Shear walls are structural elements usually employed in tall buildings to assist in resisting lateral loads. Shear walls can be solid or pierced (coupled), depending on their location in the building. In the design of tall buildings, structural engineers normally throw the entire lateral load (say wind action) to the shear walls, which means that the columns will not be relied on for lateral stability. In a more practical scenario however, the shear walls and columns interact in resisting lateral loads, which can be taken into account.
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In this post, we are going to review how shear walls can be modelled on Staad Pro software and consider the various accuracy levels with each method.
Let us consider the shear wall shown below;
Thickness of the wall = 250 mm
The wall is subjected to a lateral load of 300 kN at the top.
Model 1: Equivalent frame model
In this model, we represent the shear wall as an equivalent frame with representative stiffness for the various elements.
On modelling this Staad Pro, we obtained the result shown below;
The maximum horizontal deflection was obtained to be 158.376 mm.
Model 2: Plate modelling (Finite Element Mesh)
In this model, the shear wall is modelled as a full plate with approximate mesh size of (4 x 4) per square metre. The deflection due to the lateral load is given below;
The maximum horizontal deflection was obtained to be 156.588 mm.
It is important also to compare the deflection obtained by considering different mesh sizes in order to compare the accuracy of utilizing the equivalent frame model. Thank you for visiting Structville today, and God bless.
When structural members are subjected to compressive forces, the members may fail before the compressive resistance (A.fy) is reached. This premature failure is usually caused by secondary bending effects such as imperfections, the eccentricity of loading, asymmetry of the cross-section etc. In such cases, the failure mode is normally buckling.
The buckling of columns is usually associated with a sudden large displacement (bending) on the column following a slight increase or modification of an existing compressive load. The load at which the buckling of columns occur is usually less than the compressive resistance of the column, and it is mainly influenced by the length, cross-section, and boundary conditions of the column.
This is unlike when the member is subjected to tensile forces, where the member will generally fail when the stress in the cross-section exceeds the ultimate strength of the material. Members subjected to tensile forces are inherently stable.
In essence, for very short columns or members subjected to tensile forces, the stress is proportional to the load, and the failure load is equal to the yield stress times the area of the cross-section (A.fy). Commonly, compression members can be classified as short, intermediate, or slender depending on their slenderness ratio. Short elements will fail by crushing, slender elements will fail by excessive lateral deflection, and intermediate elements will fail by the combination of both.
In this article, we are going to focus on flexural buckling.
In the year 1757, Leonhard Euler developed a theoretical basis for the analysis of premature failure due to buckling. The theory was based on the differential equation of elastic bending of a pin-ended column, which related the applied bending moment to the curvature along the length of the column.
Note that L in this case is the effective buckling length which depends on the buckling length between the pinned supports or the points of contraflexure for members with other boundary conditions.
Subsequently, the Perry-Robertson formula was developed to take into account the deficiencies of the Euler method. The formula evolved from the assumption that all practical imperfections could be represented by a hypothetical initial curvature of the column. In the UK, the Perry-Robertson formula was modified and is referred to in BS 5950 as the Perry-Strut formula. As a matter of fact, it also forms the basis of flexural buckling in BS EN 1993 (Eurocode 3).
Worked Exampleon the Buckling of Columns
In the example below, we are going to show how the failure load of a column changes with an increase in effective length due to the phenomenon of buckling.
Let us investigate the load-carrying capacity of UC 305 x 305 x 158 according to EC3. We will consider the ends of the column to be pinned, and we will start with an initial length of 1.0 m.
Properties of UC 305 × 305 × 158
D = 327.1 mm; B = 311.2 mm; tf = 25mm; tw = 15.8mm; r = 15.2mm; d = 246.7mm, b/tf = 6.22; d/tw = 15.6; Izz = 38800 cm4; Iyy = 12600 cm4; iy-y = 13.9cm; iz-z = 7.9 cm, A = 201 cm2
Thickness of flange tf = 25mm. Since tf > 16mm, Design yield strength fy = 265 N/mm2 (Table 3.1 EC3)
Section classification
ε = √(235/fy ) = √(235/265) = 0.942
We can calculate the outstand of the flange (flange under compression) C = (b – tw – 2r) / (2 ) = (327.1 – 15.8 -2(15.2)) / (2 ) = 140.45mm. We can then verify that C/tf = 140.45/25 = 5.618 5.618 < 9ε i.e. 5.618 < 8.478.
