8.7 C
New York
Friday, September 20, 2024
Home Blog Page 57

Simple Proofs Why Shorter Spans Are More Critical in Slab Design

By
Ubani Obinna Uzodimma
-

A slab is a structural member whose depth is considerably smaller than the lateral dimensions. They provide useful surfaces for floors in buildings and support the occupancy loads. Reinforced concrete slabs may be supported by beams, columns, walls, or masonry. When a slab is supported on two opposite sides, they are referred to as one-way slabs because the load is transferred from the slab to the beam in the perpendicular direction.

one%2Bway%2Bslab
Typical one-way slab system

However, if a slab is supported by beams on the four sides, and the ratio of length to width is greater than 2, majority of the load is transferred to the short direction to the supporting beams, and one-way action is obtained in effect, even though supports are provided on all sides.

The structural action of one-way slabs may be visualized in terms of the deflected surface. This can be approximated as a cylindrical surface, and curvatures and bending moments are parallel to the short side Lx. The slab is normally analysed a beam of a unit strip, with the bending moment, shear forces, and reinforcement determined per unit strip.

56777
Cylindrical one-way bending action of a thin pate

In many design circumstances, rectangular slabs have dimensions where the ratio of the longer side to the shorter side is less than 2 (and are also supported in such a way that two-way action results). When loaded, such slabs bend into a dished surface rather than a cylindrical one. This means that at any point the slab is curved in both principal directions, and since bending moments are proportional to curvatures, moments also exist in both directions.

dttttu
Two-way bending action of a thin plate

BHXXrA jYR

To resist these moments, the slab must be reinforced in both directions, by at least two layers of bars perpendicular, respectively, to two pairs of edges. The slab must be designed to take a proportionate share of the load in each direction.

Why are Short Span Critical in Slabs?

Whenever civil engineering students enter a structural design classroom for the first time, they are told that the shorter span is more critical in the design of slabs. This is usually source of wonder for first timers because by mere instincts, the longer span should be more critical.

This article shows very simple proofs why the short span is more critical in slabs.

Deflection of centre-strip approach

This approach offers an extremely simplified concept that shows that the load transmitted to the shorter span is greater than the load transmitted to the longer span.

Let us consider the two way action of the slab shown below;

ert

Looking at the centre-strips highlighted in red, a little consideration will show that at the intersection point of the strips, the deflection is equal. Logical right?

Let the uniform load  on the longer strip be qy
Let the uniform load on the shorter strip be qx

We can therefore say that;

Deflection at centre w = 5qyLy4/384EI = 5qxLx4/384EI
5qyLy4/384EI = 5qxLx4/384EI
qyLy4 = qxLx4

We can verify from the above relationships that;
qx/qy = Ly4 / Lx4

Since Ly > Lx, we can accept that the load on the short span (qx) is greater than the load on the long span (qy) since Ly/Lx > 1.0
This is an oversimplification of the behaviour of slabs though.

Yield Line Approach

The yield line method was developed to determine the limit state of slabs by considering the yield lines that occur at the slab collapse mechanism. The yield lines are usually approximated to originate at the corners, forming at an angle of 45° until they intersect. These yield lines usually show trapezoidal loads going to the longer supporting beam of the slab, and triangular loads going to the shorter supporting beam.

Let us assume that the slab is subjected to a unit pressure load (1 kN/m2)

Ly = 6m
Lx = 5m

Area of trapezium = 8.6825 m2
Area of triangle = 6.316 m2

Therefore;
The load parallel to the short span (Px) = 2 × 8.6825m2 × 1 kN/m2 = 17.365 kN
The load parallel to the long span (Py) = 2 × 6.316m2 × 1 kN/m2 = 12.632 kN

Total load on slab = (5 × 6)m2 × 1 kN/m2 = 30 kN

Therefore, you can see why the shorter span is more heavily loaded based on the yield line pattern.

Finite Element Analysis

When we carry out finite element analysis on slabs, the result offers an insight;
Let us check out the slab investigated above using finite element analysis results from Staad Pro.

SER

Let us assume that the slab is subjected to a unit pressure load (1 kN/m2)

Ly = 6m
Lx = 5m
Thickness of slab = 150 mm
v = 0.2
E = 21.7 kN/mm2

From the result, the bending moment parallel to the short span is given below;

In the longer direction;

my

If you look at the results above, the moments parallel to the the short span is more critical than the moment parallel to the longer side.

Conclusion

The load transferred to supporting beams through the short span of the slab is heavier than the load transferred through the long span of the slab. For a one way slab, the slab is analysed as a beam of unit width , and the main reinforcement is provided parallel to the short span, while distribution bars are provided parallel to the long span. Minimum steel can be used as distribution bars.

But for two way slabs, the slab is designed for strength in the two principal directions, but the main bars are placed at the bottom, parallel to the short span, while the other is placed near to the bottom (on top of the bottom bar) parallel to the long span

An Introduction To Isogeometric Analysis (IGA)

By
Ubani Obinna Uzodimma
-

1.0 Introduction
Isogeometric Analysis (IGA) is a computational approach that integrates Finite Element Analysis (FEA) and Computer Aided Design (CAD). Isogeometric Analysis is developed for the purpose of utilizing the same data set in both design and analysis (Raknes, 2011). In today’s CAD and FEA packages one have to convert the data generated in design to a data set suitable for FEA. Converting the data is not trivial, as the computational geometric approach is different in CAD and FEA. IGA makes it possible to utilize the NURBS geometry, which is the most used basis in CAD packages, in FEA directly. Isogeometric analysis is thus a great tool for optimizing models, as one easily can make refinements and perform testing and analysis during design and development.

2.0 CAD and FEA
CAD systems are characterised by “exact” (small gaps within predefined tolerances are allowed) descriptions of the inner and outer shells of the object and 3D models are sufficiently accurate for production purposes (Ferreira, 2014). In FEA the object is represented with a description of composed structures of finite elements, where there is a cruder representation of the shape of inner and outer hulls. In addition FEA needs compact geometric descriptions, where unnecessary detail is removed to focus the analysis on essential properties of the objects, namely, the simulation is performed only in the regions of interest to study the physical problem.

