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The Role of Compression Steel in Deflection of Reinforced Concrete Structures

By
Ubani Obinna
-

Deflection is one of the most important serviceability limit state criteria in the design of reinforced concrete structures. Excessive deflection in a structure can lead to cracking of finishes, impaired functionality, and unsightly appearance. Different codes of practice have provisions for controlling deflection in concrete structures. In BS 8110-1:1997 and EN 1992-1-1 (Eurocode 2), deflection can be assessed using the simplified span/depth ratio (deemed-to-satisfy rules) or by following the rigorous method (direct calculation of deflection). However, rigorous method is lengthy and has numerous uncertainties involved in the calculation, hence the span/depth ratio approach is widely used. In both codes of practice, compression reinforcement has been shown to be beneficial to the deflection response of reinforced concrete structures.

Concrete is good in compression but weak in tension. As a result, steel reinforcement is usually provided in the tension zone of concrete structures to resist tensile stresses. Sometimes, steel is provided in the compression zone to assist concrete in resisting compression, and in this case, is normally referred to as doubly reinforced section. The major reasons for providing compression reinforcement in concrete structures are;

  • to reduce long-term deflection due to sustained loading,
  • to increase ductility,
  • to changing failure mode from compression-initiated to tension-initiated, and for
  • ease in construction (for example using hanger bars to provide support for stirrups).

However, it has been recognised that deflection in a concrete beam and slab increases with time for up to nine years[1]. The long-term deflection of a reinforced concrete member under sustained load is mainly due to creep and shrinkage, and it is usually larger than the immediate deflection[2]. It is important to realise that creep and shrinkage are time-dependent properties of concrete, which are influenced by environmental conditions and the concrete mix design[3]. While creep depends on the sustained load, shrinkage is independent of the load. Other important concrete properties used in assessing level of deflection are the modulus of elasticity, modulus of rupture, and tensile strength properties.

In normal strength concrete, compression reinforcement helps in reducing creep under sustained load in the compression zone, and this helps in reducing the final deflection of the structure. However, research has suggested that the effect of compression reinforcement may be minimal in reducing deflection in high strength concrete structures due to low creep[4]. Studies have shown that the effect of compression reinforcement in reducing long term deflection is dependent on the compressive strength of the concrete[4]. However, compression reinforcement is rarely used for the purpose of resisting deflection in concrete structures.

When the props and formwork are removed from the soffit of a slab, it undergoes immediate elastic deflection if the load is low. If the load is high, there will be cracks in the tension zone (say mid-span), and the deflection may not be entirely elastic[1]. However in the long term, the initial deflection under sustained load will increase because of shrinkage, creep, and increased cracking (note that cracking is deemed normal in concrete members).

Concrete drying shrinkage will cause warping of the slab in zones where there are unequal areas of top and bottom reinforcement. This is typical in the mid-span of a reinforced concrete slab, hence the slab will warp downwards and adds to the long-term deflection. The magnitude of warping is found to be directly proportional to the drying shrinkage of the concrete. It continues at a decreasing rate with increasing time for several years after pouring, and is independent of load. Therefore, provision of top reinforcements can curtail the effect of warping, thereby reducing the final deflection.

Read Also…
Structural Design of Cantilever Slabs
Calculation of crack width and crack spacing in reinforced concrete

Furthermore, it is important to note that the actual long-term deflection of a normally reinforced slab or beam in the typical floor of a multi-level building depends on some other factors which may not be duly paid attention to during design. Some of these factors are materials supply, construction techniques, loading history, weather and time. Loading history and method of construction is known to affect the deflection of slabs. The construction load coming from the floor above have been shown to significantly influence the deflection behaviour of slabs in a multi-story building construction[5].

Assessment of deflection in BS 8110

BS 8110-2:1985 says that the sag in a concrete member will become noticeable if the deflection exceeds span/250. According to clause 3.4.6.1 of BS 8110, deflections in a structure may be calculated and then compared with the serviceability requirements given in BS 8110-2:1985. However, in all normal cases the deflections of a beam will not be excessive if the ratio of its span to its effective depth is not greater than the appropriate ratio. The basic span to effective depth ratio is given in Table 3.9 of BS 8110-1:1997 with values of 7 (cantilever), 20 (simply supported), and 26 (continuous members). The basic span/effective depth ratio should be multiplied by modification factors given in in Tables 3.10 and 3.11 of the code. BS 8110-1:1997 acknowledges that compression reinforcement influences deflection.

The modification factor for tension reinforcement of reinforced concrete is given in Table 3.10 of BS 8110-1;

Modification factor = 0.55 + {(477 – fs)/[120(0.9 + M/bd2)]} ≤ 2.0

Where fs is the service stress in the member and it is given by;

fs = 2fyAsreq/3Asprov

Where;
Asprov = Area of steel provided
Asreq = Areas of steel required
fy = yield strength of reinforcement
M = Applied bending moment at ultimate limit state

The modification factor for compression reinforcement is given in Table 3.11 of BS 8110-1 as;

Modification factor for compression reinforcement = (1 + 100As’prov/bd)/(3 + 100As’prov/bd)

Where As’prov is the area of compression steel provided.

In BS 8110-1, the area of compression reinforcement used for the calculation may include all bars in the compression zone, even those not
effectively tied with links.

Read Also…
Formulas for calculating different properties of concrete to Eurocode 2
Structural design of flat slab to Eurocode 2

Assessment of deflection in Eurocode 2

Clause 7.4.1(4) of EC2 says that the utility and appearance of a structure may be impaired when the deflection exceeds span/250. This limit accounts for both long term and short term deflection. Eurocode 2 follows a slightly different approach in assessing the deflection of a reinforced concrete member using the span/depth ratio approach.

EC 2 allows span/effective depth limits (L/d) to be calculated from equations 7.16a, 7.16b and 7.17 of the code:

L/d = K[11+(1.5√fck x ρ0/ρ) + 3.2√fck(ρ0/ρ – 1)1.5] if ρ0 ≤  ρ               (7.16a)

L/d = K(11+(1.5√fck x ρ0/(ρ – ρ’)) + √fck √(ρ0‘/ρ))/12) if ρ0 >  ρ         (7.16b)

where K is a factor which depends on the static system of the structure (structural form). It has a value of 1.0 for simply supported structures, 1.5 for continuous structures, 1.3 for continuous – end span structures, 1.2 for flat slab, and 0.4 for cantilevers.