Therefore the flange is class 1 plastic
Web (Internal compression) d/tw = 15.6 < 33ε so that 15.6 < 31.088. Therefore the web is also class 1 plastic Resistance of the member to uniform compression (clause 6.2.4) NC,Rd = (A.Fy)/γmo = (201 × 102 × 265) / 1.0 = 5326500 N = 5326.5 kN
Buckling resistance of member (clause 6.3.1) Since the member is pinned at both ends, the critical buckling length is the same for all axis Lcr = 1000 mm
In the major axis (¯λy ) = 1000/(139 × 88.454) = 0.081 In the minor axis (¯λz ) = 1000/(79 × 88.454) = 0.143
Check h/b ratio = 327.1/311.2 = 1.0510 < 1.2, and tf < 100 mm (Table 6.2 EN 1993-1-1:2005)
Therefore buckling curve b is appropriate for the y-y axis, and buckling curve c for the z-z axis. The imperfection factor for buckling curve b, α = 0.34 and curve c, α = 0.49 (Table 6.1)
This tells us that the column is short and that at a length of 1.0 m, the column will fail by crushing. Buckling is not a significant mode of failure. If we incrementally increase the length of the column by 1.0 m and follow the steps described above, we will arrive at the following results;
Column length (m)
Failure Load (kN)
1
5326
2
3090
3
4700
4
4270
5
3810
6
3330
The graph of the relationship is given below;
Therefore, the load-carrying capacity of compression members reduces with an increase in length. When the height of the pin-ended column was increased from 1m to 6m, a reduction of about 37% was observed in the buckling load. In the design of axially loaded members such as trusses, it is, therefore, advisable to design the members in such a way that the longer members are in tension and the shorter members are in compression.
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Cantilever slabs are common features in residential and commercial buildings due to the need to have bigger spaces on upper floors. To achieve this, architects normally extend the slab beyond the ground floor building line, thereby forming a cantilever. Cantilever slabs in reinforced concrete buildings are usually characterised by a reinforced concrete slab projecting from the face of the wall and having a back span that extends into the interior panels of the building.
The design of reinforced concrete cantilever slab is usually governed by serviceability limit state requirements. Typically, the design involves the provision of adequate concrete thickness and reinforcement to prevent excessive deflection and vibration.
During the detailing of reinforced concrete cantilever slabs, the main reinforcements are provided at the top and they extend into the back span by at least 1.5 times the length of the cantilever or 0.3 times the length of the back span, whichever is greater. Furthermore, it is important to provide at least 50% of the reinforcement provided at the top, and at the bottom.
In this article, we are going to show how we can analyse and design cantilever slabs subjected to floor load and block wall load according to the requirements of EN 1992-1-1:2004 (Eurocode 2).
Cantilever slab supporting block wall load
Worked Example – Design of Cantilever Slabs
A cantilever slab 200 mm thick is 1.715m long, and it is supporting a blockwork load at 1.0m from the fixed end. Design the slab using the data given below;
Purpose of building – Residential fck = 25 Mpa fyk = 460 Mpa Concrete cover = 25 mm Height of block wall = 2.75 m Unit weight of concrete = 25 kN/m3 Unit weight of block with renderings = 3.75 kN/m2
Engineers, site managers, and quantity surveyors are always faced with the challenge of specifying as accurately as possible, the quantity of materials needed to execute a specific item of work. In this article, we are going to explain how you can estimate the quantity of mortar (cement and sand) needed to lay blocks per square metre of wall. It is well known that we need about 10 blocks to build one square metre of wall.
