Design of Slender Columns According to EC2

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July 24, 2017

Loads from slabs and beams are transferred to the foundations through the columns. In typical cases, columns are usually rectangular or circular in shape. Normally, they are usually classified as short or slender depending on their slenderness ratio, and this in turn influences their mode of failure. Slender columns are likely to fail by buckling than by crushing.

Columns are either subjected to axial, uniaxial, or biaxial loads depending on the location and/or loading condition. Eurocode 2 demands that we include the effects of imperfections in structural design of columns. Column design is covered in section 5.8 of EN 1992-1-1:2004 (EC2).

Slender Columns Requirements in EC2

Clause 5.8.2 of EN 1992-1-1 deals with members and structures in which the structural behaviour is significantly influenced by second order effects (e.g. columns, walls, piles, arches and shells). Global second order effects are more likely to occur in structures with a flexible bracing system.

Column design in EC2 generally involves determining the slenderness ratio (λ), of the member and checking if it lies below or above a critical value λ_lim. If the column slenderness ratio lies below (λ_lim), it can simply be designed to resist the axial action and moment obtained from an elastic analysis, but including the effect of geometric imperfections. These are termed first order effects. However, when the column slenderness exceeds the critical value, additional (second order) moments caused by structural deformations can occur and must also be taken into account.

So in general, second order effects may be ignored if the slenderness λ is below a certain value λ_lim.

λ_lim = (20.A.B.C)/√n ——- (1)

Where:
A = 1/(1 + 0.2ϕ_ef) (if ϕ_ef is not known, A = 0.7 may be used)
B = 1+ 2ω (if ω is not known, B = 1.1 may be used)
C = 1.7 – r_m (if r_m is not known, C = 0.7 may be used)
Where; ϕ_ef = effective creep ratio (0.7 may be used)
ω = A_sf_yd / (A_cf_cd); mechanical reinforcement ratio;
A_s is the total area of longitudinal reinforcement
n = N_Ed / (A_cf_cd); relative normal force
r_m = M₀₁/M₀₂; moment ratio
M₀₁, M₀₂ are the first order end moments, |M₀₂| ≥ |M₀₁|

If the end moments M₀₁ and M₀₂ give tension on the same side, r_m should be taken positive (i.e. C ≤ 1.7), otherwise negative (i.e. C > 1.7). For braced members in which the first order moments arise only from or predominantly due to imperfections or transverse loading r_m should be taken as 1.0 (i.e. C = 0.7).

Also, clause 5.8.3.1(2) says that for biaxial bending, the slenderness criterion may be checked separately for each direction. Depending on the outcome of this check, second order effects (a) may be ignored in both directions, (b) should be taken into account in one direction, or (c) should be taken into account in both directions.

Solved Example

Let us consider the structure shown below. The effects of actions on column member BC is as shown below. It is required to design the column using the following data;

f_ck = 25 N/mm², f_yk = 460 N/mm², Concrete cover = 35mm

Second moment of area of beam AB = (0.3 × 0.6³)/12 = 0.0054 m⁴
Stiffness of beam AB (since E is constant) = 4I/L = (4 × 0.0054) / 6 = 0.0036
Second moment of area of column BC = (0.3 × 0.6³)/12 = 0.0054 m⁴
Stiffness of column BC (since E is constant) = 4I/L = (4 × 0.0054)/7.5 = 0.00288

Remember that we will have to reduce the stiffness of the beams by half to account for cracking;

Therefore, k₁ = 0.00288/0.0018 = 1.6

Since the minimum value of k₁ and k₂ is 0.1, adopt k₁ as 1.6. Let us take k₂ as 1.0 for base designed to resist moment.

Take the unrestrained clear height of column as 7000mm

l_o = 0.5 × 7000√[(1 + (1.6/(0.45 + 1.6))) × (1 + (1.0/(0.45+ 1.0)))] = 6071 mm

Radius of gyration i = h/√12 = 600/√12 = 173.205
Slenderness ratio λ = 6071/173.205 = 35.051

Critical Slenderness for the x-direction
λ_lim = (20.A.B.C)/√n
A = 0.7
B = 1.1
C = 1.7 – M₀₁/M₀₂ = 1.7 – (-210/371) = 2.266

n = N_Ed / (A_c f_cd)
N_Ed = 3500 × 10³ N
A_c = 300 × 600= 180000 mm²
f_cd = (α_cc f_ck)/1.5 = (0.85 × 25)/1.5 = 14.167 N/mm²
n = (3500 × 10³) / (180000 × 14.167) = 1.3725

λ_lim = (20 × 0.7 × 1.1 × 2.266 )/√1.3725 = 29.786

Since 29.786 < 35.051, second order effects need to be considered in the design

Design Moments
M_Bot = 210 KNm, M_Top = 371 KNm
e_i is the geometric imperfection = (θ_i l₀/2) = [(1/200) × (6071/2)] = 15.1775 mm
e_iN_Ed = 15.1775 × 10^-3 × 3500 = 53.121 KNm

First order end moment
M₀₁ = M_Bot + e_iN_Ed = -210 + 53.121 = -156.879 KNm
M₀₂ = M_Top + e_iN_Ed = 371 + 53.121 = 424.121 KNm

Equivalent first order moment M_0Ed
M_0Ed = (0.6M₀₂ + 0.4M₀₁) ≥ 0.4M₀₂ = 0.4 × 424.121 = 169.648 KNm
M_0Ed= (0.6 × 424.121 – 0.4 × 156.879) = 191.721 KNm

Nominal second order moment M₂
Specified concrete cover = 35mm
Diameter of longitudinal steel = 32 mm
Diameter of links = 10 mm

Thus, the effective depth (d) = h – C_nom – ϕ/2 – ϕ_links
d = 600 – 35 – 16 – 10 = 539 mm

1/r₀ = ε_yd/(0.45 d)
ε_yd = f_yd/Es = (460 /1.15) / (200 × 10³) = 0.002
1/r₀ = 0.002/(0.45 × 539) = 8.2457 × 10^-6
β = 0.35 + f_ck/200 – λ/150 (λ is the slenderness ratio)
β = 0.35 + (25/200) – (35.051/150) = 0.2413

Kϕ = 1 + βϕ_ef ≥ 1.0 (ϕ_ef is the effective creep ratio, assume 0.87)
Kϕ = 1 + (0.2413 × 0.87) = 1.2099 ≥ 1.0
Assume K_r = 0.8
1/r = K_r.Kϕ. 1/r₀ = 0.8 × 1.2099 × 8.2457 × 10^-6 = 7.981 × 10^-6
e₂ is the deflection = (1/r) (l₀²) / 10 = 7.981 × 10^-6 × 6071² / 10 = 29.415 mm

M₂ = N_Ed.e₂ = 3500 × 29.415 × 10^-3 = 102.954 KNm

Design Moment M_Ed
M_Ed = maximum of {M_0Ed + M₂; M₀₂; M₀₁ + 0.5M₂}
M_Ed = maximum of {191.721 + 102.954 = 294.675 kNm; 424.121 kNm; -156.879 + (0.5 × -102.954) = -208.356 kNm}

Longitudinal Steel Area
d₂ = C_nom + ϕ/2 + ϕ_links
d₂ = 35 + 16 + 10 = 61 mm

d₂/h = 61/600 = 0.1016
Reading from chart No 2; d₂/h = 0.10;

M_Ed/(f_ck bh²) = (424.121 × 10⁶) / (25 × 300 × 600²) = 0.1571
N_Ed/(f_ck bh) = (3500 × 10³) / (25 × 300 × 600) = 0.777

From the chart, (A_sF_yk)/(bhf_ck) = 0.53
Area of longitudinal steel required (As) = (0.53 × 25 × 300 × 600)/460 = 5185 mm²

Provide 6Y32 + 2Y20 (A_Sprov = 5452 mm²)

A_s,min = (0.1 N_Ed)/f_yd = (0.1 × 3500 × 1000) / 400 = 875mm², 0.002bh = 0.002 × 300 × 600 = 360 mm²
A_s,max = 0.04bh = 0.04 × 300 × 600 = 7200 mm²

Links
Minimum size = 0.25ϕ = 0.25 × 32 = 8mm < 6mm
We are adopting Y10mm as links
Spacing adopted = 300mm less than {b, h, 20ϕ, 400mm} Provide Y10 @ 300 mm links

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Static Analysis of Suspension Bridges: A solved Example

By

Ubani Obinna Uzodimma

-

July 17, 2017

Cables are made from high-strength steel wires that are twisted together. They offer a flexible structural system, which can resist loads only by axial tension. Cables allow engineers to cover very large spans especially in bridges and other suspended structures . Structurally, cables are extremely efficient because they make the most effective use of structural material in that their loads are carried solely through tension through the wire. Therefore, there is no tendency for buckling to occur either from bending or from compressive axial loads.

