This should be a long post, but I am going to try and keep it as brief as possible. This post is more like an excerpt from the publication ‘Structural Analysis and Design of Residential Buildings using Staad Pro V8i, CSC Orion, and Manual calculations’. Here, we are going to briefly present some practical analysis and design of some reinforced concrete elements using Staad Pro software, Orion software, and manual calculations. Ultimately, we are going to make some comparisons of the results obtained based on the different methods adopted in the analysis and design. To learn how to model, design, and detail buildings from the scratch using Staad Pro, Orion, and manual methods, see the link at the end of this post.

To show how this is done, simplified architectural floor plans, elevations, and sections, for a residential two-storey building have been shown below for the purpose of structural analysis and design (see the pictures below).

The first step in the design of buildings is the preparation of the ‘general arrangement’, popularly called the G.A. The G.A. is a drawing that shows the disposition of the structural elements such as the slabs and their types, the floor beams, the columns, and their interaction at the floor level under consideration. For the architectural drawings above, the adopted G.A. is shown in Figure 7. below. There are no spelt out rules about how to prepare G.A. from architectural drawings, but there are basic guidelines that can guide someone on how to prepare a buildable and structurally efficient G.A.

**Design data**:

F_{ck} = 25 N/mm^{2}, F_{yk} = 460 N/mm^{2}, C_{nom} (slabs) = 25mm, C_{nom} (beams and columns) = 35mm, C_{nom} (foundations) = 50mm

Thickness of slab = 150mm; Dimension of floor beams = 450mm x 230mm; Dimension of columns = (230 x 230mm)

**DESIGN OF THE FLOOR SLABS****PANEL 1: MANUAL ANALYSIS**

The floor slab (PANEL 1) is spanning in two directions since the ratio (k) of the longer side (Ly) to the shorter side (Lx) is less than 2.

Hence, k = Ly/Lx = 3.825/3.625 = 1.055 (say 1.1)

Moment coefficients (α) for two adjacent edges discontinuous (pick from table);**Short Span**

Mid-span = 0.042

Continuous edge = 0.056**Long Span**

Mid-span = 0.034

Continuous edge = 0.045

**Design of short span****Mid span**

M = αnLx^{2} = 0.042 × 10.9575 × 3.6252 = 6.0475 KN.m

M_{Ed} = 6.0475 KNm

Effective Depth (d) = h – Cc – ϕ/2

Assuming ϕ12mm bars will be employed for the construction

d = 150 – 25 – 6 = 119mm; b = 1000mm (designing per unit width)

k = M_{Ed}/(f_{ck}bd^{2} )

= (6.0475 × 10^{6})/(25 × 1000 × 119^{2} ) = 0.0171

Since k < 0.167 No compression reinforcement required

z = d[0.5+ √(0.25 – 0.882k)] = z = d[0.5+ √(0.25 – (0.882 × 0.0273)] = 0.95d

A_{s1} = M_{Ed}/(0.87f_{yk} z)

A_{s1 }= (6.0475 × 10^{6})/(0.87 × 460 × 0.95 × 119) = 133.668 mm^{2}/m

Provide Y12mm @ 250mm c/c BOT (A_{Sprov} = 452 mm^{2}/m)

A little consideration will show that this provided area of steel will satisfy serviceability limit state requirements. To see how to carry out deflections and crack control verifications, see the the link at the bottom of this post.

Result from Orion showing the Short Span (mid span) design moments (Wood and Armer effects inclusive) (PANEL 1)

**Result from Staad showing the Short Span (mid span) design moments (Wood and Armer effects inclusive) (PANEL 1)**

A little observation will show that the design moment values from the different methods are very similar. The full detailing of the floor slabs is as shown below.

Figure 9: Bottom Reinforcement Detailing

Figure 10: Top Reinforcement Detailing

Figure 11: Section of the floor slab

**DESIGN OF THE BEAMS**

Let us take Beam No 1 from our GA as a design case study:

The loading of the beam has been carried out as shown below. The beam is primarily subjected to load from the slab, the weight of wall, and its own self-weight. To see how to manually calculate the loading on beams, follow the link at the end of the post.