ρ0 = reference reinforcement ratio = 10-3√fck
ρ = required tension reinforcement ratio
ρ’ = required compression reinforcement ratio
fck = concrete cylinder strength (N/mm²)

Equations 7.16 above have been derived on the assumption that the steel stress at SLS at a cracked section at the midspan of a beam or slab or at the support of a cantilever is 310 MPa (corresponding roughly to fyk = 500 MPa). Where other stress levels are used, the values obtained using Expression (7.16) should be multiplied by 310/σs. It will normally be conservative to assume that:

310/σs = 500/(fyk Asreq/Asprov)           (7.17)            

Where;

σs = tensile stress at mid-span (support for cantilevers) under the design load at SLS
Asprov = area of steel provided at the section
Asreq  = area of steel required at the section for ultimate limit state.

A little observation of equation 7.16 will show that the direct value of compression reinforcement provided cannot be used in EC2 unless the reference reinforcement ratio of 10-3√fck is greater than the actual reinforcement ratio. This goes on to say that the value to be used should be based on the area of compression steel required and not the area of compression steel provided (say H12 mm hanger bars). This is where BS 8110 greatly differs from EC2.

In BS 8110, the area of steel provided can be increased to order to reduce the service stress in the reinforcement. This usually increases the modification factor and the allowable L/d value. However, some questions have been asked on the area of reinforcement to be used in Eurocode assessment of deflection such as that from Beal[6];

It is not clear how Eq. 7.16(a) and (b) are intended to be applied when the reinforcement service stress varies from 310N/mm² as it is not clear how the reinforcement ratio ρ in eq. 7.16 is to be calculated. Is it the actual amount of reinforcement present, or is it the reinforcement which would have been required for a design stress of 310N/mm²? As currently drafted, it is not clear which of these interpretations is correct.

References

[1] Taylor J. P. (2009): The deflection of reinforced concrete. Taylor Lauder Bersten Pty Ltd
[2] Zhon W., Kokai T. (2010): Deflection calculation and control for reinforced concrete flexural members. Canadian Journal of Civil Engineering 37(1):131-134 https://doi.org/10.1139/L09-121
[3] Elaghoury Z. (2019): Long-term deflection of reinforced concrete beams. M.Sc thesis Submitted to the Department of Civil and Environmental Engineering, The University of Western Ontario
[4] Muhaisin M.H., Jawdhari A.R., Ammash H.K. (2019): Revised formula for predicting the long-term deflection ultiplier of normal and high strength concrete. Rev. IBRACON Estruct. Mater. 12(6):1345-1352 http://dx.doi.org/10.1590/s1983-41952019000600007 
[5] Vollum R. L. (2009): Comparison of deflection calculations and span-to-depth ratios in BS 8110 and Eurocode 2. Magazine of Concrete Research 61(6):465-476 https://doi.org/10.1680/macr.2009.61.6.465
[6] Beal A. N. (2009): Eurocode 2: Span/depth ratios for RC slabs and beams. The Structural Engineer, Vol. 87 No. 20, 20th October 2009

Design of Pile Foundation System for Bridge Piers

By
Ubani Obinna
-

Piles are the most preferred type of foundation for supporting the piers of overhead bridges. They offer greater resistance to vertical and horizontal actions from the bridge deck than shallow foundations. Furthermore, they extend the foundation of piers beyond scouring zones so that there will be no loss of bearing capacity. In this article, we are going to show how to design the foundation of a bridge pier using piles.

We have previously talked about the design the piers and pier cap of an overhead bridge. If you missed the post, read it below.

Structural Design of Pier and Pier Caps of Highway Bridges

Let us go ahead and design the foundation of the bridge pier using the soil investigation report shown below.

Safe working load of piles

Step 1: Determine the load from the superstructure

The loading on the pier cap at ultimate limit state and serviceability limit state are shown below;

Loading on bridge pier
Actions on the pier cap

When analysed on Staad Pro, the following support reactions were obtained;

ULS
Vertical support reaction on each pier = 6001.483 kN
Horizontal reaction = 368.443 kN
Bending Moment = 702.032 kNm

SLS
Vertical support reaction on each pier = 4259.247 kN
Horizontal reaction = 260.901 kN
Bending Moment = 497.125 kNm

Step 2: Determination of the number of piles required and the layout

We normally use the service loads to determine the number of piles required for the structure, and use ultimate loads (factored loads) for the structural design.

Using 900 diameter piles, Number of piles required per pier = 4259.247/1210 = 3.52
Therefore, provide 4 Piles per pier leg. The layout of the pile cap is shown below. However, attempts were made to combine the two piers under one pile cap, but no economical solution was reached due to the wide spacing of the piers (6 m).

pile cap layout
Adopted layout of the pile cap

Step 3: Provide longitudinal reinforcement for the piles

The reinforcements required for the piles is usually based on the minimum required for the selected pile diameter

Relationship between pile diameter and longitudinal reinforcement
Relationship between minimum area of reinforcement and pile diameter

For a pile diameter of 900 mm, the minimum area of steel required is 2500 mm2
Therefore provide 9H20 mm (Asprov = 2826 mm2)
Provide helical links of H10@250 pitch

Read also…
Reinforcement design of piles

Step 4: Analyse and design the pile cap

The required thickness of the pile cap is usually estimated from the diameter of the piles.

h = 2 x diameter of piles + 100 = 2(900) + 100 = 1900 mm deep pile cap

Different approaches can be used for analysing the pile cap such as bending theory, strut and tie method, or finite element analysis. Let us obtain the internal forces in the pile cap from finite element analysis using Staad Pro.