As load-bearing elements, cables have several unique features. One of them is that vertical loads give rise to horizontal reactions at the support, which, as in case of an arch, is called the thrust. To accommodate the thrust it is necessary to have a supporting structure. It may be a pillar of a bridge, tower, or pylon. Cables are usually employed in suspension bridges, cable-stayed bridges, tower guy wires, roofs, etc.

In this post a simplified static model of suspension bridge is presented for the purpose of structural analysis and design (linear first order analysis). The girder has an internal hinge at point G.

SUSPENSION%2BBRIDGE%2BMANUAL%2BANALYSIS%2BSOLVED%2BEXAMPLE

GEOMETRICAL PROPERTIES

We know that the thrust at the support of the cables = H = Cx = Dx

We can therefore verify that Cy = Dy = Htan⁡α
You can verify that α = tan^-1(4/7) = 29.744°
β = tan^-1(3/7) = 23.198°

SUPPORT REACTIONS

∑M_B = 0
35Ay + 35Cy – 12Cx + 12Dx – (12 × 35²)/2 – (300 × 26) = 0
35Ay + 35Cy – 7350 – 7800 = 0
35Ay + 35(0.5714H) = 15150
35Ay + 20H = 15150 ————– (1)

∑M_G^L = 0
17.5Ay + 17.5Cy – 12Cx – (12 × 17.5²)/2 – (300 × 8.5) + 5H = 0
17.5Ay + 17.5(0.5714H) – 12H + 5H – 1837.5 – 2550 = 0
17.5Ay + 3H = 4387.5 —————– (2)

Solving (1) and (2) simultaneously;
Ay = 172.653 KN; H = 455.537 KN

∑M_A = 0
35By + 35Dy – 12Dx + 12Cx – (12 × 35²)/2 – (300 × 9) = 0
35By + 35(0.5714H) = 10050
35By + 20H = 10050 ———— (3)

∑M_G^R = 0
17.5By + 17.5Dy – 12Dx + 5H – (12 × 17.5²)/2 = 0
17.5By + 17.5(0.5714H) – 12H + 5H = 1837.5
17.5By + 3H = 1837.5 ————– (4)

Solving (3) and (4) simultaneously;
By = 26.939 KN; H = 455.357 KN

Hence, Cy = Dy = H tan⁡α = 455.357 × 0.5714 = 260.195 KN

ANALYSIS OF THE JOINTS OF THE CABLE

Analysis of support C

∑F_x = 0
-455.357 + F_C1 cos29.744 = 0
F_C1 = 455.357/cos⁡29.744 = 524.453 KN

Analysis of joint 1

∑F_x = 0
– F_C1cos⁡α + F₁₂ cos⁡β = 0
-524.453cos29.744 + F₁₂ cos⁡23.198 = 0
F_C1 = (524.453 × cos29.744)/cos⁡23.198 = 495.411 KN

∑F_y = 0
F_C1sin⁡α – F₁₂ sin⁡β – P₁ = 0
(524.453sin29.744) – (495.411sin⁡23.198) – P₁ = 0
P₁ = 65.047 KN

Analysis of joint 2

∑F_y = 0
– F₁₂ sin⁡β – P₂ = 0
– 495.411sin⁡23.198 – P₂ = 0
P₂ = 195.147 KN

Therefore, the equivalent loading on the girder is given below;

INTERNAL STRESSES ON GIRDER

Bending Moment on the girder
M_A = 0
M₁ = (172.635 × 7) – (12 × 7²)/2 = 914.445 KNm
M_E = (172.635 × 9) – (12 × 9²)/2+ (65.047 × 2) = 1197.809 KNm
M₂ = (172.635 × 14) – (12 × 14²)/2 + (65.047 × 7) – (300 × 5) = 196.219 KNm
M_G = (172.635 × 17.5) – (12 × 17.5²)/2 + (65.047 × 10.5) – (300 × 8.5) + (195.147 × 3.5) = 0

Coming from the right
M_B = 0
M₄ = (26.939 × 7) – (12 × 7²)/2 = -105.427 KNm
M₃ = (26.939 × 14) – (12 × 14²)/2 + (65.047 × 7) = -343.525 KNm
M_G = (26.939 × 17.5) – (12 × 17.5₂)/2 + (65.047 × 10.5) + (195.147 × 3.5) = 0

BENDING%2BMOMENT%2BDIAGRAM%2BOF%2BSUSPENSION%2BBRIDGE%2BGIRDER

Shear force on the girder

Q_A = 172.635 KN
Q₁^L = 172.635 – (12 × 7) = 88.635 KN
Q₁^R = 172.635 – (12 × 7) + 65.047 = 153.682 KN
Q_E^L = 172.635 – (12 × 9) + 65.047 = 129.682 KN
Q_E^R = 172.635 – (12 × 9) + 65.047 – 300 = -170.318 KN
Q₂^L = 172.635 – (12 × 14) + 65.047 – 300 = -230.318 KN
Q₂^R = 172.635 – (12 × 14) + 65.047 – 300 + 195.146 = -35.172 KN
Q_G^L = 172.635 – (12 × 17.5) + 65.047 – 300 + 195.146 = -77.172 KN
Q₃^R = 172.635 – (12 × 21) + 65.047 – 300 + 195.146 = -119.172 KN
Q₃^L = 172.635 – (12 × 21) + 65.047 – 300 + 195.146 + 195.146 = 75.974 KN
Q₄^L = 172.635 – (12 × 28) + 65.047 – 300 + 195.146 + 195.146 = -8.026 KN
Q₄^R = 172.635 – (12 × 28) + 65.047 – 300 + 195.146 + 195.146 + 65.047 = 57.021 KN
Q₄^R = 172.635 – (12 × 35) + 65.047 – 300 + 195.146 + 195.146 + 65.047 = -26.939 KN

SHEAR%2BFORCE%2BDIAGRAM%2BOF%2BSUSPENSION%2BBRIDGE%2BGIRDER

Obviously, there are no axial forces in the bridge girder.

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Plastic Collapse Analysis of Propped Cantilever Beams

By

Ubani Obinna Uzodimma

-

July 13, 2017

We all know that for a propped cantilever beam, there are two possible locations of plastic hinges – which are at the span (point of maximum moment) and at the fixed support. For the propped cantilever loaded as shown above, the degree is static indeterminacy is 1.

Since the number of possible location of plastic hinges is 2, therefore the number of independent mechanisms is;

P = 2 – 1 = 1 (which is a beam mechanism)

As a result, we are going to carry out our elastic analysis in two stages, so as to determine the load factor at which the beam will form a mechanism and completely collapse. We will start by assuming a value of unity for our load factor (i.e. λ =1.0)

STAGE 1

It is very easy to verify that for the beam loaded as shown above, the fixed end moment at support A = 3PL/16 = (3 × 267 × 8) / 16 = 400.5 KNm

It is also easy to verify that the vertical support reaction at support C (R_C) = 5P/16 = (5 × 267)/16 = 83.4375 KNm

Therefore, the maximum span moment at point B (M_B) = 83.4375 × 4 = 333.75 KNm
The bending moment is as given below.

We can also obtain the vertical deflection at point B by quickly placing a unit load at point B on a basic system of the structure (could be a cantilever or a simply supported beam). Then by combining the shapes from the two states of loading using Vereschagin’s rule, we can obtain the deflection at point B. This is given below.

EIδB = 1/6 × 4 [2(400.5) – 333.75] × 4 = 1246
δ_B = (1246/EI) metres

Therefore, for support A to become plastic, the load factor λA₁= M_P/M_E = 500.625/400.5 = 1.25
Also for section B to become plastic, the load factor λB₁ = M_P/M_E = 500.625/333.75 = 1.5

It is therefore obvious that the first plastic hinge will develop at support A.

Therefore the load at failure of support A = 1.25 × 267 = 333.75 KN
λA₁M_A = 1.25 × 400.5 = 500.625 KN.m
λA₁M_B = 1.25 × 333.75 = 417.1875 KN.m

Deflection of beam at failure of support A = δ_B= (1.25 × 1246)/EI= (1557.5/EI metres)

STAGE 2
Now, support A is assumed to have failed (formed a plastic hinge). We now model it as a real hinge and carry out another elastic analysis.