Right now, numerical simulation often involves using the Finite Element Method (FEM) where the geometry model, derived from CAD, usually will suffer a reparameterization of the CAD geometry by piecewise low order polynomials (mesh), generally applying linear Lagrange polynomials to approximate the geometry. This information transfer between models suitable for design (CAD) and analysis (FEM) is considered being a bottleneck in industry (industrial applications) of today, because it introduces significant approximation errors, or makes the simulations computational costly in FEM models with fine mesh, and entails a huge amount of man-hours to generate a suitable finite element mesh (Ferreira, 2014).

Nowadays most CAD systems use spline basis functions and often Non-Uniform Rational B-Splines (NURBS) of a different polynomial order. For that reason, IGA is based on NURBS, where the main idea is to use the same mathematical description for the geometry in the design process within CAD and the numerical procedures used in FEA simulations. In other words, IGA is characterised by the use of a common spline basis for geometric modelling and Finite Element Analysis of a given object and can be considered a generalisation of FEM analysis (Ferreira, 2014). Much research in application of IGA have been done at various topics of FEA: linear and non-linear static and dynamic analysis of thin-walled structures, fluid mechanics, fluid structure interaction, shape and topology optimisation, vibration analysis, buckling and others.

Isogeometric analysis based on NURBS (Non-Uniform Rational B-splines) was introduced by Tom Hughes (2005) and further expanded by Cortrell et al, (2006). The objectives of IGA are to generalise and improve upon FEA in the following ways:

  • Get volumetric spline models having the expected behaviour according to the Partial Differential Equations system (PDEs) and according with boundary conditions and loads behind the physical phenomena in analysis;
  • Represent common engineering shapes (circles, cylinders, spheres, ellipsoids, etc.) and provide accurate modelling of complex geometries. This means that geometrical errors optimally should be eliminated;
  • Simplify mesh refinement of geometries by eliminating the need for communication with CAD geometry once the initial mesh is constructed;
  • Provide systematic refinements (h, p and k refinements) that exhibit improved accuracy and efficiency compared with classical FEA.
SPACING%2BAND%2BMAPPING%2BIN%2BIGA
3.0 Differences Between Isogeometric Analysis and Finite Element Analysis
The concept of isogeometric analysis is that the basis functions that are used to model the exact geometry are also used as a basis for the solution field. In classical FEA it is the other way around; the basis functions we choose to approximate the unknown solution field is used to approximate the already known geometry (Raknes, 2011). NURBS based Galerkin finite element method is somewhat similar to classical FEA, only now, different basis functions are being used.
For example, in finite element analysis, a circle is an ideal achieved in the limit of mesh refinement (i.e., h-refinement) but never achieved in reality, whereas a circle is achieved exactly for the coarsest mesh in isogeometric analysis, and this exact geometry, and its parameterization, are maintained for all mesh refinements. It is interesting to note that, in the limit, the isogeometric model converges to a polynomial representation on each element, but not for any finite mesh. This is the obverse of finite element analysis in which polynomial approximations exist on all meshes, and the circle is the idealized limit (Cotrell, 2006). The table below shows a comparison between FEA and IGA.


344
Example
Analysis of a CCCC (all sides clamped) thin plate (Kirchoff’s theory)
CLAMPED%2BPLATE
Data
a = 6.0 m
b = 6.0 m
Thickness of plate (h) = 150 mm
Pressure Load (q) = 6.2 kN/m2

Poisson ratio (v) = 0.2
Modulus of elasticity (E) = 21.7 kN/mm2

By Isogeometric Analysis
(a) Meshing

MESHING%2BOF%2BTHE%2BPLATE

(b) Enforcement of Boundary Conditions

ENFORCEMENT%2BOF%2BBOUNDARY%2BCONDITIONS

(c) Deflection profile of the plate (MATLAB)

DEFLECTION%2BOF%2BTHE%2BSTRUCTURE

Maximum deflection at mid span =  – 0.001593681267073 m
wmax = 1.59368 mm

Comparing this result with other methods;
Classical solution (Timoshenko) = 1.593 mm
Finite element Analysis (Staad Pro) = 1.560 mm (rectangular mesh size division  = 25 x 25)

References
[1] Ferreira J.P. (2014): Elasto-plastic analysis of structures using an Isogeometric Formulation. M.Eng thesis presented to the Department of Mechanical Engineering , University of Porto, Portugal

[2] Cottrell J, A. Reali, Y. Bazilevs, and T. Hughes (2006):  “Isogeometric analysis of structural vibrations,” Computer Methods in Applied Mechanics and Engineering, vol. 195, no. 41– 43, pp. 5257 – 5296. John H. Argyris Memorial Issue. Part II. 

[3] Raknes Siv Bente (2011): Isogeometric Analysis and Degenerated Mappings. An M.Sc thesis submitted to the Department of Mathematical Sciences, Norwegian University of Science and Technology.

[4] Timoshenko S. And S. Woinowsky-Krieger (1959): Theory of plates and shells. McGraw-Hill Book Company, Inc


Design of Bolted Beam Splice Connections | EN 1993

By
Ubani Obinna Uzodimma
-

In the construction of steel structures, there is always a need to join steel members for the purpose of continuity. It is not always possible to have the full lengths of members to span the desired length due to production, handling, and transportation issues. As a result, it is very common to bolt or weld two or more steel members, so as to achieve the desired length.

A bolted beam splice connection is a structural joint used in the construction of steel beams. It is designed to connect two or more beams end-to-end to form a longer beam, increasing the span length or overall structural capacity. The design of a bolted beam splice connection involves several considerations to ensure its strength, stability, and reliability.

The design of a bolted beam splice connection involves careful consideration of load transfer, bolted connection details, splice configuration, stiffness requirements, structural analysis, connection detailing, and material properties. By implication, all the anticipated design forces acting at the point of the beam splice must be taken into consideration during the design.

Therefore, the design involves performing structural analysis to determine the forces and moments acting on the connection. This analysis considers the applied loads, beam properties, and connection geometry. Finite element analysis or other advanced analysis methods may be employed to evaluate the connection’s behaviour under different loading conditions.