PILE CAP MODEL ON STAAD PRO
3D model of the pile cap on Staad Pro

When analysed at ULS, the maximum moment in the pile cap is shown below;

FE model of pile cap
Bending moment in the pile cap

MEd = 1796 KN.m/m

Effective depth d = 1900 – 100  = 1800 mm

k = MEd/(fckbd2) = (1796 × 106)/(40× 1000 × 18002) = 0.0138

Since k < 0.167, no compression reinforcement required
z = d[0.5+ √(0.25 – 0.882k)]
z = d[0.5+ √(0.25 – 0.882(0.0138))] = 0.95d

As1 = MEd/(0.87yk z) = (1796 × 106)/(0.87 × 500 × 0.95 × 1800) = 2414 mm2/m
Provide 20H25mm BOT (ASprov = 8380 mm2)

If we are to use simplified strut and tie method;

Ft = (NEd x l)/8d = (6001.483 x 2.7)/(8 x 1.8) = 1125.28 kN
Ast = Ft/0.87fyk = (1125.28 x 103)/(0.87 x 500) = 2587 mm2/m

You can compare the two results above.

To calculate the minimum area of steel required;

ASmin = 0.26 (fctm/fyk)bd
fctm = 0.3fck2/3 = 0.3 x 40(2/3) = 3.51 N/mm2
ASmin = 0.26 x (3.51/500) x 1000 x 1800 = 3285 mm2/m

Check if  ASmin < 0.13bd/100 = 2340 mm2/m

Provide H32 @ 225 c/c (Asprov = 3573 mm2)

Read also…
Structural aspects of pile foundation design

Step 5: Check for punching shear around the pier perimeter

VEd = 6001.483 kN

Column perimeter u = πd = π x 1200 = 3769.9 mm

v = V/ud
v = (6001.483 × 1000)/(3769.9 × 1800) = 0.8844 N/mm2
VRd,max = 0.2(1 –  fck/250)fcd
VRd,max = 0.2 (1 – 40/250) x 40/1.5 = 4.48 N/mm2

0.8844 N/mm2 > 4.48 N/mm2 This is okay

This shows that the punching shear around column perimeter is ok.

shear checks of pile caps
Critical shear checks for pile caps

Step 6: Check for critical punching shear

Consider the critical section for shear to be located at 20% of the pile diameter inside the pile cap.

Distance of this section from the column face;
av = 0.5(Spacing between piles – width of column) – 0.3(pile diameter)
av = 0.5(2700 – 1200) – 0.3(900) = 480 mm

Length of corresponding perimeter for punching shear
u = 4(2160) = 8640 mm

Perimeter of pile cap = 4(3900) = 15600 mm

The contribution of the column load to the shear force may be reduced by applying a factor β = av/2d, since av < 2d where 0.5d ≤ av ≤ 2d

But a little consideration will show that av(480 mm) < 0.5d(900 mm), therefore, take av as 0.5d (900)
Therefore β = 900/2(1800) = 0.25

v = βV/bd
V = 6001.483/4 = 1500.37 kN

v = (0.25 x 1500.37 x 103)/(1000 x 1800) = 0.208 N/mm2

VRd,c = [CRd,c.k.(100ρ1 fck)(1/3)]

Where;
CRd,c = 0.18/γc = 0.18/1.5 = 0.12
k = 1 + √(200/d) = 1 + √(200/1800) = 1.33 > 2.0, therefore, k = 1.33
ρ1 = As/bd = 3573/(1000 × 1800) = 0.001985 < 0.02

VRd,c = [0.12 × 1.33(100 × 0.001985 × 40)(1/3)]=  0.318 N/mm2

Since VRd,c (0.318 N/mm2) > VEd (0.208 N/mm2), No shear reinforcement is required Okay.

Step 7: Check for Anchorage Length

Let us calculate the design tension anchorage length of H32mm bar (fyk = 500 N/mm2, concrete cover = 100 mm, Concrete cylinder strength fck = 40 N/mm2) for a bent bar assuming good bond conditions.

lbd = α1 α2 α3 α4 α5 lb,req ≥ lb,min
lb,rqd = (ϕ/4) σsd/fbd
fbd =2.25η1 η2 fctd
η1 = 1.0 ‘Good’ bond conditions
η2 = 1.0 bar size ≤ 32

fctd = (αct fctk 0.05)/γc

where;
fctk 0.05 = characteristic tensile strength of concrete at 28 days = 2.5 N/mm2 (Table 3.1 EC2)
γc = partial (safety) factor for concrete = 1.5
αct = coefficient taking account of long-term effects on the tensile strength, this is an NDP with a recommended value of 1.

fctd = (1.0 × 2.5)/1.5 = 1.667 N/mm2
fbd = 2.25 × 1.0 × 1.0 × 1.667= 3.75 N/mm2

lb,rqd = (ϕ/4) σsd/fbd
σsd = 0.87 × 500 = 435 N/mm2

lb,rqd = (ϕ × 435 )/(4 × 3.75) = 29ϕ

Therefore;

lbd = α1 α2 α3 α4 α5(29ϕ) (see Table 8.2 EC2)

For a bent bar
α1 = 1.0 (Cd = 75 mm is ≤ 3ϕ = 3 × 32 = 96 mm)
α2 = 1.0 – 0.15 (Cd – 3ϕ)/ ϕ = 0.901 < 1.0
α3 = 1.0 conservative value with k = 0
α4 = 1.0 N/A
α5 = 1.0 conservative value
lbd = 0.9 × (29ϕ) = 26.1ϕ = 26.1 × 32 = 835 mm

With the effective depth of 1800 mm, the pile dimension will satisfy the anchorage requirements.

Design of Bridge Pier and Pier Cap

By
Ubani Obinna
-

Bridge pier caps are horizontal structural members used for transferring bridge deck loads to the piers, before they are transferred to the foundation. They can be made of different shapes or forms, depending on the required aesthetics of the bridge. However, pier caps (or heads) must be designed to resist the actions transferred from traffic, self weight of the bridge deck, wind actions, or accidental actions due to vehicle impact. By the nature of their configuration, the piers and the pier cap/head of bridges should be designed as monolithic rigid frames.

However, some of the different configurations that are obtainable in bridge piers are;

  • Solid wall pier
  • Hammer head bridge pier
  • Rigid frame bridge pier
rigid wall pier cap
Solid wall Pier
Hammer head pier cap
Hammer head Pier Cap
Rigid frame pier cap
Rigid Frame Pier Cap

In this article, we are going to show how to design reinforced concrete rigid frame bridge pier and pier cap using Staad Pro software. Different methods can be used for obtaining the reactive forces from the bridge deck such as grillage analysis, finite element analysis, or classical methods such as Courbon’s theory. We will pick the reactive forces from finite element analysis, and apply them to the bridge pier cap.