From statics, the maximum moment of the structure is given by;

PL/4 = ( 267 × 8)/4 = 534 KNm

For section B to become plastic and form a hinge;
λB₂ = (M_P– λA₁M_B)/M_E = (500.625 – 417.1875) / 534 = 0.15625

Therefore, the total load factor at collapse (λ) = λA₁ + λB₂ = 1.25 + 0.15625 = 1.40625

Also, the load at complete collapse of the beam = 1.40625 × 267 = 375.468 KN

The deflection at collapse = δ_B= (1.40625 × 1246) / EI= (1752.1875/EI) metres

Verification using the static method

From geometrical relations, you can observe that δ_B = 4θ

Internal work done due to rotations of the structure at full plastic moment = M_Pθ + M_P(2θ) = 3M_Pθ (the rotation at section C will not count because it is a natural hinge).

External work done by the collapse load = 375.468 × 4θ = 1501.871θ

But External work done = Internal work done;
Therefore, 3M_P θ = 1501.871θ

Therefore, M_P = 1501.871/3 = 500.624 KNm

This shows that the load factor we obtained from our analysis is correct.

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Design of Doubly Reinforced Beams | Worked Example

By

Ubani Obinna Uzodimma

-

July 11, 2017

Doubly reinforced beams are reinforced concrete beams that incorporate both compression steel (in the compression zone) and tension steel (in the tension zone). This design is typically employed when the required moment capacity of a beam exceeds that which can be provided by singly reinforced beams (those with only tension steel).

In singly reinforced concrete beams without moment redistribution, k = M_Ed/f_ckbd² is limited to k’ = 0.167. Therefore, once the value of k > k’, the beam must be designed as a doubly reinforced beam. In doubly reinforced beams, compression steel reinforcements are required because the concrete section (in compression) by itself cannot develop the required moment of resistance.

However, when there is a moment redistribution in the section, k > k’ where k’ is dependent on the moment redistribution ratio (δ) and is given by:

k’ = 0.6δ – 0.18δ² – 0.21 where δ ≤ 1.0

In all cases, it is recommended that k’ be limited to 0.167 to ensure ductile failure. It is important to understand that the redistribution of moments derived from elastic analysis of a reinforced concrete structure accounts for the plastic behaviour of the material as it approaches its ultimate limit state.

To facilitate this plastic response, concrete sections must be designed to permit the formation of plastic hinges upon the yielding of tensile reinforcement. This approach ensures a ductile structure characterized by a gradual failure at the ultimate limit state, rather than a sudden, catastrophic collapse due to compressive failure of the concrete.

To guarantee the formation of plastic hinges and the necessary section rotation for ductile behaviour, Eurocode 2 imposes limitations on the maximum neutral axis depth, x_bal. These restrictions ensure that the tensile steel experiences high strains and this promotes the desired plastic response.

The permissible value of x_bal is determined based on the desired degree of moment redistribution. According to Eurocode 2, the limit is x_bal≤ 0.8(δ – 0.44)d for f_ck ≤ C50. However, this limit was modified to x_bal as x_bal ≤ (δ – 0.4)d in the UK Annex to the EC2.

Design of Doubly Reinforced Beam

To ensure a ductile response and prevent sudden compressive failure of the concrete, it is required that the neutral axis depth should not exceed 0.45 times the effective depth (0.45d). This guideline is typically adopted in the design of sections incorporating compression reinforcement (doubly reinforced beam sections).

Rectangular stress block for doubly reinforced beam

In the design of reinforced concrete beams, if the design ultimate moment is greater than the ultimate moment of resistance i.e. M_Ed > M_lim, then compression reinforcement is required. Provided that d₂/x ≤ 0.38 (i.e. compression steel has yielded) where d₂ is the depth of the compression steel from the compression face and x = (d − z)/0.4

The area of steel in compression, A_s2, is given by:
A_s1 = (M_Ed – M_lim) / (f_sc(d – d₂)) —- (1)
f_sc= E_sε_sc ≤ 0.87f_yk
ε_sc = steel compressive strain = 0.0035(x – d’)/x
E_s = Modulus of elasticity of steel = 200000 N/mm²

Area of tension reinforcement:
A_s1 = M_lim / (0.87f_ykz) + A_s2 (f_sc/0.87f_yk) —- (2)

Where:
M_Ed = Design bending moment of the section
M_lim= 0.167f_ckbd²
z = d[0.5+ √(0.25 – 0.882k’)]
where
k’ = 0.167

Design Example of a Doubly Reinforced Beam

To show how this is done, let us consider the flexural design of support A of the beam loaded as shown below. The depth of the beam is 600 mm, and the width is 400 mm.
f_ck= 35 N/mm²; f_yk= 500 N/mm²; concrete cover = 40mm

DOUBLY%2BREINFORCED%2BBEAM%2B%2528QUESTION%2529

DOUBLY%2BREINFORCED%2BBEAM%2B%2528BMD%2529

Top reinforcement (Hogging moment)
Support A (No moment redistribution)
Effective depth (d) = 600 – 40 – 16 – 10 = 534 mm
d₂= 40 + 8 + 10 = 58 mm

M_Ed = 761.24 kNm

But since the flange is in tension, we use the beam width to calculate the value of k (this applies to all support hogging moments)
k = M_Ed / (f_ckb_wd²) = (761.24 × 10⁶) / (35 × 400 × 534²) = 0.1906
z = d[0.5+ √((0.25 – 0.882(0.167))] = 0.82d = 0.82 × 534 = 437.88 mm
x = (d − z)/0.4 = (534 − 437.88)/0.4 = 240.3 mm

d₂/x = 58/240.3 = 0.241 < 0.38 (therefore reinforcement bar has yielded and f_sc= 0.87f_yk)

Since k < 0.167, compression is required
Area of compression reinforcement A_S2 = (M_Ed – M_lim) / (0.87f_yk (d – d₂))

M_lim = 0.167f_ckbd² = (0.167 × 35 × 400 × 534²) × 10^-6 = 666.69 kNm

A_S2 = ((761.24 – 666.69) × 10⁶) / (0.87 × 500 × (534 – 58)) = 457 mm²
Provide 3H16 Bottom (A_sprov = 603 mm²)

Area of tension reinforcement A_s1 = M_Rd / (0.87f_yk z) + A_S2
Where z = d[0.5+ √(0.25 – 0.882K’)]
K’ = 0.167
z = d[0.5+ √((0.25 – 0.882(0.167))] = 0.82d

A_s1 = M_Rd / (0.87f_yk z) + A_S2 = ( 666.69 × 10⁶) / (0.87 × 500 × 0.82 × 534) + 457 mm² = 3957 mm²

Provide 3H32mm + 4H25 mm Top (A_Sprov = 4376 mm²)

Comparison of Shear Design to EC2 and BS 8110

By

Ubani Obinna Uzodimma

-

July 11, 2017

Eurocode 2 uses the variable strut inclination method for shear design. It is slightly more complex than the procedure in BS 8110 but can result in savings in the amount of shear reinforcement required.

Like BS 8110, Eurocode 2 models shear behaviour using the truss analogy in which the concrete acts as the diagonal struts, the stirrups act as the vertical ties, the tension reinforcement forms the bottom chord, and the compression steel/ concrete forms the top chord. Whereas in BS 8110 the strut angle has a fixed value of 45°, in EC2 it can vary between 21.8° and 45° and it is this feature which is responsible for possible reductions in the volume of shear reinforcement.

The concrete strut angle changes with respect to the magnitude of the shear force applied. The angle decreases with a reduction in shear stress, with a minimum angle of 21.8° (where cot = 2.5) which happens to be the most frequently used design strut angle.

Shear design in Eurocode 2 often involves comparing the design shear stress (v_Ed) to the concrete shear resistance (v_Rd,c). If v_Ed< v_Rd,c then in the case of one-way spanning slabs no shear reinforcement is required; in the case of beams nominal shear reinforcement is required. v_Rd,cis dependent on the amount of design tensile reinforcement and the design concrete strength.

If v_Ed < v_Rd,max (ie v_{Rd,cot θ = 2.5}), then the shear reinforcement may be calculated as:
A_sw/s ≥ v_Ed/(0.87 × f_yk × z × cot 2.5)

If v_Ed > v_Rd,max (ie v_{Rd,cot θ = 2.5}), then checks against v_{Rd,cot θ = 1.0}and determination of θ are required.
If v_Ed> v_{Rd,cot θ = 1.0} then a larger section or greater concrete strength is required.