In this article, we are going to look at a design example of a beam splice connection (beam-to-beam connection using steel plates). This joint should be able to transmit bending, shear, and axial forces.

Installation of bolted beam splice connection on site

Design and functions of Bolted Splice Connections

The primary function of a beam splice connection is to transfer the loads between the beams effectively. The connection should be designed to transfer axial forces, shear forces, and bending moments from one beam to another without significant deformation or failure.

The connection typically consists of bolts, washers, and nuts. High-strength bolts are commonly used due to their ability to withstand heavy loads. The number, size, and grade of bolts are determined based on the design requirements and the applied loads.

The configuration of the splice connection depends on the structural requirements and the type of loads being transferred. Common types include full-depth end plate splices, extended end plate splices, and flange plate splices. The choice of splice configuration is influenced by factors such as the beam profile, connection stiffness, and fabrication constraints.

The stiffness of the splice connection plays a crucial role in maintaining the overall structural integrity and load distribution. Adequate stiffness helps to limit deflections and minimize differential movement between the spliced beams. The connection should be designed to ensure that it does not introduce excessive additional stiffness or flexibility to the overall beam system.

Design Example

Let us design a bolted beam splice connection for a UB 533 x 210 x 101 kg/m section, subjected to the following ultimate limit state loads;

MEd = 610 kNm
VEd = 215 kN
NEd = 55 kN

At serviceability limit state;
MEd,ser = 445 kNm
VEd,ser = 157 kN
NEd,ser = 41 kN

Properties of UB 533 x 210 x 101
h = 536.7 mm
b = 210 mm
d = 476.5 mm
tw = 10.8 mm
tf = 17.4 mm
r = 12.7 mm
A = 129 cm2
Iy = 61500 cm4
Iz = 2690 cm4
Wel,y = 2290 cm3
Wpl,y = 2610 cm3;
hw = h – 2tf = 476.5 mm

Cover plates
Let us assume 20 mm thick cover plates for the flanges and 15 mm thick plates for the web. The thickness and dimension to be confirmed later in the post.

Bolts
M24 preloaded class 8.8 bolts
Diameter of bolt shank d = 24mm
Diameter of hole d0 = 26mm
Shear area As = 353 mm2

Materials Strength
Beam and cover plates;
fy,b = fy,wp = 275 N/mm2
fu,b = fu,wp = 410 N/mm2

Bolts
Nominal yield strength fyb = 640 N/mm2
Nominal ultimate strength fub = 800 N/mm2

Partial Factors for Resistance (BS EN 1993-1- NA.2.15)
Structural Steel
γm0 = 1.0
γm1 = 1.0
γm2 = 1.1

Parts in connections
γm2 = 1.25 (bolts, welds, plates in bearing)
γm3 = 1.0 (slip resistance at ULS)
γm3,ser = 1.1 (slip resistance as SLS)

Step 1: Internal Forces at Splice
For a splice in flexural member, the parts subject to shear (the web cover plates) must carry, in addition to the shear force and moment due to the eccentricity of the centroids of the bolt groups on each side, the proportion of moment carried by the web, without any shedding to the flanges.

The second moment of area of the web is;
Iw = [(h – 2tf)3tw/12]
Iw = [(476.53 × 10.8)/12] × 10-4 = 9737 cm4

Therefore the web will carry 9737/61500 = 15.8% of the moment in the beam (assuming an elastic stress distribution), while the flange will carry the remaining 84.2%.

The area of the web is;
Aw = 476.5 × 10.8 × 10-2 = 51.462 cm2
Therefore, the web will carry 51.462/129 = 39.8% of the axial force in the beam, while the flange will carry the remaining 60.1%.

Forces at ULS
The force in each flange due to bending is therefore given by;
Ff,M,Ed = 0.842 × [(MEd)/(h – tf)]
Ff,M,Ed = 0.842 × [(610 × 106)/(536.7 – 17.4)] × 10-3 = 989 kN

And the force in each flange due to axial force is given by;
Ff,N,Ed = 0.601 × (55/2) = 16.53 kN

Therefore
Ftf,Ed = 989 – 16.53 = 972.47 kN
Fbf,Ed = 989 + 16.53 = 1005.53 kN

The moment in the web = 0.158 × 610 = 96.38 kNm
The shear force in the web = 215 kN
The axial force in the web = 0.398 × 55 = 21.89 kN

Forces at SLS
The force in each flange due to bending is therefore given by;
Ff,M,Ed = 0.842 × [(MEd,ser)/(h – tf)]
Ff,M,Ed = 0.842 × [(445 × 106)/(536.7 – 17.4)] × 10-3 = 721.53 kN

And the force in each flange due to axial force is given by;
Ff,N,Ed = 0.601 × (41/2) = 12.3205 kN

Therefore
Ftf,Ed = 721.53 – 12.32 = 709.21 kN
Fbf,Ed = 721.53 + 12.32 = 733.85 kN

The moment in the web = 0.158 × 445 = 70.31 kNm
The shear force in the web = 157 kN
The axial force in the web = 0.398 × 41 = 16.318 kN

Choice of Bolt Number and Configuration
Resistances of Bolts
The shear resistance of bolts at ultimate limit state
For M24 bolts in single shear = 132 kN
For M24 bolts in double shear = 265 kN

Flange Splice
For the flanges, the force of 1005.53 kN at ULS can be provided by 8 M24 bolts in single shear.

The full bearing resistance of an M24 bolt in a 20 mm cover plate (without reduction for spacing and edge distance) is;

Fb,max,Rd = (2.5fudt)/γm2
Fb,max,Rd = [(2.5 × 410 × 24 × 20)/1.25] × 10-3 = 393.6 kN

This is much greater than the resistance of the bolt in single shear, and thus the spacings do not need to be such as to maximise the bearing distance. Four lines of 2 bolts at a convenient spacing may be used.