The configuration of the bridge deck is shown below;

Bridge deck Staad Pro 1
Finite element model of the bridge deck

The length of the bridge deck is 15 m, and the loading and configuration is shown below;

Bridge deck with five girder
Structural configuration of the bridge deck
Loaded bridge deck 1
Traffic action on the bridge deck (LM 1)

The maximum reactive forces from finite element analysis are as follows;

Self weight = 276.288 kN
Traffic UDL = 194.839 kN
Traffic moving load = 320.762 kN

These reactions are transmitted through the elastomeric pad bearing to the pier cap. Note that there are accompanying horizontal actions due to braking, acceleration forces, skidding of vehicles etc. Design value of braking or acceleration forces depend on the vertical loads applied on notional lane 1. The values of horizontal forces should be combined with appropriate value for vertical traffic load corresponding to its frequent value. However, when the leading traffic loads are vertical ones, the accompanying value of the braking and acceleration forces are to be defined in the National Annex and can be set to zero. But for simplicity, horizontal actions have not been considered in this article.

Therefore the ultimate load that will be transferred to the pier cap is given by;

VEd = 1.35Gk + 1.5Qk = 1.35(276.288) + 1.5(194.839 + 320.762) = 1146.39 kN

Since this is assumed to be an intermediate pier, the pier cap will be subjected to double of the load above. Therefore, the maximum ULS load from each girder will be 1146.39 x 2 = 2239 kN. Note that the factored self weight of the pier and pier cap should be included in the model.

Fro proper analysis and design of the foundation, it is important to separate the service and ultimate loads properly. At SLS, the maximum reactive force from each girder is 791.889 kN. The preliminary sizing of the rigid frame pier cap and the design data of the pier system is shown below;

Diameter of piers = 1200 mm
Dimensions of pier cap = 950 x 1200 mm
Support of frame = Fixed

Loaded Bridge Pier 1
Bridge Deck Load on the Pier Cap
3D MODEL OF PIER CAP
3D Model of bridge pier on Staad Pro

When analysed on Staad Pro, the internal forces in the rigid frame pier cap are as follows;

BMD
Bending moment diagram
SFD
Shear force diagram
AFD
Axial force diagram

The summary of the internal forces at ULS are as follows;

Pier Cap
Sagging moment = 3298.257 kNm
Hogging moment = 2911.468 kNm
Maximum shear force = 3468.85 kN
Maximum axial force = 368 kN (compression)

Piers
Maximum moment = 1508.62 kNm
Maximum shear force = 368.443 kN
Maximum axial force = 6001.483 kN

Structural design of the Pier Cap

Let cover to reinforcements = 50 mm
Yield strength of reinforcement fyk = 500 MPa
Compressive strength of concrete fck = 40 MPa

Span – Sagging

MEd = 3298.257 KN.m

Effective depth (d) = h – Cnom – ϕ/2 – ϕlinks
Assuming ϕ25 mm bars will be employed for the main bars, and ϕ12 mm bars for the stirrups (links)
d = 1200 – 50 – 12  – 12 = 1126 mm

k = MEd/(fckbd2) = (3298.257 × 106)/(40× 950 × 11262) = 0.0684

Since k < 0.167, no compression reinforcement required
z = d[0.5+ √(0.25 – 0.882k) ]
z = d[0.5+ √(0.25 – 0.882(0.0684))] = 0.935d

As1 = MEd/(0.87yk z) = (3298.257 × 106)/(0.87 × 500 × 0.934 × 1126) = 7209 mm2
Provide 20H25mm BOT (ASprov = 8380 mm2)

Hogging
MEd = 2911.468 KN.m

k = MEd/(fckbd2) = (2911.468 × 106)/(40 × 950 × 11262) = 0.0604
Since k < 0.167, no compression reinforcement required
z = d[0.5+ √(0.25 – 0.882K)]
z = d[0.5+ √(0.25 – 0.882(0.0604))] = 0.943d

As1 = MEd/(0.87fyk z) = (2911.468 × 106)/(0.87 × 500 × 0.943 × 1126) = 6303 mm2
Provide 16H25mm TOP (ASprov = 6704 mm2)

Shear Design (Support A)
VEd = 3468.85 kN
NEd = 368 KN (compression)

We are going to anchor the 16No of H25mm 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 = (368 × 1000) / (950 × 1126) = 0.344 N/mm2
k = 1 + √(200/d) = 1 + √(200/1126) = 1.42> 2.0, therefore, k = 1.42
ρ1 = As/bd = 6704/(950 × 1126) = 0.00627 < 0.02; K1 = 0.15

VRd,c = [0.12 × 1.42(100 × 0.00627 × 40 )(1/3) + (0.15 × 0.344)]  × 950 × 1126 =  588745.259 N = 588.745 KN

Since VRd,c (588.745 KN) < VEd (3468.85 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 – 40/250) = 0.504
fcd = (αcc ) fck)/γc = (1 × 40)/1.5 = 26.667 N/mm2
Let z = 0.9d

VRd,max = [(950 × 0.9 × 1126 × 0.504 × 26.667) / (2.5 + 0.4)] × 10-3 = 4461.811 kN

Since VRd,c < VEd < VRd,max
Hence, Asw/S = VEd/(0.87fykzcot θ) = 3468850/(0.87 × 500 × 0.9 × 1126 × 2.5) = 3.147

Minimum shear reinforcement;
Asw/S = ρw,min × bw × sinα (α = 90° for vertical links)
ρw,min = (0.08 × √(fck))/fyk = (0.08 × √40)/500 = 0.001012
Asw/S (min) = 0.001012 × 950 × 1 = 0.961

Maximum spacing of shear links = 0.75d = 0.75 × 1126 = 844.5 mm
Provide 6H12mm @ 200mm c/c (Asw/S = 3.39) Ok

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

Side Bars
Provide H16@200 c/c

The typical detailing of the pier cap is shown below. Note that bar marks and layer information were not included.