Simple Proof; Area of wall = 1 m2 Planar area of one standard block (Nigeria) = 450mm x 225mm = 101250 mm2 = 0.10125 m2 Therefore, the number of blocks required (disregarding mortar) = 1/0.10125 = 9.87 (say 10 blocks)
Now, how do we estimate the quantity of mortar needed to lay the blocks? First of all, let us look at the dimensions of a typical 9 inches block with holes (the most popular block for building in Nigeria).
Typical dimensions for sandcrete blocks
From the image of the sandcrete, we can say that the cross-sectional area of the block that receives horizontal mortar is;
Ab = (0.45 x 0.225) – 2(0.15 x 0.125) = 0.0637 m2
Now, let us assume that the mortar will be 25 mm (1 inch) thick.
The typical arrangement of blocks in a one square metre wall is shown below;
Therefore, we can estimate the volume of mortar required to build one square metre (1 m2) of wall as follows;
Vertical mortar = 8 x (0.025 x 0.225 x 0.275) = 0.0123 m3 Horizontal mortar = 10 x (0.0637 x 0.025) = 0.0159 m3 Total = 0.0282 m3
Therefore the volume of mortar required to lay one square metre (1 m2) of 9-inch block (with holes) can be taken as 0.03 m3 for all practical purposes.
Cement requirement The typical mix ratio for mortar (laying of blocks) is 1:6 Quantity of cement required = 1/7 x 1440 kg/m3 = 205.71 kg Making allowance for shrinkage between fresh and wet concrete = 1.54 x 205.71 = 316.79 kg Quantity of cement required in bags = 316.79/50 = 6.33 bags/m3 of mortar
Therefore, the quantity of cement required to lay 1m2 of wall = 0.03 × 6.33 = 0.1899 bags If 0.1899 bags of cement = 10 blocks Therefore, 1 bag of cement = 52 blocks
Therefore, 1 bag of cement is required to lay 52 blocks using a 1:6 mix ratio of mortar.
Sand Quantity of sand required = 6/7 x 1600 kg/m3 = 1371.428 kg Making allowance for shrinkage between fresh and wet concrete = 1.54 x 1371.428 = 2112 kg Quantity of sand required in tonnes = 2.112 tonnes/m3 of mortar
Therefore, the quantity of sand for mortar required to lay 1m2 of wall = 0.03 × 2.112 = 0.06336 tonnes of sand If 0.06336 tonnes of sand = 10 blocks Therefore, 1 tonne of sand = 157 blocks
Therefore, 1 tonne (1000 kg) of sand is required to lay 157 blocks using a 1:6 mix ratio of mortar.
Further Example To go further, let us assume that we have 150 m2 (about 1500 blocks) of wall and want to estimate the quantity of cement and sand needed to lay the blocks.
The volume of mortar required = 0.03 x 150 = 4.5 m3
For 4.5m3 of mortar; Provide = 4.5 x 6.33 = 29 bags of cement
For 4.5m3 of mortar; Provide = 4.5 x 2112 = 9504 kg of sharp sand
Therefore you need about 10 tonnes of sharp sand and 29 bags of cement for laying 150m2 of 9 inches block wall.
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Pedestrian bridges (footbridges) are structures designed to enable human beings cross over obstacles such as busy highways, water bodies, gullies, etc. There are several variations of foot bridges based on structural configuration and materials. Modern footbridges are increasingly becoming elements of street beautification, with a view on sustainability and environmental friendliness. In this post, a simple pedestrian bridge has been modelled on Staad Pro software, and the result of internal stresses due to crowd load on the bridge presented. Actions on Bridges
The actions on footbridges is explicitly covered in section 5 of EN 1991-2. The load models presented in the code and their representative values (which include dynamic amplification effects), should be used for ultimate and serviceability limit state static calculations. However, adhoc studies are required when vibration assessment based on specific dynamic analysis is necessary.
There are three mutually exclusive envisaged vertical load models for footbridges (5.3.1(2) EN 1991-2:2003);
(1) A uniformly distributed load representing static effect of a dense crowd,
(2) One concentrated load representing the effect of maintenance load,
(3) One or more, mutually exclusive standard vehicles to be taken into account when maintenance or emergency vehicles are expected to cross the footbridge itself (not applicable in this post).