Where applied shear stresses are higher, one has to check against the maximum allowable shear stress at the maximum strut angle of θ = 45º (e.g. 5.28 MPa for a C30/37 concrete when cot θ = 1.0) and where necessary determine the actual intermediate strut angle, θ.

Solved Example for Shear Design

For the beam loaded as shown below and disregarding load factors, we are going to obtain the shear reinforcement required for section A. The shear force at the centreline of support has been adopted for this design, and the area of tension steel provided at this section A_s1 = 4825 mm².

Design Data:
Concrete Strength = 35 N/mm²
Grade of steel = 460 N/mm²
Width of beam = 400mm
Depth of beam = 600mm
Effective depth = 543mm
Area of tension steel provided at section A_s1 = 4825 mm²

Shear Design to EC2

Support A; V_Ed = 500.46 kN

V_Rd,c = [C_Rd,c.k.(100ρ₁ f_ck)^(1/3) + k₁.σ_cp]b_w.d ≥ (V_min + k₁.σ_cp) b_w.d

C_Rd,c = 0.18/γ_c = 0.18/1.5 = 0.12
k = 1 + √(200/d) = 1 + √(200/543) = 1.606 > 2.0, therefore, k = 1.606
V_min = 0.035k^(3/2) f_ck^0.5
V_min = 0.035 × (1.606)^1.5 × 35^0.5 = 0.421 N/mm²
ρ₁ = A_s/bd = 4825/(400 × 543) = 0.022 > 0.02; Therefore take 0.02

σ_cp = N_Ed/Ac < 0.2f_cd (Where N_Ed is the axial force at the section, A_c = cross-sectional area of the concrete), f_cd = design compressive strength of the concrete.) Take N_Ed = 0

V_Rd,c = [0.12 × 1.606 (100 × 0.02 × 35 )^(1/3)] 400 × 543 = 172511.992 N = 172.511 kN

Since V_Rd,c (172.511 kN) < V_Ed (500.46 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.v₁.f_cd) / (cot⁡θ + tanθ)
V₁ = 0.6(1 – f_ck/250) = 0.6(1 – 35/250) = 0.516
f_cd = (α_cc f_ck) / γ_c = (0.85 × 35) / 1.5 = 19.833 N/mm²
Let z = 0.9d

V_Rd,max= [(400 × 0.9 × 543 × 0.516 × 19.833) / (2.5 + 0.4)] × 10^-3 = 689.83 kN

Since V_Rd,c < V_Ed < V_Rd,max
Hence A_sw / S = V_Ed / (0.87.F_yk.z.cot θ) = 500460 / (0.87 × 460 × 0.9 × 543 × 2.5 ) = 1.0235

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) / 460 = 0.001028
A_sw/S_min = 0.001028 × 400 × 1 = 0.411
Maximum spacing of shear links = 0.75d = 0.75 × 543 = 407.25

Provide X10mm @ 150mm c/c as shear links (A_sw/S = 1.0467) Ok!!!!

Shear Design to BS 8110-1:1997

Design of support A
Ultimate shear force at the centerline of support
V = 500.46 kN

Using the shear force at the centreline of support;
Shear stress v = V/(bd ) = (500.46 × 10³) / (400 × 543) = 2.304 N/mm²

v(2.304 N/mm² ) < 0.8√f_cu (4.732 N/mm² ). Hence, the dimensions of the cross-section are adequate for shear.

Concrete resistance shear stress
v_c = 0.632 × (100A_s/bd)^1/3 (400/d)^1/4
(100A_s/bd) = (100 × 4825) / (400 × 543) = 2.221 < 3 (See Table 3.8 BS 8110-1;1997)
(400/d)^1/4 = (400/543)^1/4 = 0.926; But for members with shear reinforcement, this value should not be less than 1. Therefore take the value as 1.0

v_c = 0.632 × (2.221)^1/3 × 1.0 = 0.8245 N/mm²
For f_cu = 35 N/mm², v_c = 0.8245 × (35/25)^1/3 = 0.922 N/mm²

Let us check;
(V_c + 0.4)1.322 N/mm² < v(2.304 N/mm²) < 0.8√f_cu (4.732 N/mm²)

Therefore, provide shear reinforcement links.
Let us try 2 legs of Y10mm bars (Area of steel provided = 157 mm²

A_sv/Sv = (0.95 × F_yv) / (b_v (v – v_c)) = [400 × (2.304 – 0.922)] / (0.95 × 460) = 1.264

Maximum spacing = 0.75d = 0.75 × 543 = 407.25 mm
Provide Y10mm @ 100 mm c/c links as shear reinforcement (A_sv/Sv = 1.57)

We can therefore see that disregarding load factor and flexural design requirements, EC2 is more economical than BS 8110 in shear design, and in this case study by about 19%.

Free MATLAB Code for Flexural Design of R.C. Sections According to EC2

By

Ubani Obinna Uzodimma

-

July 5, 2017

In this post, I am going to present an extremely simplified MATLAB code for carrying out flexural design of reinforced concrete beams (rectangular or flanged) according to Eurocode 2. Before you can use this code, you must carry out your analysis and obtain your design forces and moments. You can just copy the code and paste on your MATLAB script. There are no external functions to be called for the program.

% ANALYSIS AND DESIGN OF RECTANGULAR AND FLANGED BEAMS PER EUROCODE 2
% THIS PROGRAM WILL CALCULATE THE AREA OF REINFORCEMENT REQUIRED, AND ALSO DO THE DEFLECTION VERIFICATION USING THE DEEMED TO SATISFY RULES OF EC2

clc
disp(‘THIS PROFORMA WAS WRITTEN BY O.U.R UBANI’)
disp(‘DOWNLOADED FROM WWW.STRUCTVILLE.COM’)
disp(‘ANALYSIS AND DESIGN OF BEAMS PER EUROCODE 2’)

% INPUT MATERIAL PROPERTIES
Fck = input(‘Enter the grade of concrete (N/mm^2)Fck = ‘);
Fyk = input(‘Enter the yield strength of steel (N/mm^2)Fyk = ‘);

% INPUT DESIGN MOMENT
MEd = input(‘Enter the ultimate design moment (KNm) MEd = ‘);

% INPUT SECTION AND DESIGN PROPERTIES
h = input(‘Enter the depth of beam (mm)h = ‘);
b = input(‘Enter the effective flange width of the beam (mm)b = ‘);
bw = input(‘Enter the beam width (mm)bw = ‘);
Cc = input(‘Enter concrete cover (mm) = ‘);
dr = input(‘Enter the diameter of reinforcement (mm) = ‘);
dl = input(‘Enter the diameter of links = ‘);

% CALCULATION OF EFFECTIVE DEPTH
disp(‘Effective depth d (mm)’)
d = h-Cc-(dr/2)-dl % Effective depth
disp(‘Depth of reinforcement from the face of concrete’)
do = Cc+dl+(dr/2)

% ANALYSIS FOR INTERNAL STRESSES
ko = 0.167;
k = (MEd*10^6)/(Fck*b*d^2)
if k>ko
disp(‘Since k > ko, Compression reinforcement is required’)
Mcd = (Fck*b*d^2*(k-ko))*10^(-6)
As2 = (Mcd*10^6)/(0.87*Fyk*(d-do))
z = 0.5*d*(1+sqrt(1-3.53*ko))
As1 = ((ko*Fck*b*d^2)/(0.87*Fyk*z))+ As2
else
disp(‘Since k < ko, No Compression reinforcement is required’)
disp(‘Lever arm (la)’)
la = 0.5+sqrt(0.25-0.882*k)
if la>0.95
disp(‘Since la > 0.95,’)
la = 0.95
else
la = 0.5+sqrt(0.25-0.882*k)
end
As1 =(MEd*10^6)/(0.87*Fyk*la*d)
end

% MINIMUM AREA OF STEEL REQUIRED
fctm = 0.3*(Fck^(2/3)) %MEAN TENSILE STRENGTH OF CONCRETE (TABLE 3.1 EC2)
ASmin = 0.26*(fctm/Fyk)*bw*d
if ASmin < 0.0013*bw*d
ASmin = 0.0013*bw*d
end
if As1<ASmin
As1 = ASmin
else
disp(‘Since As1 > Asmin, provide As1 which is the area of steel required’)
end

Asprov1 = input(‘Enter area of tension steel provided (mm^2) = ‘);
Asprov2 = input(‘Enter area of compression steel provided if any (mm^2) = ‘);

% CHECK FOR DEFLECTION

disp(‘DO YOU WANT TO CHECK FOR DEFLECTION?’)