Detailing of a pier cap
Typical detailing sketch of the pier cap

Design of the piers

You are expected to go through the process of designing columns according to Eurocode 2. You will need to check if the column is short or slender. If the column is slender, you will need to obtain the additional moments due to second order effects.

However for simplicity in this post, let us add the effect of imperfection to the column moment. You can go through the rigourous process of ensuring that you are designing for the worst effect.

Let us assume that effective length of the pier is 0.85L = (0.85 x 6000) = 5100 mm (note that Eurocode requires a more rigourous approach for calculating the effective length of columns)

The geometric imperfection of the column ei = [(1/200) x (5100/2)] = 12.75 mm

eiNEd = 6001.483 x 0.01275 = 76.5189 kNm

Let us therefore take the column design moment MEd to be = 1508.62 + 76.5189 = 1585.13 kNm

Hence,
NEd = 6001.483 kN
MEd = 1585.13 kNm
d = 1200 – 50 – 50 = 1100
d/h = 0.9 (say)

Interaction diagram for circular columns

Let fcd = (0.85 x 40)/1.5 = 22.667 N/mm2

NEd/h2fcd = (6001.483 x 1000)/(12002 x 22.667) = 0.183 (Conservatively say 0.2)
MEd/h3fcd = (1585.13 x 106)/(12003 x 22.667) = 0.04

Therefore, Asfyd/h2fcd = 0.1
As.req = (0.1 x 12002 x 22.667)/(0.87 x 500) = 7504 mm2

The minimum area steel required is 0.1NEd/fyd = 1379 mm2 or 0.002Ac = 2262 mm2 whichever is greater.

Therefore provide 20H25 (Asprov = 8380 mm2)

To determine the shear reinforcement, shear design should be carried out. The reinforcement provided should be used to check the pier against vehicle collision. Note that this accidental load case should be unfactored and considered to be acting alone.

Structural Stability of a Ten-Storey Braced Steel Frame

By
Ubani Obinna
-

Structural stability is broadly defined as the capacity of a structure to recover equilibrium. As a topic in structural engineering, it is concerned with structural members that are subjected to external loading that induces compressive stresses in the body of the structure. Emphasis is on understanding the behavior of structures in terms of load displacement characteristics; on formulation of the governing equations; and on calculation of the critical load. An approach based on continuum method has been presented in this article for the evaluation of critical load of a ten-storey braced steel frame.

This is based on the work of Zalka (2013) and the assumptions in the analysis are as follows;

  • The structure at least four storeys high with identical storey heights
  • The frame is regular in the sense that their characteristics do not vary over the height
  • Sway structures with built-in lower end at ground floor level and free upper end
  • The floor slabs have great in-plane and small out-of-plane stiffness
  • The deformations are small and the material of the structures is linearly elastic
  • P-delta effects are negligible
  • The frameworks are subjected to uniformly distributed vertical load at storey levels
  • The critical load defines the bifurcation point

Analysis Example

Calculate the critical load of the ten-storey steel frame work shown below. The height of each floor is 3m, and the properties of the members are given below. Take the modulus of elasticity of steel as 210,000 N/mm2.

Columns – UC 305 x 305 x 158 (Area = 201 cm2, Iyy = 38800 cm4)
Beams – UB 406 x 178 x 60 (Area = 76.5 cm2; Iyy = 21600 cm4)
Diagonal bracing = UA 100 x 100 x 10 (Area = 19.2 cm2; Iyy = 177 cm4)

MULTI STOREY BRACED STEEL WORK

The shear stiffness of the structure (for single braced frames) is shown below;

Bracing type

d = √(32 + 42) = 5 m
l = 4 m
h = 3 m
Ad = 19.2 cm2 = 19.2 x 10-4 m2
Ah = 76.5 cm2 = 76.5 x 10-4 m2
Eh = Ed = 210 x 106 kN/m2

K = {[53/(19.2 x 10-4 x 210 x 106 x 3 x 42)] + [4/(76.5 x 10-4 x 210 x 106 x 3)]}-1 = 137198.521 kN

The global second moment of area is;

Ig = ∑Ac,iti2 = 2 x (201 x 10-4 x 22) = 0.1608 m4

Load distribution factor rs is obtained from Table 1 as rs = 0.863.

Table 1: Load distribution factor rs as a function of n (the number of storeys) (Zalka, 2013)

Load distribution factor table

The global bending critical load is;

Ng = (7.837rsEIg)/H2 = (7.837 x 0.863 x 210 x 106 x 0.1608)/302 = 253760.179 kN

As a function of βs = K/Ng = 137198.521/253760.179 = 0.541

The critical load parameter αs is obtained by interpolating from the Table 2.
αs = 0.9

Table 2: Critical load parameter αs as a function of parameter βs (Zalka, 2013)

Critical load parameter 1

Finally, the critical load of the framework Ncr = αsK

Ncr = 0.9 x 137198.521 = 123485.25 kN

As a comparison, let us model the frame in Staad Pro software and carry out buckling analysis on the structure.

Staad Model

Based on the assumptions made in the analysis, the frame has been subjected to a load of 5 kN/m at each level. From the analysis result;

Total vertical load on the structure = 200 kN
Buckling amplification factor αs for Mode 1 = 687.439

Therefore, the critical buckling load Ncr = 200 x 687.439 = 137489 kN

Buckling mode 1

The difference obtained in the analysis result is 10.1%, but the continuum method appears to be more conservative than finite element analysis. According to Zalka (2013), the maximum error (difference obtained from finite element analysis result) expected from using this method is 17%.

References:
Zalka K.A. (2013): Structural Analysis of Regular Multi-storey Buildings. CRC Press Taylor and Francis Group

What is the immediate solution to these geohazards?

By
Ubani Obinna
-

Erosion, landslides, earthquakes, etc are identified are geohazards which normally require significant engineering efforts to put their effects under control. While the destructive effects of earthquakes can only be prevented by designing earthquake resistant structures, erosions and landslides are slightly unique.

erosion 1
erosion 3
ero

Depending on their stage of development, erosion control structures can be built to stop the expansion of gullies. This can accompanied by slope stability solutions, use of geogrids, geotextiles, etc. Since we are committed to learning and development at Structville, let us say that you are invited to site to offer solution to the problems shown in these pictures. Which solution will offer and what procedure will you follow?