The effect of crowd action on a footbridge is represented by a uniformly distributed load which depends on the loaded length of the footbridge. However, when a road bridge is supporting a footway, a UDL value of 5.0 kN/m2 is recommended by the code for that section. This value is synonymous with continuous dense crowd action which is given by Load Model 4 of road bridges. However, when there is no such risk of dense crowd action, the load on the pedestrian bridge is given by;
qfk = 2.5 kN/m2 ≤ 120/(L + 30) ≤ 5.0 kN/m2
For local assessment, a concentrated load of 10 kN is is considered to be acting on a square surface of sides 0.1 m. This load is not to be combined with other variable non-traffic load.
For horizontal forces, a horizontal force acts simultaneously with the corresponding vertical load whose characteristic value is equal to 10% of the total load corresponding to the uniformly distributed load. This horizontal force does not coexist with the concentrated load, and acts along the bridge deck axis at the pavement level on square surface of sides 0.1 m. This force is usually sufficient to ensure the horizontal longitudinal stability of the bridge.
However, there are non-traffic actions that are possible on footbridges such as thermal action, wind action, snow action, indirect actions such as support settlement, etc. It is also important to consider accidental actions on the bridge especially vehicle collision on the substructure of the bridge. For stiff piers, EN 1991-2:2003 recommends a minimum force of 1000 in the direction of vehicle travel or 500 kN in the perpendicular direction. These collision forces are supposed to act at 1.25 m above the level of the ground surface.
N/B: Vehicle collision on the piers is more common on footbridges than highway bridges. Therefore, it is highly recommended that the piers of the footbridge be protected using any reasonable and effective approach. Road restraint system can be installed at a distance from the piers. The deck should be high enough to prevent collision also.
Analysed Example
A pedestrian bridge with the following configuration is shown below;
With these dimensions shown above, the pedestrian bridge has been modelled on Staad Pro…
For the sake of clarity, the cross-sectional dimensions of the members are as follows;
Main longitudinal girders – 400 x 750 mm
Transverse girders at the supports – 400 x 750 mm
Intermediate transverse girders – 250 x 450mm
All columns – 400 x 400 mm
Thickness of slab – 200 mm
The self weight of the structure will be calculated automatically by Staad Pro. The pedestrian crowd action has been calculated as;
qfk = 120/(27 + 30) = 2.10 kN/m2 ≤ 2.5 kN/m2
But let us envisage that there is a possibility of continuous dense crowd action (e.g the pedestrian bridge at Ojota, Lagos, or lets say the pedestrian bridge will be located as Oshodi Bus Stop, Lagos). In this case, it is very reasonable to take qfk as 5.0 kN/m2
The total force due to UDL on the bridge can be taken as 5 kN/m2 x 2m x 27m = 270 kN
Therefore, an equivalent horizontal force that can be applied on the bridge deck is 27 kN
The images below show the internal stresses diagram due to crowd load;
The maximum moment at the intermediate supports of the main longitudinal beam due to characteristic crowd load is 62.7 kNm.
The other internal forces diagrams due to crowd load alone are shown below;
So we are going to stop here for now. Remember that you have to consider other load cases and combinations in order to arrive at the design moment.
The modelling, analysis, and drawings you see on this post was produced by Ubani Obinna for Structville Integrated Services. We are creative, we think critically, we research, and we solve problems. You can trust us with yours, so feel free to contact us. WhatsApp: +2347053638996 E-mail: ubani@structville.com
In construction, there always comes a time when there is a need to bond hardened concrete (substrate) with fresh concrete topping/overlay. The aim of this post is to explain how to bond old and fresh concrete successfully. Furthermore, we will review the strength of the interfacial bond between old and new concrete based on laboratory studies.
Proper bonding is very important for adequate performance when the fresh concrete topping is used to overlay an existing hardened concrete. This construction feature is usually found in bridge deck construction, concrete pavements, precast filigree slab, pile caps (in some cases) etc. Proper bonding between the substrate and the topping is not always guaranteed unless simple precautions are taken.