Web Splice
For the web splice, consider one or two lines of 4 bolts on either side of the centreline. The full bearing resistance on the 10.8 mm web is;
Fb,max,Rd = (2.5fudt)/γm2
Fb,max,Rd = [(2.5 × 410 × 24 × 10.8)/1.25] × 10-3 = 212.54 kN

This is less than the resistance in double shear, and will therefore determine the resistance at ULS. To achieve this value, the spacings will need to be;
e1 ≥ 3d0 = 3 × 26 = 78mm
p1 ≥ 15d0/4 = (15 × 26)/4 = 97.5 mm
e2 ≥ 1.5d0 = 1.5 × 26 = 39 mm
p2 ≥ 3d0 = 3 × 26 = 78mm

Initially, try 4 bolts at a vertical spacing of 120mm at a distance of 80mm from the centreline of the splice.

e56

The additional moment due to the eccentricity of the bolt group is;
Madd = 215 × 0.08 = 17.2 kNm


Bolt forces at ULS
Force/bolt due to vertical shear = 215/4 = 53.75 kN
Force/bolt due to axial compression = 21.89/4 = 5.47 kN
Force/bolt due to moment (considering top and bottom bolts only) = (96.38 + 17.2)/0.36 = 315.5 kN
Thus, a single row is adequate.

Considering the second line of bolts at a horizontal pitch of 100mm
Force/bolt due to vertical shear = 215/4 = 53.75 kN
Force/bolt due to axial compression = 21.89/4 = 5.47 kN
Force/bolt due to moment (considering top and bottom bolts only) = (96.38 + 17.2)/0.36 = 315.5 kN

e57

The additional moment due to eccentricity of this bolt group is;
Madd = 215 × (0.08 + 0.1/2) = 27.95 kN.m

The polar moment of inertia of the blot group is given by;
Ibolts = 6 × 1202 + 8 × (100/2)2 = 106400 mm2

The horizontal component of the force on each top and bottom bolt is;
FM,horiz = {[(96.38 + 27.95) × 120]/106400} × 103 = 140.22 kN

The vertical component of the force on each top and bottom bolt is;
FM,vert = {[(96.38 + 27.95) × 50]/106400} × 103 = 58.425 kN

Therefore, the resultant force on the most highly loaded bolt is;
Fv,Ed = sqrt[(53.75 + 58.425)2 + (140.22 + 5.47)2] = 183.87 kN

This is less than the full bearing resistance, and is therefore satisfactory for such a bolt spacing.

Chosen Joint Configuration

e58

Summary of cover plate dimensions and bolt spacing

Flange cover plates; 
Thickness tfp = 20 mm
Length hfp = 660 mm
Width bfp = 200 mm
End distance e1,fp = 70 mm
Edge distance e2,fp = 30 mm

Spacing
In the direction of the force   p1,f  = 100 mm
Transverse to direction of force   p2,f  = 120 mm
Across the joint in direction of force   p1,f,j  = 120 mm

Web cover plates; 
Thickness twp = 12 mm
Length hwp = 460 mm
Width bwp = 460 mm
End distance e1,wp = 50 mm
Edge distance e2,wp = 50 mm


Spacing
Vertically   p1,w  = 120 mm
Horizontally  p2,w = 100 mm
Horizontally across the joint   p1,w,j  = 160 mm

Resistance of flange splices
Resistance of net section
The resistance of a flange cover plate in tension (Nt,fp,Rd) is the lesser of Npl,Rd and Nu,Rd.

Here;
Nu,Rd = (0.9Anet,tpfu,fp)/γM2

Where;
Anet,fp = (bfp – 2d0)tfp = [200 – (2 × 26)] × 20 = 2960 mm
Therefore;
Nu,Rd = [(0.9 × 2960 × 410 )/1.1] × 10-3 = 992.945 kN

Npl,Rd = (Afpfy,fp)/γM0 = [(200 × 20 × 275)/1.0] × 10-3 = 1100 kN

Since 1100 kN > 992.945 KN
Nt,fp,Rd = 992.945 kN

For the flange in tension;
Ftf,Ed = 989 – 16.53 = 972.47 kN

Therefore, the tension resistance of a flange cover plate is adequate.

Block Tearing Resistance;
n1,fp = 3 and n2,fp = 2;

The resistance to block tearing (Nt,Rd,fp) is given by:
Nt,Rd,fp = {(fu,fpAnt,fp)/γm2} + {[Anv,fp(fy,fp/√3)]/γm0}

Where:
Ant,fp = tfp(2e2,fp – d0)
Ant,fp = 20 × [(2 × 40) – 26] = 1080 mm2

Anv,fp = 2tfp[(n1,fp – 1)p1,fp + e1,fp – (n1,fp – 0.5)d0]
Anv,fp = 2 × 20[(3 – 1) × 100 + 70 – (3 – 0.5) × 26] =  8200 mm2

Therefore;
Nt,fp,Rd = {[(410 × 1080)/1.1] + [(8200 × 265/√3)/1.0]} × 10-3 = 1657.127 kN

Resistance of cover plate to compression flange
Local buckling resistance between the bolts need not be considered if:
p1/t ≤ 9ε
ε = sqrt(235/fy,fp) = sqrt(235/265) = 0.94
9ε = 9 × 0.94 = 8.46

Here, the maximum spacing of bolt across the centreline of the splice p1,f,j  = 120 mm
p1,f,j/tfp = 120/20 = 6

Since 8.46 > 6, no local buckling verification required.

RESISTANCE OF WEB SPLICE
Resistance of web cover plate in shear;
The gross shear area is given by:
Vwp,g,Rd = (hwptwpfy,wp/1.27γm0√3)

For two web cover plates;
Vwp,g,Rd = 2{[(460 × 12)/1.27] × [275/√3]} × 10-3 = 1380.185 kN
VEd = 215 kN. Therefore, the shear resistance is adequate.

The net shear area is given by:

e59

Vwp,net,Rd = Av,wp,net(fu,wp/√3)/γm2
Av,net = (hwp – 3d0)twp
= (460 – 3 × 26) × 12 = 4584 mm2

For two web cover plates;
Vn,Rd = {2 × [4584 × (410/√3)]/1.1} × 10-3 = 1972.9 kN
Therefore, shear resistance is adequate.