Grand River

Human Factors in Civil Engineering Design and Construction

By
Ubani Obinna
-

Human factors or ergonomics is understood to be a body of knowledge that deals with the interaction of human beings with systems and devices taking into cognizance information from physiological and psychological characteristics. Human factors engineering can be seen as a process, as a body of knowledge, and/or as a discipline.

The primary aim of ergonomics is to minimise human error, reduce risks, enhance safety, and improve productivity when a human being is interacting with a system or using a device. For instance, in software engineering and application development, understanding the difference between usability, user interface (UI), and user experience (UX), and optimizing between them is an important factor in ergonomics. While this aspect of engineering is well studied in the field of software/mechanical/industrial/production engineering, it has not received significant attention in civil engineering designs.

In civil engineering and architecture, human beings interact with a building and the facilities provided in it. This also comes to bear in the usage of infrastructures such as bridges, walk ways, ramps, parks, and other public infrastructures designed for human use. It is therefore very important that the spaces and facilities in a building and infrastructures be optimised so that they will offer safety, comfort, and good experience to the end user – an aspect different from structural design.

Construction site safety and productivity of workers also comes to mind when we talk of ergonomics in civil engineering. This relates to the provision of adequate man space for working, having good work environment, optimised placement of scaffolds and platforms, motivating workers, lifting of weights, operation of construction machines etc. Human factor has been attributed as the cause of major construction accidents.

Apart from the aspects that are usually taken care of by mechanical and software engineers in the design of machines and tools, civil engineers should also watch out for the outcome of their own designs by adopting a human factor approach. In ergonomics, it is understood that engineers or designers should not rely on logic, intuition, or common sense in developing how humans interact with systems, but should use rigourous scientific methods. Human-system mismatches should be approached using methods that are well developed in behavioural sciences.

Human factors in Civil Engineering Designs

For instance, the image above shows a typical walkway design, and a completely different user experience. This aspect of design cannot be gotten right without ergonomics. In another instance, if pedestrian bridges are poorly positioned, humans will prefer to cross the busy highway instead of making use of it, thereby exposing themselves to avoidable hazards.

This type of information can therefore inform the design of walkways in streets such that it flows naturally to the pedestrian bridge without the user feeling that his time is being wasted. Therefore human performance monitoring, behaviour, and user experiences observed in many engineering designs should be developed into a framework that will be part of civil engineering designs.

avoiding use of pedestrian bridges

The major fields of research in human factors engineering are identified as physical, cognitive, and organisational. Physical ergonomics has to do with anatomy, physiological, and bio-mechanical characteristics relating with human beings and physical activities.

Cognitive ergonomics is concerned with mental processes, such as perception, memory, reasoning, and motor response, as they affect interactions among humans and other elements of a system, while organisation ergonomics has to do with the optimization of socio-technical systems, including their organizational structures, policies, and processes. All these should be incorporated into the framework for design of public infrastructures.

The Role of Haunches in Portal Frames

By
Ubani Obinna
-

Haunches are usually provided at the eaves of rigid portal frames due to a lot of beneficial reasons. They are usually cut from the same section as the rafter, or can be fabricated by welding different plates together. After fabrication, they are welded to the underside of the rafter at the eaves. Knowing that maximum bending moment occurs at the eaves of portal frames, haunches play the major function of increasing the bending resistance of the rafter.

Read Also…
Structural analysis of portal frames subjected to gravity load
Preliminary plastic analysis of portal frames

A portal frame is a continuous frame with moment-resisting connections. If the connection between the column and the rafter is not rigid, the frame will be unstable in-plane. The continuous nature of the frame provides in-plane stability, and resistance to lateral loads such as wind load. As a result, the stability of the frame and its resistance to deformation depends on the stiffness of the columns and rafters, which are the primary members of the frame. Hot rolled steel sections are normally used as the primary members, with the resistance of the rafters enhanced locally by a haunch at the eaves, where the bending moments are greatest. Usually, the frame is assumed to have nominally pinned bases, even if the actual base details possess appreciable stiffness. The main frames are generally fabricated from UKB sections. The eaves haunch is created by adding a tapered length cut from a rolled section, or fabricated from plate.

Read Also …
Design of bolted beam splice connections to Eurocode 3
Design of roof purlins

Consequently, haunches enhance the economical design of rafters of portal frames, by reducing the depth of the section required. Haunches also assist in the reduction of deflection and facilitate efficient rigid bolted connection between the rafter and the column. The typical length of the eaves haunch is generally 10% of the frame span. When this is done, the value of the hogging moment at the end of the haunch will be approximately equal to the sagging moment in the rafter, which can lead to a better design of the portal frame. Note that the depth of haunch below the rafter is approximately equal to the depth of the rafter section.

Recommended length of haunches in a portal frame
Recommended length of haunches in a portal frame

The apex haunch may be cut from a rolled section – often from the same size as the rafter, or fabricated from plate. This member is not usually modelled in the frame analysis and is not needed to enhance the bending resistance; it is only used to facilitate a bolted connection.

Engineers React to Large Span Cantilever Slab Construction

By
Ivy Ugorji
-

A structural engineer posted a picture of a large cantilever slab construction on LinkedIn, and so many reactions have followed the picture due to its size and dimensions. Cantilever structures are generally structures that are fixed (or continuous) at one end, and free (unsupported at the other end). As a matter of fact, they are one of the most challenging structures to design given that the bending moment increases with the square of the length, while the deflection increases to the fourth power of the length. As a result, controlling deflection and dealing with stability of large span cantilever structural systems is always a challenge in design.

Read Also…
Design of large span cantilever structures
Structural design of cantilever slabs

We have posted the comments from various professionals on the post to enable us review, scrutinize, and assess the possibility of designing and constructing such type of structure. However the exact span, design data, calculation sheet, and drawings of the structure were not made available, but a lot of people still managed to lend their voices to the construction.

