{-6} = 56.485 KNm

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

k = 1 + √(200/d) = 1 + √(200/404) = 1.704 > 2.0, therefore, k = 1.702

ρ_{1} = As/bd = 402/(300 × 404) = 0.003317 < 0.02; K_{1} = 0.15

V_{Rd,c} = [0.12 × 1.704(100 × 0.003317 × 35 )^{(1/3)}] 300 × 404 = 65469.358 N = 65.469 KN

Since V_{Rd,c} (65.469 KN) < V_{Ed} (157.7 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 × 404 × 0.516 × 23.333)/(2.5 + 0.4)] × 10^{-3} = 452.863 KN

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

Hence, A_{sw}/S = V_{Ed}/(0.87F_{yk}zcot θ) = 157700/(0.87 × 500 × 0.9 × 404 × 2.5 ) = 0.3988

**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 < 0.3988, adopt 0.3988

Maximum spacing of shear links = 0.75d = 0.75 × 404 = 303mm

Provide H8mm @ 250mm c/c (A_{sw}/S = 0.402) Ok

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

**Design Considerations for Torsion**

T_{Ed} /T_{Rd,max} + V_{Ed} /V_{Rd,max} ≤ 1

(4.891/56.485) + (157.7/453.863) = 0.434 < 1.0 Hence it is ok

However, note that the actual shear force at the point where torsion is maximum is actually less than the shear force at the the support. The relationship above is an error, but on the safe side.

The maximum torsion occurs at an angle of 9.5° from the support (see Table above). Therefore, the actual shear force at that section ( V_{Ed}) = Shear force at support – wrα

V_{Ed} = 157.7 – (133.862 × 3 × (9.5/180) × π) = 91.114 KN

Therefore for educational purposes, this is the shear force that should be used to check the shear-torsion interaction. A little consideration will show that V_{Rd,max } is constant all through the section, but V_{Rd,c} might vary depending on the longitudinal reinforcement provided at the section.

Rechecking the interaction above;

(4.891/56.485) + (91.114/453.863) = 0.287< 1.0 Hence it is ok

**Area of transverse reinforcement to resist torsion**

A_{sw}/s = T_{Ed}/2A_{k} f_{yw,d} cotθ

A_{sw}/s = (4.819 × 10^{6}) / (2 × 135000 × 0.87 × 500 × 2.5) = 0.0164 < A_{sw}/S (min)

Therefore, links provided for shear will be adequate for resisting torsion.

**Area of longitudinal reinforcement to resist torsion**

A_{s1} = T_{Ed}U_{k} cot θ / 2A_{k} f_{yd}

A_{s1} = (4.819 × 10^{6} × 1140 × 2.5) / (2 × 75600 × 0.87 × 500) = 208 mm^{2}

According to clause 6.3.2(4), in compressive chords, the longitudinal reinforcement may be reduced in proportion to the available compressive force. In tensile chords the longitudinal reinforcement for torsion should be added to the other reinforcement. The longitudinal reinforcement should generally be distributed over the length of side, but for smaller sections it may be concentrated at the ends of this length.

However, for the avoidance of doubt since no definition was given for what could be regarded as a ‘smaller section’, Provide 1H12mm bar at the middle of the section at both faces. The tensile and longitudinal reinforcement provided at the top and bottom of the section should be able to take care of the rest.

**Further information**

Nominal shear reinforcement is required in rectangular sections when;

T_{Ed} /T_{Rd,c} + V_{Ed} /V_{Rd,c} ≤ 1 ————– Equation (6.31 of EC2)

Where;

T_{Rd,c} is the value of the torsion cracking moment:

V_{Rd,c} is as defined above.

T_{Rd,c} = f_{ctd}⋅t⋅2A_{k}

_{
}τ = f_{ctd} = f_{ctk} /γ_{c} = 2.2/1.5 = 1.466 MPa (f_{ctk} deducted from Table [3.1 – EC2]).

It results therefore:

T_{Rd,c} = f_{ctd}⋅t⋅2A_{k} = 1.466 × 90 × 2 × 75600 × 10^{-6} = 19.949 kNm

From above calculations;

V_{Rd,c} = [0.12 × 1.704(100 × 0.003317 × 35 )^{(1/3)}] 300 × 404 = 65469.358 N = 65.469 KN

(4.819/19.949) + 91.114/65.469) = 1.633 > 1.0 (Shear force at point of maximum torsion was used)

Therefore obviously this showed that shear reinforcement calculations were required.

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You are doing a great job Ubani. #thumbs up

Thanks for the effort

Sir

Thank you Engr. Apele

Thank you so much

Dear Ubani, i just throughly look to your post. The design of curved beam with moment coefficients O.K. But it is not good to assume the storage tank weight as uniform load. If the circular beam and the tank is casted together, the 5m depth tank shell will support the circular beam. Even concrete tank is seperate and freely supported , the tank wt will be supported by columns mostly. Another comment, the liquid content is LL. U may check ur assumtions with FEM analysis..Good Luck…….

I totally agree with you on both counts. Except the tank will be permanently filled with water. It is safe to consider the water as a live load.

Thank you for your intelligent contributions. As for the real load transfer pattern, I need more expert opinion.

On the second issue, I initially wanted to treat water as live load because the volume of water in the tank will fluctuate. But on the other hand, I do not need any statistical analysis to predict the maximum load the tank can ever experience due to water storage, so why should I use a higher factor of safety?

Dear Ubani, i just saw ur last comment… First of all ur tank ht 5 m and water ht 4,5m. What is dictating this level? say ..overflow pipe.. that means prone to blocking or valve closing and water content will reach to max level and then spills.. Another issue, unit wt of content. If u are sure for the content will be only water and if u assume the max level can be reached , u can use indeed γQ = 1.2 . Good luck…

Theta = 2π÷n not π÷n

dear Ubani, how do you calculated and get the coefficient in the table for -/+ moment and torsional moment ?

Kindly furnish reference material from which this procedure is taken

• This is the first time I am visiting your website. I would like to tell you that I am genuinely engaged by reading your blogs. Keep up the good work!Fabrication Drawings preparation in USA

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Thank you