G = input(‘ENTER (1) FOR YES OR (0) FOR NO = ‘);

if G == 1;

disp(‘CHECK FOR DEFLECTION’)

disp(‘BASIC SPAN/EFFECTIVE DEPTH RATIO (K)’)

disp(‘CANTILEVER = 0.4’)

disp(‘SIMPLY SUPPORTED = 1.0’)

disp(‘SIMPLY SUPPORTED AND FIXED AT ONE END = 1.3’)

disp(‘FIXED AT BOTH ENDS = 1.5’)

K = input(‘Enter the selected value of k = ‘);

L = input(‘Enter the deflection critical length of the member (mm) = ‘);

disp(‘Fs is the stress in tensile reinforcement under service loading’)

Fs = (310*Fyk*As1)/(500*Asprov1)

disp(‘Bs = 310/Fs’)

Bs = 310/Fs

disp(‘P is the reinforcement ratio in the section (Asprov1)/(b*d)’)

P = (Asprov1)/(b*d)

disp(‘Po = sqrt(Fck)/1000’)

Po = sqrt(Fck)/1000

P1 = (Asprov2)/(b*d)

if

P <= Po

I_deflection = K*(11+(1.5*sqrt(Fck)*(Po/P)) + 3.2*sqrt(Fck)*((Po/P)-1)^1.5)

else

I_deflection = K*(11+(1.5*sqrt(Fck)*(Po/(P-P1))) + 3.2*sqrt(Fck)*((Po/P)-1)^1.5)

end

CV = input(‘Is the beam is flanged? (1) for YES and (0) for NO = ‘)

if CV == 1

disp(‘S is the ratio of beff/bw’)

S = b/bw

bn = (11-(b/bw))/10

if S >= 3.0;

K_deflection = Bs*0.8* I_deflection

else

K_deflection = Bs*bn* I_deflection

end

if L>7000

Limiting_deflection = (7000/L)*K_deflection

else

Limiting_deflection = K_deflection

end

else

Limiting_deflection = Bs*I_deflection

end

Actual_deflection = L/d

if

Limiting_deflection > Actual_deflection

disp(‘Deflection is satisfactory’)

else

disp(‘DEFLECTION IS NOT SATISFACTORY !!!!! Increase depth of beam, or increase area of steel, or both. Then rerun proforma’)

end

2.0 MATLAB CODE FOR SHEAR DESIGN ACCORDING TO EC2

% SHEAR DESIGN IN EUROCODE 2
clc
disp(‘SHEAR DESIGN ACCORDING TO EC2′)
disp(‘THIS PROFORMA WAS WRITTEN BY O.U.R. UBANI’)

% MATERIALS PROPERTIES
Fck = input(‘Enter the grade of concrete (N/mm^2) = ‘);
Fyk = input(‘Enter the yield strength of steel (N/mm^2) = ‘);

VEd = input(‘Enter the value of shear force at ULS (KN) = ‘);

% SECTION PROPERTIES
h = input(‘Enter the depth of beam (mm) = ‘);
bw = input(‘Enter the beam width (mm) = ‘);
Cc = input(‘Enter concrete cover (mm) = ‘);
dr = input(‘Enter the diameter of reinforcement (mm) = ‘);
dl = input(‘Enter the diameter of links = ‘)
d = h-Cc-(dr/2)-dl % Effective depth
do = Cc+dl+(dr/2)

% CALCULATION OF THE SHEAR CAPACITY OF THE SECTION WITH NO SHEAR REINFORCEMENT

Asprov1= input(‘Enter the area of steel provided in the shear zone (mm^2) = ‘) % The reinforcement must exceed the design anchorage length by at least the effective depth
Crd = 0.12
k1= 1+sqrt(200/d)
if k1>2
k1=2
end
disp(‘Reinforcement ratio’)
P1 = (Asprov1/(bw*d))
if P1>0.02
P1 = 0.02
end

% THIS SECTION IS TO BE CONSIDERED IF THERE IS AXIAL FORCE IN THE SECTION
disp(‘Axial force in the section’)
N = input(‘Enter the value of AXIAL FORCE IF ANY(+VE FOR COMP, AND -VE FOR TENSION)(kN) = ‘)
disp(‘Axial stress in the section (Ds)’)
Ds = (N*1000)/(bw*h)
if Ds > (0.2*0.85*Fck)/1.5
Ds = (0.2*0.85*Fck)/1.5
end
k2 = 0.15;
disp(‘Minimum shear stress in the section (N/mm^2)’)
Vmin = 0.035*k1^(1.5)*sqrt(Fck)
disp(‘Concrete resistance shear stress (VRd) (N/mm^2)’)
VRdQ = ((Crd*k1)*(100*P1*Fck)^(1/3)*(bw*d))/1000 % SHEAR FORCE CONTRIBUTION
VRdN = ((k2*Ds)*(bw*d))/1000 % AXIAL FORCE CONTRIBUTION
VRd = VRdQ + VRdN % TOTAL SHEAR RESISTANCE
if VRd < ((Vmin+(k2*Ds))*(b*d))/1000
VRd = ((Vmin+(k2*Ds))*(b*d))/1000
end
if VRd>VEd
disp(‘Since VRd > VEd’)
disp(‘NO SHEAR REINFORCEMENT REQUIRED, PROVIDE NOMINAL LINKS’)
else
if VRd<VEd
disp(‘Since VRd < VEd’)
disp(‘SHEAR REINFORCEMENT REQUIRED, CALCULATE COMPRESSION STRUT CAPACITY’)
disp(‘Assume strut angle (thetha) = 21.8 deg, cot(thetha)= 2.5’)
thetha = 21.8
V1 = 0.6*(1-(Fck/250))
disp(‘Design compressive strength of concrete (fcd) (N/mm^2)’)
fcd = 0.567*Fck
z = 0.9*d
disp(‘Maximum capacity of compression strut (KN)’)
VRDmax = ((b*z*V1*fcd)/(2.9))/1000
if VRDmax > VEd
disp(‘Since VRDmax > VEd’)
disp(‘OK! Calculate diameter and spacing of links’)
ASMINlinks_to_spacing = ((0.08*sqrt(Fck))/Fyk)*b
ASlinks_to_spacing = (VEd*1000)/(z*0.87*Fyk*2.5)
else
disp(‘Since VRDmax > VEd, it means we need a higher strut angle (beta)’)
disp(‘By calculating the strut angle’)
disp(‘Shear stress at the section (N/mm^2)’)
v = (VEd*1000)/(b*d)
beta = (0.5)*asind((v)/(0.153*Fck*(1-(Fck/250))))
if beta>45
disp(‘Since beta < 45 deg, SECTION INADEQUATE FOR SHEAR, INCREASE DEPTH !!!!!’)
else
disp(‘The ratio of Area of steel/spacing of links’)
ASlinks_to_spacing = (VEd*1000)/(z*0.87*Fyk*cotd(beta))
end
end
end
Area_of_legs = input(‘Enter the area of number of legs selected = ‘)
Spacing = input(‘Enter the spacing = ‘)
disp(‘CHECK’)
T = Area_of_legs/Spacing
if T > ASlinks_to_spacing
disp(‘Shear reinforcement is ok’)
else
disp(‘Increase area of steel, or reduce spacing’)
end
end

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ALSO IN THIS BLOG
Analysis and Design of Continuous R.C. Raker Beam for Stadium using EC2
Two Approaches to the Calculation of Deflection in EC2; Choose a Faster Approach for Yourself

Two Approaches to the Calculation of Deflection in Eurocode 2

By

Ubani Obinna Uzodimma

-

July 5, 2017

The verification of deflection is an important serviceability check in reinforced concrete design. Most designs codes adopt the ‘deemed to satisfy’ rules which limit the deflection to an acceptable depth using Basic Span/Effective depth ratio. This is generally more frequently employed than carrying out the rigorous exact calculation of deflection which is however not
very reliable.

Eurocode 2 has its own set of deemed to satisfy rules, and there are two approaches to calculation of deflection which yields the same result. It is not like that they are two different formulars, but they are just one formular expressed in two different ways. This post shows the two approaches, so that you can select the one you deem fast enough for carrying out your manual calculations.