The internal forces from the loading is as shown below;

The internal forces from Orion software for Beam No 1 is as shown below. Load decomposition using finite element analysis was used for the load transfer.

The internal forces from Staad software for Beam No 1 is as shown below.

A_{s1} = M_{Ed}/(0.87f_{yk}z)

= (36.66 × 10^{6})/(0.87 × 460 × 0.95 × 399) = 241.667 mm^{2}

Provide 2Y16 mm BOT (A_{Sprov} = 402 mm^{2})

The detailing of Beam No 1 is as shown below;

**DESIGN OF THE COLUMNS**

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. 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 the structural design of columns. The design of columns is covered in section 5.8 of EC2.

The column axial loads have been obtained by summing up the reactions from all the beams supported by the columns, including the self weight of the column.

Let us use column A1 as example.

At the roof level, the column is supporting beam No 2 (Support Reaction V_{1} = 13.27 KN) and Beam No 3 (Support Reaction V_{A} = 12.99 KN). At the first floor level (see Analysis and Design of Beam No 1 and 2), the column is supporting Beam No 1 (Support Reaction V_{1} = 41.38 KN), and Beam No 2 (Support Reaction V_{A} = 42.49 KN). Therefore the summation of all these loads gives the axial load transferred from the beams. For intermediate supports, note that the summation of the shear forces at the support gives the total support reaction (neglect the signs and use absolute value. Another method of calculating Column Axial Load is by *Tributary Area Method*. This method has not been adopted in this work.

**COLUMN A1**

Total Columns Self weight = 12.14 KN

Load from roof beams = 13.27 + 12.99 = 26.26 KN

Load from floor beams = 46.21 + 42.49 = 88.70 KN**Total = 127.13 KN**** **

Axial Load from Orion (A1) = 126.6 KN

Axial Load from Staad (A1) = 130.684 KN

**COLUMN A3**

Total Columns Self weight = 12.14 KN

Load from roof beams = 35.41 + 11.46 = 46.87 KN

Load from floor beams = 105.33 + 60.85 = 166.18 KN**Total = 225.19 KN**

Axial Load from Orion (A3) = 202.3 KN

Axial Load from Staad (A3) = 201.632 KN

**COLUMN A5**

Total Columns Self weight = 12.14 KN

Load from roof beams = 17.19 + 5.70 = 22.89 KN

Load from floor beams = 83.64 + 37.91 = 121.55 KN**Total = 156.58 KN**

Axial Load from Orion (A5) = 155.9 KN

Axial Load from Staad (A5) = 163.207 KN

**COLUMN A7**

Total Columns Self weight = 12.14 KN

Load from roof beams = 43.15 + 9.48 = 52.63 KN

Load from floor beams = 38.26 + 62.45 = 100.71 KN**Total = 165.48 KN**

Axial Load from Orion (A7) = 133.9 KN

Axial Load from Staad (A7) = 140.392 KN

As you can see, for design purposes, the axial loads from the three methods are very comparable. To see how to obtain the column design moments from the use of sub-frames, follow the link at the end of the post.

**Design of Column E5**

Reading from chart; d_{2}/h = 0.2;

M_{Ed}/(f_{ck} bh^{2} )

= (10.002 × 10^{6})/(25 × 230 × 230^{2} ) = 0.03288

N_{Ed}/(f_{ck}bh)

= (399.88 × 10^{3})/(25 ×230 × 230) = 0.302

From the chart:

(A_{s}F_{yk})/(bhf_{ck} ) = 0.05

Area of longitudinal steel required (As) = (0.05 × 25 × 230 × 230)/460 = 143.75 mm^{2}

A_{s,min} = 0.10 NEd/fyd

= (0.1 × 399.887)/400 = 0.099 mm^{2} < 0.002 × 230 × 230 = 105.8 mm^{2}

Provide 4Y16mm (Asprov = 804 mm^{2})