For adequate bonding, it is very important to prepare the surface of the substrate adequately. The preparation of the surface usually involves roughening the surface, and removal of all dirt, oil, grease, and loosened or unbonded portions of the existing concrete.
By implication, the surface of the substrate should be hard, firm, clean, and free from loosened particles. This can be achieved by the use of chipping hammers, wire brushing the surface etc. After this is done, the exposed concrete surface can be cleaned by using pressurised clean water, air, etc. The man hours involved depend on the area of the surface, location, and the ease of cleaning (e.g reinforcement obstruction).
After surface preparation, there is usually a need to apply a bonding agent on the surface of the existing concrete in order to facilitate the bonding. Epoxy-based bonding agents are very popular for such operations. It is recommended that a bonding agent is applied prior to casting the fresh concrete. In essence, the procedure should be ‘wet-to-wet’ as the bonding agent should not be allowed to dry before the fresh concrete topping is placed.
Hardened Concrete With Bonding Agent Ready for Topping/Overlay
In a research carried out by Vandhiyan and Kathiravan (2017), the compressive strength of monolithic and bonded concrete was compared using 150mm x 150mm cube specimens at 28 days. With an epoxy-based bonding agent, the compressive strength of the bonded concrete was about 5% less than the monolithic strength, while without the bonding agent, the compressive strength was about 28% less than the monolithic compressive strength.
Research has also shown that the moisture condition of the substrate affects the shear bond strength of bonded concrete. Shin and Wan (2010) investigated the interfacial bond strength of old and new concrete considering saturated surface dry (SSD) and air dry conditions. Saturated surface dry is a condition in which the concrete contains moisture that is equal to its potential absorption, without the surface being wet or damp.
At a water/cement ratio of 0.45 (for the topping concrete), the shear bond strength at the interface was about 44% greater when the substrate was at SSD condition than when it was air dry. At a water/cement ratio of 0.6 for the topping layer, an increase in shear bond strength was recorded, but there was a reduction in the compressive strength of the concrete.
So the recommendation in this article is that when casting a topping layer of fresh concrete on old concrete, adhere to the following guidelines;
(1) Prepare the surface properly (2) Make sure that the substrate is at saturated surface dry condition (3) Use a bonding agent and follow the manufacturer’s technical recommendation properly.
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References Vandhiyan R., Kathiravan M. (2017): Effect Of Bonding Chemical On Bond Strength Between Old And New Concrete. SSRG International Journal of Civil Engineering- (ICRTCETM-2017) – Special Issue – April 2017 ISSN : 2348 – 8352 pp 129-134
H-C. Shin, Z. Wan (2010): Interfacial shear bond strength between old and new concrete. Fracture Mechanics of Concrete and Concrete Structures – Assessment, Durability, Monitoring and Retrofitting of Concrete Structures- B. H. Oh, et al. (eds) ⓒ 2010 Korea Concrete Institute, Seoul, ISBN 978-89-5708-181-5 pp 1195 – 1200
In the month of May, we announced the commencement of Structville Design Competition, where civil engineering students and serving NYSC members from various universities and polytechnics in Nigeria competed for small tokens in the design of reinforced concrete structures. The exercise was aimed at developing the interest of students in Structural Design, and preparing them for excellence in the field of structural engineering. You can view the details of the competition below; Structville Design Competition for Students To view the result sheet of the competition, click the link below; Structville Design Challenge Results Let us meet the outstanding performers…… 1st Position – USMAN UMAR (Ahmadu Bello University, Zaria)
Usman Umar is a 500 level student from the Department of Civil Engineering, Faculty of Engineering, Ahmadu Bello University, Zaria, and he’s an indigene of Ankpa LGA, Kogi State, Nigeria. While speaking to Structville during an interview, Usman dreams of becoming a structural engineer with vast experience in both consultancy and construction field, and he is quite willing to work with any consultancy or construction firm. On what motivated him to study civil engineering, he said,
I grew up with the curiosity of learning the principles which guide the operations of machines and systems. By the time I was in secondary school I already knew I had to go for studying an engineering course. Although I almost opted for electrical engineering while preparing for UTME, the awesomeness in the the construction of structures such as bridges and skyscrapers couldn’t let me. So I went for Civil Engineering.