Let us look at their responses below;

Cantilever slab

(1) A senior engineer from New Delhi area of India doubted the stability of the structure citing inadequate support conditions. He suggested that cables could have been used to support the slab further. According to him,

Doesn’t look like it is safe. The slab is too thick whereas the supporting structures are too thin. Also, normally such thick slabs are designed as flat slabs but I don’t see the flat slab connection type being followed at the support junctions. It looks like to take care of the cantilever load and moments coming on the slab the thickness has been increased. Hopefully the designer has checked all the loads and moments properly otherwise it is going to fail in the long run. The best option would have been to provide some sort of pulling arrangements at the top corner of the slab via cables or pin jointed stereo frames at the outmost corner where there is no support from the bottom of the slab and that way the designer would have succeeded in decreasing the slab thickness as well as given an aesthetic look to it.

(2) Another structural engineer from India attributed it to the power of post tensioned concrete design as he wrote;

Power of PT Design

(3) A registered Professional Engineer (P.E) from the USA commented that the loading on the slab may not be high enough to cause serious problems. According to him,

There are some issues, but if the math and science can be proved it can be done. Usually in situations like this, the loading is majorly under developed so when you compare the loading to the strength, it will be sufficient. But in reality the loading is not high enough. Expect 250 psf live load to 500 psf live load and use a heavy concrete weight with 2000 pci strength or 1500 pci strength. They tend to screw up the mix design on these.

(4) A retired engineer from India wrote;

This is possible. The slab must have been designed as two adjecent edges supported slab. I did come across such a design in early stages of my career as structural engineer

(5) A registered civil engineer from Nigeria expressed doubts over the safety of the slab and said,

The span of this cantilever slab seems to be much, and may be unsafe.

(6) A senior structural engineer (M.Tech) from India also considered the safety of the structure to be doubtful due to the lack of a back span and said,

That seems to be a dangerous cantilever. I don’t see a continuous backing portion either. I hope it must have been designed and checked well.

(7) The idea of lack of back-span was also raised by another engineer who said that it does not matter whether the slab was post tensioned or not. According to him;

No back span, seems to be very dangerous. Whether it is PT or NonPT the cantilever behaviour is same. Looks very huge self weight due to higher thickness

(8) A civil engineer from Nigeria pointed out that a blend of lightweight materials on a steel framing can be used to achieve the structural system. According to him,

It depends on the material used to achieve the cantilever. POP or any lightweight material can be used to achieve this cantilever then the frame can be metal. The rest will be a good aesthetic work to produce a good look. No load should be on the artificial slab.

However, materials like POP may not be adequate or durable given the exposure of the structure. However, the use of polystyrene can work, after which it will be screeded with mortar to give a finished appearance of concrete. Polystyrene has be extensively used for doing concrete fascia and other building elements.

(9) Another Engineer pointed out that the image could be edited, because the support system provided for the structure (walls and columns from the picture) looked too small to support the massive weight of the cantilever structure. According to him,

Looks to me like a edited image of the structure but if it’s true I don’t know how this structure is safe. Supports looks weaker than the heavy cantilever slab resting on it.


(10) Another engineer who is rounding up his PhD in Spain pointed out that the structure must be post tensioned, however, it seems unsafe especially in the presence of any seismic event. According to him;

Post-tension! But even if it’s not behaving similarly in cantilever spans. The thickness! Huge Self weight! with moderate seismic load or any lateral load, it will make severe damages! Please, treat Infrastructure in high responsibilities.

(11) An independent Structural Engineer with over 39 years of experience from India suggested that the most important thing is to balance the weight of the structure and the moment generated at the supports. According to him,

Main thing is to see the balancing weight and Moment at the support and the section of slab should be with max. thickness at supports and minimum thickness at free end , we have already designed and constructed a stadium at Dehradun in 2001 & standing safe till date.

However, in as much as the exact details of the structure are not available, what do you think?

Understanding Human-Structure Interaction

By
Ubani Obinna
-

Recent researches have shown that human beings interact with structural systems undergoing vibration. Initially it was thought that human body is an inert mass during vibration of structures, but studies have proved otherwise. Human-structure interaction is a study which describes the independent human system and structural system working together as a whole. It also studies the vibration of structural systems where people are involved, and how human body responds to structural vibration.

Vibration serviceability issues are becoming increasingly important in structures subjected to crowd action such as foot bridges, galleries, stadiums, places of assembly etc. This is important given the increasing use of slender and aesthetically pleasing elements in structures. It is understood that the excess lateral vibration of the Millennium bridge which caused the bridge to ‘wobble’ on its opening day gave the impetus for researches on how human beings interact with structural vibration.

According to B.R. Ellis [1] who researched on human-structure interaction in the year 1997;

It has occasionally been assumed that the effect of the people is simply that of an added mass on the system, however, both
site and laboratory tests demonstrate that human bodies do not act solely as mass on the structure and show that the problem is somewhat more complex

In an experiment reported by Ji and Ellis [2], the frequency of a beam was tested when it was empty, and compared with when someone was standing on it. The frequency of the bare beam was observed to be 18.68 Hz, while the frequency of the occupied beam was observed to be 20.02 Hz. If the human body acted as an inert mass, the frequency of the human occupied beam would be smaller rather than bigger. It was therefore observed that the human body did not act as an inert mass but as a mass-spring-damper system. This particular discovery formed the basis for the new topic called human-structure interaction.

Modelling human-structure interaction

Two approaches are usually adopted when studying the dynamic property of a human body based on the bio-mechanical method. One approach is to develop the part-body model and the other approach is to develop the whole-body model [3]. In the body part model, the attention is on the local dynamic properties of body parts such as the hands, heads, or the vibration of the soft tissues, under the assumption that the forces acting on the model are already known. However, the whole-body model is concerned with the holistic dynamic property of the body (for example a human standing on a beam subjected to external excitation). In this case, the force acting on the body is unknown in advance and the interaction between the body and the structure occurs [3].