Resistance to block tearing

e60

Vb,Rd = (fu,wpAnt)/γm2 + (fy,wpAnv.Anv)/γm0√3

For a single vertical line of bolts;
Ant = twp(e2 – d0/2)
Ant = 12 × (50 – 26/2) = 444 mm2
Anv = twp – e1 – (n1 – 0,5)d0)
Anv = 12 × [460 – 50 – (3 – 0.5)26] = 4140 mm2

For two web plates
Vb,Rd = 2 × {[(410 × 460)/1.1] + [(275 × 4140)/(√3 × 1.0)]} × 10-3 = 1657.53 kN

VRd = min{1657.53,  1972.9, 1380.185 }
VRd = 1380.185 kN

VEd/VRd = 215/1380.185 = 0.1557 < 1.0
This is ok.

Resistance of beam web
Resistance of net shear area
Vn,w,Rd = [Av,net(fu/√3)]/γm2

Where;
Av,net = Av – 3d0tw
Av = A – 2btf + (tw + 2r)tf but not less than ηhwtw

Av = 12900 – (2 × 210 × 17.4) + (10.8 + 2 × 12.7) × 17.4 = 6221.88 mm2
Av = 6221.88 – (3 × 26 × 10.8) = 5379.48 mm2

Thus;
Vn,w,Rd = {[5379.48 × (410/√3)]/1.1} × 10-3 = 1157.632 kN
This is ok.

Resistance of web cover plate to combined bending, shear, and axial force

Vwp,Rd = 1380.185 kN
Vsub>Ed = 215 kN < 1380.185 kN

Therefore, the effects of shear can be neglected
Awp = 12 × 460 = 5520 mm2

Modulus of the cover plate = (12 × 4602)/6 = 423200 mm3

Nwp,Rd = 12 × 460 × 275 × 10-3 = 1518 kN

Therefore, for two web cover plates
Mc,wp,Rd = [(2 × 423200 × 275)/1.0] × 10-6 = 232.76 kNm

For two web cover plates;
Npl,Rd = 2 × 1518 = 3036 kN

Mwp,Ed = 96.38 + 27.95 = 124.33 kNm
Nwp,Ed = 21.89 kN

Mwp,Ed/Mc,wp,Rd + Nwp,Ed/Nwp,Rd
124.33/232.76 + (21.89/3036) = 0.541 < 1.0

Therefore, the bending resistance of the web cover plates is adequate.

Thank you for visiting Structville today, and God bless.


Use of Rice Husk Ash in Concrete: Literature Review

By
Ubani Obinna Uzodimma
-
1.0 Introduction
A pozzolan is a siliceous or aluminosiliceous material that, in finely divided form and in the presence of moisture, chemically reacts with the calcium hydroxide released by the hydration of portland cement to form calcium silicate hydrate and other cementitious compounds. Pozzolans are generally categorized as supplementary cementitious materials or mineral admixtures (Kosmatka et al, 2003). On their own, pozzolans possess little or no cementitious value, but when ground in fine form and in the presence of water will react with calcium hydroxide and develop cementitious properties.



Rice husk ash (RHA) has been used as a highly reactive pozzolanic material to improve the microstructure of the interfacial transition zone (ITZ) between the cement paste and the aggregate in high-performance concrete (Biu et al, 2005). Rice husk ash has also been reported to improve the properties of concrete or cement paste due to the pozzolanic reaction and its role as a micro-filler. It is often thought that the first function (pozzolanic reaction) is most important. The partial replacement of cement by rice husk ash in cement paste and mortar would provide micro-structure improvement, pore filling effect and better packing characteristics of the mixer (Kondraivendhan, 2012).
rice husks
One of the major highlights in the use RHA as cementitious material is based on environmental considerations. Rice milling industry generates a lot of rice husk during milling of paddy which comes from the fields. Rice husk ash (RHA) is about 25% by weight of rice husk when burnt in boilers.  It is estimated that about 70 million tones of RHA is produced annually worldwide. This RHA is a great environment threat causing damage to the land and the surrounding area in which it is dumped (http://www.ricehuskash.com/product.htm).
Studies have shown that RHA resulting from the burning of rice husks at control temperatures have physical and chemical properties that meet ASTM (American Society for Testing and Materials) Standard C 618-94a. At burning temperatures of 550 °C – 800 °C, amorphous silica is formed, but at higher temperatures crystalline silica is produced (Reddy and Alvarez, 2006). Amorphous silica is best for the production of concrete.  Rice husk contains about 75 % organic volatile matter and the balance 25% of the weight of this husk is converted into ash during the firing process, and is known as rice husk ash (RHA). This RHA in turn contains around 85% – 90% amorphous silica. So for every 1000 kgs of paddy milled , about 220 kgs (22 %) of husk is produced, and when this husk is burnt in the boilers, about 55 kgs (25 %) of RHA is generated (http://www.ricehuskash.com/product.htm).


2.0 Chemical Properties of RHA
According to Chandrasekhar et al (2003), the chemical composition of rice husk is found to vary from one sample to another due to the differences in the type of paddy, crop year, climate and geographical conditions. Here are samples of variation in chemical properties of RHA from various research works;
According to Habeeb and Mamoud (2010) (Malaysia);
tyue
According to Ganiron (2013) (Saudi – Arabia);
tyue2
According to Egbe-Ngu Ntui Ogork et al (2015) (Nigeria);
nty


According to Oyewumi et al (2014) (Nigeria):
NTYY
According to research done by Naveen et al (2015) (India):
rew
 Thus with all these results, you can verify that the chemical properties of RHA varies, so it is very important that before any RHA sample is used, chemical analysis should be carried out.
 
3.0 Effect of RHA on Compressive Strength of Concrete
In this section, we are going to present some research works that have been carried out by various scholars on RHA.
 