Usman believes that a lot is being taught in Nigerian classrooms, but the technical know-how and the method of teaching adopted by a lecturer determine how much is impacted unto the students. Therefore, he believes that the Nigerian tertiary institutions are doing fairly good, but a lot more should be invested in order to raise the standard higher. Usman revealed that he has no definite studying pattern, but makes sure that he takes his courseworks seriously, and by consulting a lot of design textbooks and studying the architectural drawings carefully, he was able to do well in the Structville design competition. So we all at Structville say a big congratulations to Usman Umar once again. 2nd Position – Ogungbire Adedolapo (Osun State University, Osogbo)
Ogungbire Adedolapo Mojed, is a final year Civil Engineering student at Osun State University, Osogbo. He hails from Odo-Otin Local Government, Osun state. It has always been his childhood dream to grow up to become a Civil Engineer. Adedolapo would like to own his own consultancy firm one day, and he believes that much experience will be needed in order to achieve that. He is aspiring to work with experienced Engineers and gain as much relevant experience as he can after graduating from the university. He is currently on the lookout for opportunities for pre-NYSC internship, and will be rounding off his degree programme by September 2018. On the state of teaching and learning in Nigeria, he is an advocate of more practical approach to learning. According to him;
I believe a more practical approach will be more welcomed for training Civil Engineering students in the country. We should not only be bound to the theoretical examples that we have in textbooks but relate them to practical applications in Nigeria.
Adedolapo is curious to learn new things, and he flying high academically in his school. He said there was no special approach to the design competition on his own part. As an individual, he prefers self-studying and understanding the concepts behind what he is studying. That has helped him come a long way. Congratulations Adedolapo once again. 3rd Position – Olajide Bukoye (Federal Polytechnic Offa, Kwara State)
Bukoye, Issa Abiola is a graduate of Federal Polytechnic, Offa in Kwara State. He is currently observing his mandatory National Service in Oyo state. Bukoye boasts of some years of field experience field experience and and have executed some practical designs in his career so far. According to him,
Civil Engineering has always being my childhood dream not because it’s a lucrative job but because I want to make a difference. 70% of science students in high school wants to be a Doctor, but I’ve never for once dreamt of that because I’m good with numbers and I don’t want to waste the talent. I guess Civil Engineering has always been in my blood.
Bukoye believes that the quality of teaching in Nigeria is far from standard, but he is optimistic and hopeful of positive turnarounds. If he were to make changes in Nigeria, he would focus on the education system.
Bukoye is a goal driven person, and loves to take on any challenges because he believes that every challenge is an experience and once done and dusted, it goes to one’s achievements archive. After his NYSC program, he plans to secure a job and as well go back to school not just for the certificates but to learn more in the field of civil/structural engineering.
Ground beams are employed in reinforced concrete substructures for a lot of reasons. They can be differentiated from plinth beams due to a slight variation in the purpose of their construction. Plinth beams are used to connect (chain) separate pad bases together, and blockwork can be built off the plinth beams. On the other hand, ground beams are designed mainly for the purpose of receiving load from the ground floor slab or raft, alongside other functions as envisaged by the designer.
In the design of beam and raft foundation, ground beams receive ground floor/raft slab pressure loads. These loads could be from earth pressure reactions or occupancy loads/dead loads.
Let us consider two cases as shown below:
Case 1: Raft Slab on Ground Beam Foundation
Fig 1: A typical section of a ground beam supporting a raft foundation
In this case, as shown in Fig 1, a raft foundation supported on ground beams is used to overcome the low bearing capacity of the soil. This is the cheapest alternative for constructing foundations on weak soils carrying relatively low/moderate superstructure load. The advantages of this method are;
(1) The volume of excavation to be done is reduced to a minimum since excavation will be done for the ground beams only (2) It is relatively easy to construct (3) The slab performs the triple function of acting as a structural raft slab, ground floor slab (oversite concrete), and damp proof course.
However, this construction method can demand beams that are slightly deeper than normal in order to raise the ground floor to a height above the compound level.