**Links**

Minimum size = 0.25ϕ = 0.25 × 16 = 4mm < 6mm

We are adopting Y8mm as links

Spacing adopted = 200mm less than min{b, h, 20ϕ, 400mm}

**Result from Orion for column E5**

**Result from Staad for column E5**

Staad Provided Y8@225mm links

The column detailing is as shown below;

**DESIGN OF FOUNDATIONS**

All loads from the superstructure of a building are transferred to the ground. If the foundation of a building is poorly designed, then all the efforts input in designing the superstructure is in vain. It is therefore imperative that adequate care is taken in the design of foundations. Foundation design starts with a detailed field and soil investigation. It is very important to know the index and geotechnical properties of the soil, including the soil chemistry, so that the performance of the foundation can be guaranteed.

**Analysis and Design of footing E8**

Bearing Capacity of the foundation = 150 KN/m^{2};

Effective depth

Concrete cover = 50mm

AssumingY12mm bars,

d = 400 – 50 – 6 = 344mm

The ultimate limit state design moment can be obtained by considering the figure below;

k = M_{Ed}/(f_{ck} bd^{2})

= (37.518 × 10^{6})/(25 × 1000 × 344^{2} ) = 0.01268 (designing per metre strip)

Since k < 0.167 No compression reinforcement required

z = d[0.5+ √(0.25 – 0.882k)] = z = d[0.5+ √(0.25 – (0.882 × 0.0273)] = 0.95d

A_{s1} = M_{Ed}/(0.87f_{yk}z)

= (37.518 × 10^{6})/(0.87 × 460 × 0.95 × 344) = 286.869 mm^{2}/m

**To calculate the minimum area of steel required**;

f_{ctm} = 0.3 × (f_{ck})^{(2⁄3)} = 0.3 × 25 ^{(2⁄3)} = 2.5649 N/mm^{2 } (Table 3.1 EC2)

A_{Smin} = 0.26 × f_{ctm}/F_{yk} × b × d = 0.26 × 2.5649/460 ×1000 × 344 = 498.7 mm^{2}

Check if A_{Smin} < 0.0013 × b × d (447.2 mm^{2})

Since, A_{Smin} = 498.7 mm^{2}, the provided reinforcement is adequate.

Provide Y12 @ 200mm c/c (A_{Sprov} = 565 mm^{2}/m) each way

**Shear at the column face**

Ultimate Load on footing from column = 399.887 kN

Design shear stress at the column perimeter v_{Ed} = βV_{Ed}/(u_{0}d)

β is the eccentricity factor (see section 6.4.3 of EC2)

β = 1+ 1.8√[(16.48/230)^{2}+(8.99/230)^{2}] = 1.146

Where u_{o} is the column perimeter and d is the effective depth

v_{Ed} = βV_{Ed}/(u_{0}d)

= (1.15 × 399.887 × 10^{3})/(4(230) × 344) = 1.452N/mm^{2}

V_{Rd,max} = 0.5vf_{cd}

v = 0.6[1 – (f_{ck}/250) ] = 0.6[1 – (25/250) ] = 0.54 N/mm^{2}

f_{cd} = (α_{cc} f_{ck})/γ_{c} = (0.85 × 25)/1.5 = 14.167 N/mm^{2}

V_{Rd,max} = 0.5 × 0.54 × 14.167 = 3.825 N/mm^{2}v_{Ed} < V_{Rd,max}. This is very ok

**Transverse shear at ‘d’ from the face of column**

Width of shaded area = a – d = 0.635 – 0.344 = 0.291m

Area of shaded area = (1.5m × 0.291m) = 0.4365 m^{2}

Therefore, ΔV_{Ed} = (189.386 + 175.939)/2 × 0.4365 m^{2} = 79.077 KN

v_{Ed} = V_{Ed}/bd = (79.077 × 10^{3})/(1500 × 344) = 0.15325 N/mm^{2}

V_{Rd,c} = [C_{Rd,c} k (100ρ_{1} f_{ck} )^{(1/3)} + k^{1}.σ^{cp}] × (2d/a) ≥ (V_{min} + k_{1}.σ,sub>cp) bw.d