Emmanuel and Akaangee (2015) [Nigeria] collected 4.7kg of RH and weighed it using Sartorius-2 weighing scale. 1.085kg of rice hush ash was obtained after an open burning of the rice husk in a local furnace for two hours at a temperature range of 600°C to 700°C. The finely divided ash was left to cool for 24 hours inside the furnace. It was then grounded forfour minutes to obtain a finer particle size with the aid of a disc-mill, sieved manually using a 45 micro meter sieve to ensure proper fineness of the ash. From their result, the compressive strength of the concrete were as follows;
tyd
From their result,  compressive strength of cubes increases with age of curing and decreases as the percentage of rice husk ash content increases. The major setback in their research work was that 40mm x 40mm x 160mm moulds were used to prepare the concrete samples, and the target design strength was not specified, including the mix ratio adopted. But the major highlight of the research work was that maximum compressive strength was obtained at 15% replacement of OPC with RHA.


Dahiya et al (2015) [India] carried out partial replacement of grade 42.5 Portland cement with 20% RHA. In their result, they discovered that the initial setting time increased from 30 minutes to 60 minutes. The concrete samples were cast using 150mm x 150mm x 150mm mould, and the target strength was M20. The Compressive Strength of M20 (0% RHA) concrete at 3, 7 and 28 days are 14.50, 20.50 and 30.3 respectively. Whereas on replacing cement with 20% of RHA it comes out to be 13.40, 21.60 and 30.70 respectively. In the highlight of his research, water demand increased from 0.6 to 0.8 to achieve a slump 75mm – 100mm, but strength gain was almost the same at 20% replacement.
Naveen et al (2015) (India) carried out a research on the effect of RHA on compressive strength of concrete. He worked on target strengths of M30 and M60. The summary of his mix design for M30 concrete is given below;
gtyo
Grade 53 ordinary portland cement was used for this research, and water to cement ratio of 0.43 to 0.35 was used. The mould size employed was 150mm x 150mm x 150mm, and 60 specimens of this was prepared. The result of the compressive strength is given below;
rq
From his research, the maximum compressive strength was obtained at 10% replacement. The result was also the same considering M60 grade of concrete.
Oyewumi et al (2014) [Nigeria] investigated the effect of RHA in concrete, and the proportioning of the materials is as shown below;
grt
The result of the test is as given below;
t666
From the result, it can be seen that none of the partial replacement met or the control target strength of 27.47 Mpa.  The closest for 28 day strength however came at 10% partial replacement. This is in contrast with two previous results.


Bolla et al (2015) [India] carried out a research on the effect of  partial replacement of cement with RHA on concrete. The cement has been replaced by rice husk ash accordingly in the range of 0%, 5%, 10%, 15%, and 20% by weight of cement for mix. Concrete mixtures were produced, tested and compared in terms of compressive strengths with the conventional concrete. These tests were carried out to evaluate the mechanical properties for the test results of 7, 28, 60 days for compressive strengths. The cement used for this test was grade 53, and cube sizes of 150mm x 150mm x 150mm were used.
Results of the test are given below;
edd
From this result, the maximum compressive strength was attained at 10% replacement. The author reported that the RHA has a minimum silicon dioxide of 90%.
Tsado et al (2014) [Nigeria] carried out the comparative analysis of properties of some artificial pozzolana in concrete production, which included rice husk ash, corn cub ash, and sheanut shell ash. However, we will focus on the result from RHA. A little consideration from his research showed that the silicon dioxide content of the RHA was too low at 48.44% (Bida, Niger state Nigeria). The chemical analysis result of the samples is given below;
drrr
The sample was prepared as 1:2:4 mix ratio, with a water to cement ratio of 0.6. 60 number of 150mm cubes were prepared, and the result of the test is given below;
tryy
From the result, none of the partial replacement met the 28 days strength of the control mix design. The closest with considerable value came at 10% partial replacement.


Zareei et al (2017) [Iran] evaluated the durability and mechanical properties of rice husk ash as a partial replacement of cement in high strength concrete containing micro silica. The research presented resulted from various ratios of rice husk ash (RHA) on concrete indicators through 5 mixture plans with proportions of 5, 10, 15, 20 and 25% RHA by weight of cement in addition to 10% micro-silica (MS). This was compared with a reference mixture with 100% Portland cement. Tests results indicated the positive relationship between 15% replacement of RHA with increase in compressive strengths by about 20%. The optimum level of strength and durability properties generally gain with addition up to 20%, beyond that is associated with slight decrease in strength parameters by about 4.5%.
The chemical properties of the RHA used for the research is given below;
456
In the batching, 8 cubes of 100mm x 100mm x 100mm samples were prepared, and a water to cement ratio of 0.4 was maintained throughout the test. The mix design of the test is given below;
rt5
The compressive strength of the entire batch is given below;
rrr
From the result, you can see that the maximum compressive strength was obtained at 20% replacement. Note that this research is aimed at high strength concrete, and Micro-Silica (MS) has been added.
Abalaka and Okoli (2013) [Nigeria] carried out test on strength development and durability of concrete containing pre-soaked rice husk ash. The aim of the research was to determine the optimum ordinary Portland cement (OPC) replacement with RHA resulting from the reactivity of pre-soaked RHA and the effects of presoaked RHA on durability properties (coefficient of water absorption and sorptivity) of concrete at the age of 90 days.
The mix proportion of the experiment is as given below;
The characteristic chemical composition of the RHA is given below;
dff
100mm x 100mm x 100mm steel moulds were used for the experiment, and the resulting compressive strength of the mixture is as given below;
frtt
From the above test, the maximum compressive strength was achieved at 15% replacement. Note that FOSROC’s Conplast P505 (plasticizer conforrming to BS EN 934) was used to improve workability.