This type of foundation can be analysed and designed and using finite element analysis or rigid theory. When using the rigid theory, the arrangement of the foundation and the loadings have to be fairly symmetrical, and the footing will have to be stiff enough in order to assume full rigidity. Furthermore, we usually assume that the stiffness of the raft slab covers the weak patches of the soil adequately.
The implication of the rigid theory in design is that the settlement of the soil will even out under the rigid footing, with high internal forces developing in the slab. When a flexible approach is used, there would be a differential settlement, but lower stresses on the elements, thereby leading to a more economical design. Note that rigidity criteria are a function of superstructure stiffness and soil stiffness. Different codes of practice have their own definition of rigidity.
In the rigid method, usually, the soil pressure on the slab is evaluated, and transferred to the beams by assuming a sort of combined footing approach.
Case 2: Ground Floor Slab on Ground Beam
Fig 2: Ground beams on pile caps
In this approach as shown in Figure 2, the ground floor slabs are supported directly by ground beams, which in turn are supported by pile caps. In this case, the slabs are designed as suspended slabs, and ground pressure reactions are not taken into account.
Worked Example
Let us assume that you have a raft foundation on ground beams to design, and you have been able to obtain the soil pressure. Note that the variation of soil pressure is usually a function of location and the applied loading. See an example of how to determine it below;
Let us assume that in the example above, we will be introducing ground beams (1200mm x 225mm) along the column axes as shown in Figure 3. Let the thickness of the slab be 200 mm.
Fig 3: Sketch of how ground beams were introduced in the model
From the analysis, the maximum pressure on the raft slab was discovered to be 49.975 kN/m2. Note that with the introduction of the ground beams, the 1000 mm projection is no longer necessary. So a simplified way of analysing and designing the ground beams using Staad Pro is as follows.
Step 1: Model the ground beam as appropriate and support it with pinned support at the column points. Also, assign the properties of the beam (1200 mm x 225 mm).
Fig 4: Modelling of the ground beams on Staad Pro
Step 2:Create/add a ‘floor load’ and assign it to the beams as shown below. This uses the tributary area method to transfer the load to the beams;
Fig 5: Application of ground pressure load
Remember that the load is assigned a positive value because it is an upward pressure.
Step 3: Run the analysis to obtain the bending moment given below.
Let us consider the beam along grid line B.
Bending Moment Diagram
Shear Force Diagram
The result from Staad Pro software indicates that the maximum span moment is 463 kNm. With this, you can provide the reinforcements at ULS depending on the code of practice you are using. Note that where applicable, you might need to consider wall loads, self-weight, etc, but note that these gravity forces will be likely beneficial because they will reduce the resultant load on the beam due to earth pressure. This is just a simple and straightforward approach that would always yield conservative results.
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Staircases provide simple solutions for vertical circulation in a building. In this article, we will present the different methods of modelling a simple reinforced concrete staircase for the purpose of analysis and structural design.
The different methods considered in this article are as follows;
(1) Modelling the staircase using finite element plates (2) Modelling the staircase as a simple 2D frame (3) Modelling the staircase as a linear simply supported beam
The aim of this post is to compare the results obtained from the different modelling alternatives and give reinforced concrete designers a little idea about the results to expect from their assumptions.
Let us consider a section of a staircase with dimensions as shown below;
Fig 2: Section of Staircase
The loading on the staircase is as follows;
Ultimate load on the flight = 15.719 kN/m2 Ultimate load on the landing = 14.370 kN/m2 Width of staircase = 1000 mm
We are going to model and analyse the staircase section shown using the three different methods described above.
Finite Element Plate Model (Staad Pro)
In this approach, the full dimensions and geometry of the staircase will be modelled using finite element plates. The thickness of the waist of the staircase will be assigned as the thickness of the plates, while the rise and threads will be ignored. The size of the finite element plates can be chosen to be square or rectangular but is preferable to keep the model uniform.
The 3D model of the staircase is given below;
Fig 3: Rendered 3D Model of the Staircase
To consider the effect of the support condition on the behaviour of the staircase, it will be modelled as pinned-pinned condition and pinned-roller support condition.