According to Clause 7.4.2 of EN 1992-1-1, we can verify deflections in the absence of exact calculations using deemed to satisfy basic span/effective depth (limiting deflection to depth/250);

Lateral Deflection of Multi-Storey Buildings Under the Action of Wind Load

(a) BY USING THE FIRST APPROACH
Actual L/d must be ≤ Limiting L/d × β_s

The limiting basic span/ effective depth ratio is given by;
L/d = K [11 + 1.5√(f_ck) ρ₀/ρ + 3.2√(f_ck ) [(ρ₀ / ρ) – 1]^(3⁄2) ] if ρ ≤ ρ₀ ———– (1)

L/d = K [11 + 1.5√(f_ck) ρ₀/(ρ – ρ’) + 1/12 √(f_ck) (ρ₀/ρ)^(1⁄2) ] if ρ > ρ₀ ———— (2)
Where;
L/d is the limiting span/depth ratio
K = Factor to take into account different structural systems
ρ₀ = reference reinforcement ratio = 10^-3√(f_ck)
ρ = Tension reinforcement ratio to resist moment due to design load
ρ’ = Compression reinforcement ratio

The value of K depends on the structural configuration of the member, and relates the basic span/depth ratio of reinforced concrete members. This is given in the Table below;

Table%2Bof%2Bbasic%2Bspan%2Bto%2Beffective%2Bdepth%2Bratio%2BEC2

β_s = (500 As_prov)/(F_yk As_req) ———- (3)

(b) BY USING THE SECOND APPROACH
L/d ≤ Basic ratio × α_s × β_s
Where the basic ratio = 20K

Modification factor α_s which depends on the concrete strength f_ck and reinforcement percentages, is given by:

For 100A_s/bd < 0.1f_ck^0.5;
α_s = 0.55 + 0.0075f_ck/(100A_s/bd) + 0.005f_ck^0.5[ f_ck^0.5/ (100A_s/bd) – 10]^1.5 —————- (4)

For 100A_s/bd ≥ 0.1f_ck^0.5;
α_s = 0.55 + 0.0075f_ck/[100(A_s – A_s’)/bd) + 0.013f_ck^0.25 (100A_s’/bd)]^0.5 (for doubly-reinforced sections) —————- (6)

α_s = 0.55 + 0.0075f_ck/(100A_s/bd) (for singly-reinforced sections) —————- (7)

Modification factor β_s = 310/σ_s —————- (8)
Where:
σ_s is the stress in the tension reinforcement under the characteristic loading. It will normally be conservative to assume;
β_s = (500/f_yk)(A_s,prov/A_s,req) for values of (A_s,prov/A_s,req)≤ 1.5 —————- (9)

Important hints on the calculation of deflection using both approaches;

(1) For flanged sections with b/b_w ≥ 3, the basic ratios for rectangular sections should be multiplied by 0.8. For values of b/b_w < 3, the basic ratios for rectangular sections should be multiplied by (11 – b/b_w)/10

(2) The ratio should be based on the shorter span for two-way spanning slabs, and the longer span for flat slabs.

(3) For beams and slabs, other than flat slabs, with spans exceeding 7 m, which support partitions liable to be damaged by excessive deflections, the basic ratio should be multiplied by 7/span. For flat slabs, where one or both spans exceeds 8.5 m, which support partitions liable to be damaged by excessive deflections, the basic ratio should be multiplied by 8.5/span.

SOLVED EXAMPLE

For the example above we check the deflection for span D-E.
f_ck = 35 N/mm²; f_yk = 460 N/mm²; Effective depth = 840mm; Effective width of flange (B_eff) = 1650mm; A_s,req = 1850 mm²; A_s,prov = 2101 mm²
BY APPROACH 1

ρ = A_s,prov /B_effd = 2101 / (1650 × 840) = 0.0015158;
ρ₀ = reference reinforcement ratio = 10^-3√(f_ck) = 10^-3√(35) = 0.005916
Since if ρ ≤ ρ₀;

L/d = K [11 + 1.5√(f_ck) ρ₀/ρ + 3.2√(f_ck) (ρ₀ / ρ – 1)^(3⁄2)
k = 1.3 (One end continuous)

L/d = 1.3 [11 + 1.5√(35) × (0.005916/0.001516) + 3.2√(35) × [(0.005916 / 0.001516) – 1]^(3⁄2)
L/d = 1.3[11 + 34.630 + 93.608] = 181.00

β_s = (500 As_prov)/(F_yk As_req) = (500 × 2101) / (460 × 1850 ) = 1.234

Check b/b_w = 1650/300 = 5.5
Since b/b_w ≥ 3, multiply L/d by 0.8
Since span is greater than 7m, also multiply by 7/span

Therefore limiting L/d = 1.234 × 0.8 × (7/8) × 181.04 = 156.382
Actual L/d = 8000/840 = 9.523

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

BY APPROACH 2

100A_s/bd = (100 × 2101) / (1650 × 840) = 0.1516
Check 0.1f_ck^0.5 = 0.1 × (35)^0.5 = 0.5916

Since For 100A_s/bd < 0.1f_ck^0.5

α_s = 0.55 + 0.0075f_ck/(100A_s/bd) + 0.005f_ck^0.5[ f_ck^0.5/ (100A_s/bd) – 10]^1.5
α_s = 0.55 + [(0.0075 × 35) /(0.1516)] + [0.005 × (35)^0.5 × [(35^0.5/0.1516) – 10]^1.5 = 0.55 + 1.7315 + 4.625 = 6.9065

Basic ratio = 20k = 20 × 1.3 = 26

β_s = (500/f_yk)(A_s,prov/A_s,req) = (500/460) × (2101/1850) = 1.234

Check b/b_w = 1650/300 = 5.5
Since b/b_w ≥ 3, multiply L/d by 0.8
Since span is greater than 7m, also multiply by 7/span

Therefore limiting L/d = 1.234 × 0.8 × (7/8) × 26 × 6.9065 = 155.111
Actual L/d = 8000/840 = 9.523

You can verify that the two methods can be used, with the second method being more approximate.

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Shear Deformation of Beams Using Virtual Work Method

By

Ubani Obinna Uzodimma

-

July 1, 2017

The predominant cause of deformation in beams subjected to externally applied load is bending, and that is one of the most popular analytical methods for beams with small deflection. However, additional deformation is induced in beams due to shear forces in the form of mutual sliding of adjacent sections across each other. As a result of the non-uniform distribution of shear stresses, the sections previously plane now become curved due to shearing alone.

Beams that are analysed by considering elastic deformation due to bending are known as Euler-Bernoulli beams, while beams that are analysed considering bending and shear deformations are known as Timoshenko beams. Timoshenko beam theory is useful in the analysis of thick beams.

Difference between Euler beam and Timoshenko beam

In this article, we are going to present some solved examples on the shear deformation of one-span beams due to externally applied load. Using the simple virtual work method and employing Verecshagin’s combination rule, we are going to calculate the deflection at the critical points of some beams due to bending and due to shear forces. You will discover why shear deflection is neglected in some cases, but in sections that are significantly deeper, shear forces can be quite influential.

The displacement equation due to shear forces is given below;

Shear%2BDeformation%2BEquation%2BUsing%2BVirtual%2BWork%2BMethod

Where;
k = factor that accounts for non-uniform distribution of shearing stresses. For rectangular sections, k = 3/2, and for circular sections, k = 3/4
G = Shear modulus of the section
A = Area of the cross-section
Q = Shear force due to externally applied load
Q ̅ = Shear force due to a unit virtual load at the point where the deflection is sought

Similarly, the displacement equation due to bending moment is given below;

Beam%2BDeflection%2BEquation%2BUsing%2BVirtual%2BWork%2BMethod

Where;
E = Elastic modulus of the section
I = Moment of inertia of the cross-section
M = Bending moment due to externally applied load
M ̅ = Bending moment due to a unit virtual load at the point where the deflection is sought

Solved Examples
For all examples shown below, the section shown below will be used for all calculations.

Geometrical Properties
Area (A) = bh = 0.2m × 0.4m = 0.08 m²
Moment of inertia (I) = bh³/12 = (0.2 × 0.4³)/12 = 0.0010667 m⁴
E = 21.7 KN/mm² = 21.7 × 10⁶ KN/m²
G = E / 2(1 + ν) = (21.7 × 10⁶)/2(1 + 0.2) = 9.0416 × 10⁶
Flexural rigidity EI = (21.7 × 10⁶) × 0.0010667 = 23147.39 KN.m²
Shear stiffness GA = (9.0416 × 10⁶) × 0.08 = 723333.333 KN

Example 1: Shear deflection of a simply supported beam carrying a uniformly distributed load

The internal forces diagram due to externally applied load is shown below;

Beam%2BNumber%2B1%2BInternal%2BStresses%2Bdiagram%2B%2528UDL%2529

Removing the external load and placing a unit load at the mid-span, we can plot the internal stresses diagram as shown below;

Beam%2BNumber%2B1%2Band%2B2%2BVirtual%2BLoad

We can now obtain the deflection at the mid-span by diagram combination;

(a) Deflection due to bending;

Influence%2Bcoefficient%2BBending%2B%2528Beam%2BNumber%2B1%2529

EIδ_b1 = 2 [5/12 × 62.5 × 1.25 × 2.5] = 162.760
δ_b1 = 162.760/23147.39 = 0.0070314 m = 7.0314mm

(b) Deflection due to shear;

Influence%2Bcoefficient%2BShear%2BBeam%2BNumber%2B1

GAδ_s1 = 2 [3/2 × 1/2 × 50 × 0.5 × 2.5] = 93.75
δ_s1 = 93.75/723333.333 = 0.0001296 m = 0.1296 mm

Example 2: Shear deformation of a simply supported beam carrying a concentrated load at the mid-span.