Ujene and Achuenu (2013) [Nigeria] carried out research to determine the compressive strength of concrete prepared with various agricultural wastes across Nigeria. To ensure uniformity of the data, the research is restricted to percentage replacement of 10%, 20% and 30% of cement with local binders using design strengths of 25 Mpa and 30 Mpa at 7, 14 and 28 days curing.
The designation of the alternative binders used in the research is given below;
de
The results of the findings are as follows;
drt
drrr
From the result, we can see that for grade 25, partial replacement of 10% exceeded the target design strength. This was also the same situation for grade 30 concrete.
Summary of research works
(1) Optimum replacement of cement with RHA occurs between 10 – 20% replacement.
(2) Performance in terms of increased compressive strength increases when plasticizers are used.
(3) Water requirement increases as the percentage of RHA increases. From the literature reviewed in this post, best performance was observed at water to cement ratio less than 0.4
(4) Plasticizer is therefore recommended when using RHA for concrete production.
References
[1] Bolla R.K.,  Ratnam M.K.M.V.,  Raju U. R. (2015): Experimental Studies on Concrete with Rice Husk Ash as a Partial Replacement of Cement Using Magnesium Sulphate Solution. IJIRST –International Journal for Innovative Research in Science & Technology| Volume 1 | Issue 7 | December 2015 ISSN (online): 2349-6010
 
[2] Bui DD, Hu J, and Stroeven P, (2005): Particle size effect on the strength of rice husk ash blended gap-graded Portland cement concrete. Cement and Concrete Composites, 27( 2005) pp357-66.
 
[3]Dahiya A., Himanshu,  Kumar N.,  Yadav D. (2015): Effects of Rice Husk Ash on Properties of Cement Concrete. International Journal of Advanced Technology in Engineering and Science Vol. No. 3, Issue No 12 ISSN 2348-7550 pp 59 – 63
[4] Egbe-Ngu Ntui Ogork, Okorie A.U., Augustine U.E. (2015): Hydrochloric Acid Aggression in Groundnut Shell Ash (GSA)-Rice Husk Ash (RHA) Modified Concrete. Scholars Journal of Engineering and Technology (SJET) ISSN 2321-435X (Online) Sch. J. Eng. Tech., 2015; 3(2A):129-133
 
[5] Emmanuel A., Akaangee N.C. (2015): Evaluation of the properties of Rice Hush Ash as a Partial Replacement for Ordinary Portland cement. International Journal of Scientific Research Engineering & Technology (IJSRET), ISSN 2278 – 0882 Volume 4, Issue 7, July 2015
 
[6] Ganiron Jr T.U. (2013): Effects of Rice Hush as Substitute for Fine Aggregate in Concrete Mixture. International Journal of Advanced Science and Technology Vol.58, (2013), pp.29-40 http://dx.doi.org/10.14257/ijast.2013.58.03
 
[7] Habeeb G.A.,  Mahmud H.B. (2010): Study on properties of rice husk ash and its use as cement replacement material. Materials Research Print version ISSN 1516-1439 Mat. Res. vol.13 no.2 São Carlos Apr./June 2010 http://dx.doi.org/10.1590/S1516-14392010000200011
 
[8] Kondraivendhan B. (2012): Strength and Flow Behaviour of Rice Husk Ash
Blended Cement Paste and Mortar. Asian Journal of Civil Engineering (BHRC) VOL. 14, NO. 3 (2013) pp 405-416
 
[9] Kosmatka, Steven H.; Kerkhoff, Beatrix; and Panarese, William C. (2003): Design and Control of Concrete Mixtures. EB001, 14th edition, Portland Cement Association, Skokie, Illinois, USA 
 
[10] Naveen S.B., Antil Y. (2015): Effect of Rice Husk on Compressive Strength of Concrete. International Journal on Emerging Technologies 6(1): 144-150(2015) ISSN No. (Print) : 0975-8364 ISSN No. (Online) : 2249-3255
[11] Oyewumi O.D., Abdulkadir T.S., and Ajibola V.M. (2014): Investigation of rice husk ash cementitious constituent in concrete. Journal of Agricultural Technology 2014 Vol. 10(3): 533-542 Available online http://www.ijat-aatsea.com ISSN 1686-9141
 
[12] Reddy D.V., Alvarez, M.B.S. (2006): Marine Durability Characteristics of Rice Husk Ash-Modified Reinforced Concrete. Fourth LACCEI International Latin American and Caribbean Conference for Engineering and Technology (LACCET’2006) “Breaking Frontiers and Barriers in Engineering: Education, Research and Practice” 21-23 June 2006, Mayagüez, Puerto Rico.
 
[13] Tsado T.Y., Yewa M., Yaman S., Yewa F. (2014): Comparative Analysis of Properties of Some Artificial Pozzolana in Concrete Production. International Journal of Engineering and Technology Volume 4 No. 5, May, 2014 ISSN: 2049-3444 pp 251 – 255
[14] Ujene A. O. and Achuenu E. (2013): Comparative Assessment of Compressive Strength of Concrete Containing Agricultural and Environmental Cementitious Wastes in Nigeria. Nigerian Journal of Agriculture, Food and Environment. Vol. 9No (4): pp 37-42
[15] Zareeia S.A., Amerib F., Dorostkarc F., Ahmadic M. (2017): Rice husk ash as a partial replacement of cement in high strength concrete containing micro silica: Evaluating durability and mechanical properties. Case Studies in Construction Materials 7 (2017) 73–81 ISSN 2214-5095 Published by Elsevier Ltd. http://dx.doi.org/10.1016/j.cscm.2017.05.001


Do Commercial Software Make Engineers Better?

By
Ubani Obinna Uzodimma
-

In the 1950s, a method of analysis involving discretization of a continuous domain into a set of discrete sub-domains called elements was developed in the field of engineering. This method called the Finite Element Analysis (FEA) is a numerical solution to problems in engineering and mathematics and has been applied extensively in structural engineering.

In the year 1965, NASA issued a request for the development of a structural analysis software tool.  The result of this was NASTRAN (NASA Structural Analysis), which implemented the available FEA technology to solve structural problems.

In the 1970s, the commercialisation of finite element analysis began in full force and gave birth to finite element modelling. Apart from FEA, design packages also sprang up within that period. For instance, DECIDE (DEsk-top Computers in DEsign) package was written by A.W. Beeby and H.P.J. Taylor in the mid-1970s, when they were both in the Cement and Concrete Association. The program was covering all aspects of design in accordance with CP110, from structural analysis to bar curtailment.

Also around 1980, OASYS software was developed by Ove Arup Partnership for Hewlett-Packard 9845S system. The program was produced to include a large number of various aspects of structural analysis and design, and other aspects like surveying, drainage, road works, thermal behavior of sections etc. The CP110 design package of the program contained programs for analysis and design of continuous beams, rectangular and irregular columns, flat slabs, foundation, etc.