Pinned-pinned supports
Fig 4: Model of the staircase with pinned-pinned support
The static analysis of the structure gave the result below;
Fig 5: Transverse Bending Moment on the Staircase (Mx)
Fig 6: Longitudinal Bending Moment (My) on the Staircase
Pinned-roller support
Fig 7: Modelling the Staircase with Pinned-Roller Support
The analysis result gave the following;
Fig 8: Transverse Bending Moment (Mx)
Fig 9: Longitudinal Bending Moment (My)
Static 2D Frame Model
In this case, the geometry of the staircase (height, landing, and angle of flight) is modelled as line elements and analysed as a 2D frame subjected to static load. The typical model is shown below.
Pinned-Pinned Support
The result is as given below;
Fig 10: Bending moment on the Staircase (Frame Model; pinned-pinned support)
Pinned-Roller Support
Fig 11: Bending Moment on the staircase (pinned-roller model)
Simply Supported Beam Model
In this case, the entire geometry of the staircase is idealised and analysed as a simply supported linear beam model as shown below.
Figure 12: Bending Moment on the stair (simply supported beam model – horizontal length)
Discussion of Results
The analysis results using the above-named methods gave all the bending moments in the staircase to be sagging.
The finite element plate model gave the maximum sagging at the flight to be 41.1 kNm (see Fig 5), while the maximum sagging moment on the landing was found to be 32.6 kNm (see Fig 6). This result was found in both the pinned-pinned and pinned-roller model, therefore, there is no difference in using any of the support conditions in Staad Pro.
The static 2D frame model of the staircase gave the maximum sagging moment on the flight to be 41.1 kNm, while the maximum sagging moment on the flight was 33.82 kNm (see Fig 10 and 11). In principle, it can be said there is no difference in modelling the staircase as a simple 2D frame or as a 3D plate model. The former is quicker and less complicated.
The model of span dimension adopted for the manual equivalent beam calculation was picked from SCALE software, and the maximum moment on the flight was found to be 36.04 kNm (see Fig 12). This is about 12.3% less than other models. A little consideration will show that this effect stems from the span dimensions.
To take care of this, let us modify the flight dimensions as follows; Lf,m = Lf/cos ϕ Where Lf is the horizontal span of flight, Lf,m is the actual length of the flight, and ϕ is the angle of inclination of the flight.
Thus; Lf,m = 1.75m/cos 34.4 = 2.12m
This gave the result below;
Fig 13: Bending moment on the stair (simply supported beam model, developed length)
We can see that is more comparable and conservative when compared with results from computer models. Therefore, for all simple staircases, we can use the developed length of the flight instead of the horizontal length to analyse the staircase as a simply supported beam. However, when checking for deflection, we should use the horizontal length.
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Castigliano’s method for calculating deflections is an application of his second theorem, which states that if the strain energy of a linearly elastic structure can be expressed as a function of generalised force Pi then the partial derivative of the strain energy with respect to generalised force gives the generalised displacement wi in the direction of Pi.
The second theorem of Castigliano is applicable to linearly elastic (Hookean material) structures with constant temperature and unyielding supports.
In general, this is given by;
wi = ∂U/∂Pi
The strain energy stored in a linear elastic system due to bending is given by;
Solved Example For the frame loaded as shown below, let us find the vertical deflection at point C due to bending using Castigliano’s theorem.
Solution Section BC Mx = -20x
U1 = ∫[(-20x)2/2EI] dx = ∫ -400x2/2EI = -400x3/6EI Knowing that the limit x = 1.5m; U1 = 225/EI
Section AB My = (-20 × 1.5m) = 30 kNm
U2 = ∫[(-30)2/2EI] dy = ∫ -900/2EI = -900y/2EI Knowing that the limit y = 2.5m; U2 = 1125/EI
Total strain energy = U1 + U2 = (225/EI) + (1125/EI) = 1350/EI
Let the vertical deflection at point C be δv
Work done by the externally applied load = 1/2(P × δ) Work done = Strain energy stored in the system