The internal forces diagram due to externally applied load is shown below;

Beam%2BNumber%2B2%2BInternal%2BStresses%2BDiagram%2B%2528Point%2BLoad%2529

Placing a unit load at the mid-span and plotting the internal stresses diagram;

We can now obtain the deflection at the mid-span by diagram combination;

(a) Deflection due to bending;

Influence%2Bcoefficient%2BBending%2B%2528Beam%2BNumber%2B2%2529

EIδ_b2 = 2 [1/3 × 125 × 1.25 × 2.5] = 260.417
δ_b2 = 260.417/23147.39 = 0.01125 m = 11.25 mm

(b) Deflection due to shear;

Influence%2Bcoefficient%2BShear%2BBeam%2BNumber%2B2

GAδ_s2 = 2 [3/2 × 50 × 0.5 × 2.5] = 187.5
δ_s2 = 187.5/723333.333 = 0.0002592 m = 0.2592 mm

Example 3: Shear deformation of Cantilever beam carrying a uniformly distributed load

bending and shear force diagram of a cantilever

Placing a unit load at the free end and plotting the internal stresses diagram;

We can now obtain the deflection at the mid-span by diagram combination;

(a) Deflection due to bending;

Influence%2Bcoefficient%2BBending%2B%2528Beam%2BNumber%2B3%2529

EIδ_b3 = [1/4 × 250 × 5 × 5] = 1562.5
δ_b3 = 1562.5/23147.39 = 0.0675 m = 67.502 mm

(b) Deflection due to shear;

GAδ_s3= [3/2 × 1/2 × 100 × 5 × 1.0] = 375
δ_s3 = 375/723333.333 = 0.00051843 m = 0.5184 mm

As you can see in all the examples we considered, there was no case where the additional deflection due to shear was up to 1 mm, and for practical purposes, they are very negligible for normal one span beams. Watch out for future posts where we will present cases where shear deflection becomes significant.

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Elastic Settlement of Shallow Foundations

By

Ubani Obinna Uzodimma

-

June 30, 2017

Shallow foundations on natural soils will undergo settlement when loaded. Foundation settlement is mainly made up of elastic (or immediate) settlement, S_e, consolidation settlement, S_c. Immediate settlement takes place as the load is applied, or within a period of about 7 days. It is used for all fine-grained soils including silts and clays with a degree of saturation less than or approximately less than 90%, and for all coarse-grained soils with a large coefficient of permeability, say above 10^-3 m/s. This article presents a solved example of the elastic settlement of shallow foundations.

It is important to point out that, theoretically at least, a foundation could be considered fully flexible or fully rigid. A uniformly loaded, perfectly flexible foundation resting on an elastic material such as saturated clay will have a sagging profile, as shown in the figure above, because of elastic settlement. However, if the foundation is rigid and is resting on an elastic material such as clay, it will undergo uniform settlement and the contact pressure will be redistributed. The settlement at the centre of a rigid foundation can be estimated as, S_e(rigid) ≈ 0.93S_e(flexible).

To assess the immediate settlement of a shallow foundation, it is imperative to obtain reliable values of elastic parameters of the soil. Values from the laboratory can contain errors of about 50%, and in-situ tests are often more preferred, with close attention being paid to the anisotropic behaviour of soils.

A range of values for the elastic properties of soils can be seen in Table 1;

Type of Soil	Modulus of Elasticity E_s(MN/m²)	Poisson’s Ratio
Loose sand	10 – 25	0.20 – 0.40
Medium dense sand	15 – 30	0.25 – 0.40
Dense sand	35 – 55	0.30 – 0.45
Silty sand	10 – 20	0.20 – 0.40
Sand and gravel	70 – 170	0.15 – 0.35
Soft clay	4 – 20
Medium clay	20 – 40	0.20 – 0.50
Stiff clay	40 – 100

Table 1: Elastic properties of soils

It is also necessary to obtain a reliable stress profile from the applied load. We normally have the problem of computing both correct numerical values and the effective depth H of the influence zone. Theory of elasticity equations are used for the stress computation, with the influence depth (H) below the loaded area taken as H = 0, to H = ∞.

However, note that the stratum depth actually causing settlement is not at H/B = ∞, but is at any of the following;

(a) Depth (z) = 5B; where B is the least total lateral dimension of the footing
(b) Depth to where a hard stratum is encountered. Take hard as that stratum where the modulus of elasticity is ten times the modulus of elasticity of the upper layer.

The equation for computing the elastic settlement of a shallow footing is given below;

Where;
q_o = net applied pressure on the foundation
μ_s = Poisson’s ratio of soil
E_S = Average modulus of elasticity of the soil under the foundation from z = 0 to about z = 4B or 5B

Due to the non-homogeneous nature of soil deposits, the magnitude of E_s may vary with depth. For that reason, Bowles (1987) recommended using a weighted average value of E_s as given in equation (2) below;

Weighted%2BAverage%2Bof%2BModulus%2Bof%2BElasticity

Where;
E_S(i) = Soil modulus of elasticity within a depth ∆z
z ̅ = H or 5B, whichever is smaller
B’ = B/2 for the centre of the foundation, B for the corner of the foundation
I_S = Shape factor according to Steinbrenner (1934)

Steinbrenner%2527s%2BInfluence%2Bfactor%2Bequation

Where;

Steinbrenner%2527s%2BInfluence%2Bfactor%2Bequation%2Bparameters

I_f= Depth factor which depends on (D_f/B, μ_s, L/B)

Charts for computing I_f are available in standard geotechnical engineering textbooks.
α = factor that depends on the location on the foundation where settlement is being calculated

For the calculation of settlement at the centre of the foundation;
α = 4; m’= L/B; n’ = H/0.5B
For the calculation of settlement at the corner of the foundation;
α = 1; m’= L/B; n’ = H/B

Worked Example

For the example given below, compute the settlement at the centre of the footing, assuming that a net pressure of 145 kN/m² is exerted by the foundation.

Given that B = 1.8m, L = 2.7m; z ̅ = (5 × 1.8) = 9.0 m = H (9.0 m)

E_s = [12000(5) + 7500(3) + 10200(1.2)] / 9 = 10526.667 KN/m²
At the centre of the foundation;
α = 4; m’= L/B =2.7/1.8 =1.5 ; n’ = H/0.5B=9/(0.5(1.8)) = 10

A₀ = 1.5 In ((1+ (1.5² + 1)^0.5) × (1.5² + 10²)) / 1.5 (1 + (1.5²+ 10² + 1)^0.5)
= 1.5 In(2.803 × 10.112) / (1.5 × 11.161) = 0.7896

A₁ = In ((1.5 + (1.5² + 1)) × (1 + 10²)) / (1.5 + (1.5² + 10² + 1)^0.5)
= In[(3.3027 × 10.049) / (1.5 + 10.161)] = 1.0459

A₂ = 1.5 / [10 (1.5² + 10² + 1)^0.5] = 1.5 / (10 × 10.161) = 0.0147

F₁ = 1/π (A₀ + A₁ ) = 1/π (0.7896 + 1.0459) = 0.5842
F₂ = (n’/2π) tan^-1⁡A₂ = 10/2π tan^-1⁡(0.0147) = 0.0233 (calculated in radians)

Therefore, the shape factor according to Steinbrenner;

I_S = F₁ + [(1 – 2μ_s)/(1 – μ_s)] × F₂ = 0.5842 + [(1 – (2 × 0.3) / (1 – 0.3)] × 0.0233 = 0.5975
From chart, let us obtain the foundation embedment factor (I_f);
D_f/B = 0.9/1.8 = 0.5
L/B = 2.7/1.8 = 1.5

From the chart, say I_f = 0.79
Inputting all the parameters into the elastic settlement equation given in (1)

S_e = q_o α × B’ × ((1- μ_s²)/E_s) × I_s × I_f
S_e = 145 × 4 × 0.9 × ((1 – 0.3²)/10526.667) × 0.5975 × 0.79 = 0.0213m = 21.3 mm

This gives the elastic settlement at the centre of the footing. A lot of modifications and other proposals have been made by scholars over years due to improved research, and we will keep posting them as time goes on.