C_{Rd,c} = 0.18/γ_{c} = 0.18/1.5 = 0.12

k = 1 + √(200/d) = 1+√(200/344) = 1.7624 > 2.0, therefore, k = 1.7624

V_{min} = 0.035k^{(3/2)} f_{ck}^{(1/2)}

V_{min} = 0.035 × (1.7624)^{(3/2)} × (25)^{(1/2)} = 0.4094 N/mm^{2}

ρ_{1} = As/bd = 565/(1000 × 344) = 0.001642 < 0.02;

V_{Rd,c} = [0.12 × 1.7624 (100 × 0.001642 ×25 )^{(1/3)} ] = 0.3386 N/mm^{2} × (2d/a) < V_{min}

But in this case, d = a

Hence, V_{Rd,c} = 2 × 0.3386 = 0.6772 N/mm^{2}

Since V_{Rd,c} (0.6772) > V_{Ed} (0.1503 KN), No shear reinforcement is required.

**Punching Shear at 2d from the face of column**

Punching shear lies outside the footing dimensions. No further check required.

**Design Result from Orion**

The detailing of the footing is as shown below;

The structural analysis and design of all members have been fully done, including a step by step tutorial on how to model and design on Orion and Staad Pro, and how to manually design. See the completed models below;

To download the simplified e-book where all the members have been designed and completely detailed, including bar bending schedule and quantification of materials, click HERE.

Thank you, and God bless you.

Well done sir

What is c nom

Nominal cover to reinforcement (concrete cover)

In short God Bless you my able Engr., I really love your effort and I also pray for myself to have a such zeal to help other like this.

Thank you very much… We will keep working hard

your have great zeal ….i wish you goodlucky in future….how can i get your practical book…..

my broda God bless u. u r doing a great work, d lord is ur strength.

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Regards..

Well-done sir, but I can't find any link in the end

Sir I finished in a school were by we cram to pass now am a graduate what will I show for outside I really need to be good at something

You now have to discover yourself based on what you really want, and give your all towards getting it.

good job, but one thing still unclear, how did you arrive at 12.147KN has column self weight?? by my calculation its 0.23*0.23*3*24=3.8KN

Multiply by the column hieght

Nice Content ..

Great job!! 😎

God bless you Sir. U did it all .

Thanks for this straight forward analysis,but sir how can someone load a Column of more than 2 floors

Great Job.

Thank You Sir

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In addition, the load requirements, fixtures, and even materials will vary significantly between commercial and residential roofs. For example, commercial roofs must have a larger load-bearing capacity than residential roofs often just because of the materials involved in their construction.

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NICE CONTENT….

GOOD WORK

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I calculated my soil pressure as follow:

soil pressure = 1.5 x total load coming on the column/area of footing

= 1.5 x (39.9887+ 399.887)/(1.5 x 1.5)

= (1.5 x 439.8757)/2.25

= 659.81355/2.25

= 293 kN/m2

instead of 189.386 kN/m2.

Please enlighten us on this condition.

Thanks waiting for your reply.

Where did you get the 1.5 factor from?

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One of the nice document you have made about structural design. I really like this one. Thanks for sharing with us.

God bless you for this great work. more blessings

Nice work. How can one extract design information/calculations from Orio.

Thank you for taking time to educate especially our elite who always run away from detail building drawings and do short cut at their own perils.

Engr Ubani, Great work! My only comment is that you loaded all spans with max load. I think it is necessary to do a critical load arrangement on the continous beam where the longer span is loaded with max load and short span with minimum load. This will result in larger hogging moment in the midspan of the short span. This will also help in drawing a proper BM envelope for the beam and guide in the placement of reinforcement. Clearly, the main reinforcement will be placed at the top of the mid span for the short span while main reinforcement for the longer span will be placed at the bottom.

I am still trying to get this calculation