Instead of the single degree of freedom (S-DOF models that have been extensively investigated, it is recommended that human-structure interaction be modelled with two degrees of freedom (2-DOF), where one degree of freedom is for the structure, and the other for the human being. To develop the two degree of freedom system, one approach is to model two separate single degree of freedom systems (one for the structure and one for the human being), and join the two models together to form a 2-DOF. This method is called separative modelling, and has the advantage of being easily reproducible in the laboratory for experiment [3]. On the other hand, when the system is modelled inseparably as a whole, it called integrative modelling.

human structure interaction
Fig. 1. The separative modelling of human-structure interaction. (a) Structure model, (b) Body model, (c) Coupled model [3]
Integrative modelling of human body
Fig. 2. The integrative modelling of human-structure interaction. (a) Body on SDOF structure, (b) Coupled model [3]

In a parametric research work that was carried out to compare the two models, the natural frequency, damped natural frequency and decrement coefficient of the first mode from the integrative model were found to be smaller than those from the separative model. The authors further concluded that integrative model is more reasonable than the conventional separative model, stating that the separative model can introduce up to 20% error in the calculation of the natural frequency [3].

Applications of human-structure interaction

(1) Prediction of human body response to structural vibration
(2) Evaluate structural vibration where people are involved
(3) Identification of some dynamic characteristics of the human body
(4) Reduction of human induced structural vibrations
(5) Improvement of human comfort in a working environment
(6) Improvement of animal welfare and meat quality

References
[1] Ellis B. R. (1997): Human-structure interactions in vertical vibrations. Proceedings to Institution of Civil Engineers, Structures and Buildings (1997)122 Panel Paper 11023
[2] T. Ji, Ellia B.R: Human-Structure Interaction
[3] Zhou D., Han H., T. Ji, Xu X. (2016): Comparison of two models for human-structure interaction. Applied Mathematical Modelling 40(2016): 3738 – 3748 http://dx.doi.org/10.1016/j.apm.2015.10.049

How to Plaster a Block Wall

By
Ubani Obinna
-

Plastering is the process of covering a wall surface with a thin layer of mortar for protection and decorative purposes. It is usually the first stage in the finishing works of a building, and it is categorised under ‘wet works’. The mortar for plaster works comprises of a mixture of fine sand, cement and water. The requirements for mortar and rendering is given in BS EN 998-1:2003.

Download…
A guide to BS EN 998-1 and BS EN 998-2
How to determine the quantity of mortar for laying block

Plastering of a wall is a skilled labour, and requires the services of an experienced mason. Poorly done plastering work can ruin the aesthetics of a building, and lead to high repair/maintenance costs. Plaster surfaces can be made relatively smooth or rough, depending on the next stage finishes to be applied. The typical thickness of plasters ranges from 12 mm to 15 mm. If the thickness of plaster exceeds 15 mm, it will be important to plaster in two coats. Some common plaster defects are:

(1) Cracks
(2) Spalling/falling off of plasters from the wall
(3) Undulating surfaces
(4) Efflorescence
(5) Popping etc

In order to carry out a good plastering job, here are some important steps to follow especially where accuracy is important.

(1) Check for straightness of the wall

Before the commencement of any plastering work, it is important to be very familiar with the wall to be plastered by carrying out simple checks. It is important to check the edges of the wall for squareness, and the length and height for perfect horizontality and verticality. If the walls are in conformity with the construction drawings, then there is no problem. If there are deviations, then know that the thickness of the plastering work will be affected, and that will affect the cost and duration of the work. A decision will have to be taken before the next stage of the plastering work can commence.

check for straightness

(2) Preparation of the surface

It is a good practice to lay blockwork with recessed joints to receive plaster, but this is not always the case. Either way, the walls shall be brushed clean of all dust, thoroughly wetted and surface dried before plaster is applied. If there are bonding agents specified by consultants or manufacturers, then it shall be applied according to the instructions before the plastering can commence. All put-log-holes (i.e. holes left for scaffolding) shall be properly filled in advance of the plastering.

(3) Installation of gauges

To ensure an even thickness and a true surface, gauges of plaster 15 mm x 15 mm, or broken clay tiles set in mortar shall be first established on the entire surface at about 2 metre intervals both vertically and horizontally. The thickness of the plaster specified excludes the key i.e. the grooves or open joints in the block work (if any). The minimum thickness of the plaster over any portion of the surface shall not vary from the specified thickness by more than 3 mm. However, if the wall is not straight, there will be very large variations in the thickness of the plaster. That is why step one is very important.

(4) Material

The specified mortar material in the drawing should be followed to the later. The mix ratio often recommended is 1:3, which means that one bag of 50 kg cement to 6 head pans of plaster sand should be used.

(5) Application of plaster

Plastering of walls shall commence after completion of ceiling plastering if any. The plastering shall be started from the top and worked down towards the floor. Mortar shall be applied between the gauges to slightly more than the require thickness. Furthermore, the plaster shall be well pressed into the joints, levelled and brought to a true surface by working on a straight edge (range) reaching across gauges, with small upward and sideways movement. Finally the surface shall be finished true with a wood float or trowel according to the type of finish required.

If a sandy granular texture is needed (for example plastering to receive wall tiles), the surface shall be wood floated. If a smooth finish is needed, trowelling shall be done to the extent required to produce the desired smoothness.

(6) Finishes

The plaster shall be finished to a true and plumb surface and to the degree of smoothness required. The work shall be tested frequently as the work proceeds with a true straight edge not less than 2.5m long and with plumb bobs. The gap between the straight edge and any point on the plastered surface shall not exceed 3 mm. All horizontal lines and surface shall be tested with a level and all jambs and corners with a plumb bob as the work proceeds. At corners, plastic edge trims (corner angle beads) could be used to avoid wavy edges.

corner bead

(7) Stopping work at the end of the day

In stopping work at the end of the day, the plaster shall be left cut clean to line both horizontally and vertically. When recommencing the plastering, the edge of the old work shall be scraped, cleaned and wetted before plaster is applied to the adjacent areas, to enable the two to be properly jointed together. Plastering work should be closed at the end of the day on the body of the wall not nearer than 150 mm to any corners or arrises. Horizontal joints in plaster work shall not be formed on parapet tops and copings, as these invariably lead to leakages.

(8) Curing

Curing shall be started 24 hours after finishing the plaster. The plaster shall be kept wet for a period of seven days. During this period it shall be suitably protected for all damages, at the contractor’s expense.

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