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Analysis of Three-Hinged Arch Structures

By

Ubani Obinna Uzodimma

-

June 27, 2017

Arches are important structural elements in engineering that provide economical solutions in buildings and bridges. Three-hinged arch structures are pinned at the supports (springings) and somewhere along the barrel, which is usually at the crown. The structural analysis and design of three-hinged arches involve the determination of the internal stresses (bending moment, shear force, axial force, and torsion in the structure due to externally applied load, and providing adequate sections to resist the applied load.

In horizontal beams supporting uniformly distributed load, the bending moment increases with the square of the span and hence they become uneconomical for long-span structures. In such situations, arches could be advantageously employed, since they would develop horizontal reactions, which in turn reduce the bending moment.

A three-hinged arch, which is usually made from steel or timber, is statically determinate. Unlike statically indeterminate arches, they are not affected by differential settlement or temperature changes. Three-hinged arch structures have three natural hinges as the name implies. The two supports are hinged, and another internal hinge is usually located at the crown.

A three-hinged arch has four unknown reactions, i.e., two vertical reactions and two horizontal reactions at the supports. For their determination, three equilibrium equations can be formulated considering the whole of the structure.

Since it is well known that the bending moment at any internal hinge is zero, the internal hinge in the barrel of the arch provides an additional equation for the equilibrium of any part of the system. This means that the sum of the moments of all external forces, which are located on the right (or on the left) part of the structure with respect to the internal hinge is zero. Therefore, a three-hinged arch is a geometrically unchangeable and statically determinate structure. The figure below shows a three-hinged Bolt-laminated Ekki Timber bridge at Finowfurt, Germany.

The peculiar feature of arched structures is that horizontal reactions are induced even when the structure is subjected to vertical load only. These horizontal reactions under vertical loading A_x = B_x = H are called the thrust of the structure. At any cross-section of the arch, bending moments, shear, and axial forces are developed. However, the bending moments and shear forces are smaller than corresponding internal forces in a simply supported beam covering the same span and subjected to the same load.

This fundamental property of the arch is thanks to the thrust developed. Thrusts in both supports are pointed towards each other and consequently reduce the bending moments that would arise in beams of the same span and load configuration. The two parts of an arch may be connected by a tie. In this case in order for the structure to remain statically determinate, one of the supports of the arch should be supported on a roller.

Solved Example on Analysis of a Three-hinged Arch

For the parabolic arch that is loaded as shown below, compute the support reactions and plot the internal stresses diagram for the identified sections. The arch is hinged at points A, B, and C.

SOLUTION
Geometrical properties of the arch
The ordinate (y) at any point along a parabolic arch is given by;

y = [4y_c (Lx – x²)] / L²
Where;
y_c = Height of the crown of the arch from the base
L = Length of arch
x = Horizontal ordinate of interest
Hence, y = [4 × 10 (45x – x²)] / 45²

The general equation of the arch now becomes;
y = (8/9)x – (8/405)x² —————– (1)

Differentiating equation (1) with respect to x
dy/dx = y’ = (8/9) – (16/405)x —————— (2)

From trigonometric relations, we can verify that;
Sin θ = y’/[1 + (y’)²]^0.5 —————- (3)
Cos θ = 1/[1+ (y’)²]^0.5 —————- (4)

From the above relations, we can carry out the calculations for obtaining the geometrical properties of the arch structure.

Let us consider point A (support A of the structure);

We can verify that at point A, x = 0, and y = 0;
From equation (2) above, y’ = 8/9;
Thus,
Sin θ = (8/9)/[1 + (8/9)²]^0.5 = 0.664
Cos θ = 1/[1 + (8/9)²]^0.5 = 0.747

Similarly, let us consider point 3 of the structure;

At point 3, x = 27.5m
From equation (1), we can obtain the value of y as; y = [(8/9) × 27.5] – [(8/405) × 27.5²] = 9.5061m;
The tangent at that point can be obtained from from equation (2); dy/dx = y’ = (8/9) – [(16/405) × 27.5] = – 0.1975
Thus,
Sin θ = (-0.1975)/[1 + (-0.1975)²^0.5 = -0.1937
Cos θ = 1/[1 + (-0.1975)²]^0.5 = 0.9813

For the entire section, it is more convenient to set out the geometrical properties in a tabular form. See the picture below;

Support Reactions

Let ∑M_B = 0; anticlockwise negative
(Ay × 45) – (12 × 22.5 × 33.75) – (25 × 10) – (15 × 6.913) = 0
Therefore, Ay = 210.36 kN

Let ∑M_A = 0; clockwise negative
(By × 45) – (25 × 35) + (15 × 6.913) – (12 × 22.5 × 11.25) = 0
Therefore, By = 84.64 kN

Let ∑M_C^L = 0; anticlockwise negative
(Ay × 22.5) – (Ax × 10) – (12 × 22.5 × 11.25) = 0
22.5Ay – 10Ax = 3037.5 ——— (a)

Substituting, Ay = 210.36 kN into equation (a)
Hence, Ax = 169.56 kN

Let ∑M_C^R = 0; clockwise negative
(By × 22.5) – (Bx × 10) – (25 × 12.5) – (15 × 3.087) = 0
22.5By – 10Bx = 358.8 ——— (b)

Substituting, By = 84.64 kN into equation (b)
Hence, Bx = 154.56 kN

Internal Stresses

Bending Moment
M_A = 0 (hinged support)
M₁ = (210.36 × 7.5) – (169.56 × 5.555) – (12 × 7.5 × 3.75) = 298.294 kN.m
M₂ = (210.36 × 15) – (169.56 × 8.889) – (12 × 15 × 7.5) = 298.181 kN.m
M_C = (210.36 × 22.5) – (169.56 × 10.000) – (12 × 22.5 × 11.25) = 0

Coming from the right hand side;

M_C = (84.64 × 22.5) – (154.56 × 10) – (25 × 12.5) – (15 × 3.087) = 0
M₃ = (84.64 × 17.5) – (154.56 × 9.506) – (25 × 7.5) – (15 × 2.593) = -214.442 kN.m
M₄ = (84.64 × 10) – (154.56 × 6.913) = -222.073 kN.m
M_B = 0 (hinged support)

Shear
Q = ∑V cosθ – ∑H sinθ

Q_A = (210.36 × 0.747) – (169.56 × 0.664) = 44.551 kN
Q₁ = [210.36 – (12 × 7.5)] 0.860 – (169.56 × 0.509) = 17.203 kN
Q₂ = [210.36 – (12 × 15)] 0.959 – (169.56 × 0.284) = -19.040 kN
Q_C^R = [210.36 – (12 × 22.5)] 1.000 – (169.56 × 0.000) = – 59.64 kN
Q₃ = [210.36 – (12 × 22.5)] 0.981 – (169.56 × – 0.193) = – 25.782 kN
Q₄^L = [210.36 – (12 × 22.5)] 0.897 – (169.56 × – 0.443) = – 21.618 kN
Q₄^R = {[210.36 – (12 × 22.5) – 25] × 0.897} – [(169.56 – 15) × (- 0.443)] = 7.452 kN
Q_B = (- 84.64 × 0.747) – [154.56 × (- 0.664)] = 39.402 kN

Axial
N = -∑V sinθ – ∑H cosθ

N_A = (-210.36 × 0.664) – (169.56 × 0.747) = – 266.340 kN
N₁ = – [210.36 – (12 × 7.5)] 0.509 – (169.56 × 0.860) = – 207.085 kN
N₂ = – [210.36 – (12 × 15)] 0.284 – (169.56 × 0.959) = -171.230 kN
N_C = – [210.36 – (12 × 22.5)] 0.000 – (169.56 × 1.000) = -169.560 kN
N₃ = -{[210.36 – (12 × 22.5)] × – 0.193} – (169.56 × 0.981) = -177.849 kN
N₄^L = -{[210.36 – (12 × 22.5)] × – 0.443} – (169.56 × 0.897) = – 178.516 kN
N₄^R = -{[210.36 – (12 × 22.5) – 25] × – 0.443} – [(169.56 – 15) × (0.897)] = – 176.136 kN
N_A = – (-84.64 × – 0.664) – (154.56 × 0.747) = – 171.657 kN

Internal Stresses Diagram (Not to scale)

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