While all these were going on, in 1982, Autodesk made Autocad commercially available for drafting and computer-aided designs. From that moment till now, numerous computer packages have been developed to assist engineers in analysis, design, simulation, and drafting.

We have ANSYS, Staad Pro, Etabs, SAP, Safe, STRAP, Tekla, Orion, and so many packages too numerous to mention. There are also many computer programs written with packages like FORTRAN, MS Excel VBA, MATLAB etc, which serve the purpose of helping the engineer solve a specific problem.

structural design

These packages have made life easier for design engineers in the office by considerably lowering the man-hours involved on the analysis, design, and drawing of structures. This means that working drawings, costing, and presentations can be done faster prior to kicking off of constructions. On the other hand, it has made it easier for corrections and modifications to be made to a structure, with their effects appropriately accounted for.

But the question still remains, ‘Do all these programs and packages make a design engineer better?‘

The answer is relative. An engineer is an individual who has been trained in an accredited institution, with appropriate years of experience and professional license to practice the profession of engineering. An engineer is expected to improve on the job, by following up with relevant technological advancements in the field, attending seminars and conferences, and networking with his colleagues.

For instance, a new construction procedure might demand a new method of reinforcement detailing and it expected that a modern engineer keeps up with all these developments.

Back to the issue of engineering software, if an engineer is proficient in the use of say computer programs, does that make him better than an engineer who can use just two programs? The first answer that will come to mind is probably ‘NO’. This is because the knowledge of computer programs does not make an engineer.

What makes a complete engineer is his personality, competence, knowledge, training, background on engineering practice, and years of experience. For engineers, the knowledge of software is not an end itself, but an alternative means to an end. These packages are here to help engineers work faster and more efficiently, and that is why they are programmed to request our input before they can give us an output.

I was involved in a project in Lagos Nigeria, and when I looked at the structural drawing, it was a complete waste of resources. The residential building was to go four storeys high, and no span was longer than 7m. At the ground floor, a column section of 900mm x 300mm was provided with 10Y25 (Asprov = 4910 mm2) as reinforcement. On the first floor of the same column, 28Y25 (Asprov = 13748 mm2) was provided!!

The drawing in question has been approved and construction has commenced already. Based on the design results, the designer would have asked questions. But a certain software gave him the result, and he submitted it just like that without probably reviewing the output. Even if there were unbalanced moments due to eccentricity or structural arrangement that is causing heavy bending on the column, it is his duty to scrutinize the arrangement. But he is an ‘engineer’ who can use design software, yet he has shown that he has no solid foundation in engineering as far as that output was concerned, or he doesn’t care about cost.

In terms of job opportunities and employability in the industry, the knowledge of a certain software can be an advantage to a candidate. Yes, this is because a design firm might be users of a particular computer program, and they might prefer to hire someone who already has knowledge of that software. But on the other hand, it would still be very disastrous if the desired candidate does not understand the basic theory of structures. It is easier to train someone on how to use a computer program than to train someone on the fundamentals of structural analysis and design. This calls for consideration on the part of employers.

In summary, commercial design software are helpers. We are supposed to do the thinking and design, while computer programs will help us carry out the lengthy calculations. When they produce an inconsistent result, we are supposed to ask why, and review our inputs.

Computer programs do not make better engineers, but they have made engineering as a profession better. That is why Structville remains committed to original and sound knowledge from the first principles. The current curriculum of our universities is sufficient to produce engineers who can handle simple designs after graduation, and lecturers should be willing to ‘teach’ design, and not ‘lecture’ it.

From my experience in the industry so far, the quality of teaching received in the university contributes immensely to the capacity of fresh graduates to carry out structural design independently. If they were ‘lectured‘ and not ‘taught‘, they might spend the rest of their professional career relying solely on computer programs.

Structural design at the undergraduate level should not be an issue of, “take this handout or textbook and study the examples” only. Structural design should be taught passionately from scratch and demonstrated with practical examples to the fullest extent possible.

I will ever be grateful to Prof. C. H. Aginam of the Department of Civil Engineering, Nnamdi Azikiwe University Awka. In the first semester of our 3rd year, he loaded us with knowledge on the deflection of elastic systems, analysis of continuous beams (Clapeyron’s theorem of three moments, slope deflection method, force method), Maxwell’s theorem, Betti’s law, Vereschagin’s rule, Castigliano’s theorem, analysis of portal frames, etc.

When we found ourselves mid-way in the 2nd semester, we were already designing three-storey buildings using pen and calculator from the roof beam to the foundation. I still remember the way he shouts and sweats, writing on the board, and cleaning over and over again, inspecting our notes, making sure every one downloaded and printed a copy of BS 8110-1:1997, ensuring that everyone purchased at least one design textbook of his/her choice. He is a phenomenal teacher.

Benefiting from his wealth of knowledge, I did all my designs with pen and calculator until I went for my Industrial Training in 4th year. It was during my industrial training that I had my first laptop and learned how to use AUTOCAD and a few other design software.

It is the fundamental knowledge of engineering and ingenuity that makes a good engineer; computer programs are here to make our work faster and eliminate human errors from lengthy computations. Once again, it is worth acknowledging that the availability of these design software have made engineering as a discipline better and more efficient. Let us keep getting it right.




How to Calculate the Settlement of Spread Foundation (EC7 Part 2)

By
Ubani Obinna Uzodimma
-

In this article, we will discuss an approach described in Annex E of EN 1997-2:2006 (Eurocode 7 Part 2) for determining the elastic settlement of spread foundations. This is based on a semi-empirical method and the result of the MPM test (Menard Pressuremeter Test).

The formula for evaluating the settlement of a spread footing is given below;

settlement%2Bequation


where;
Bo is a reference width of 0.6 m;
B is the width of the foundation;
λd , λc are shape factors given in Table E.2;
α is a rheological factor given in Table E.3;
Ec is the weighted value of EM immediately below the foundation;
Ed is the harmonic mean of EM in all layers up to 8 × B below the foundation;
σv0 is the total (initial) vertical stress at the level of the foundation base;
q is the design normal pressure applied on the foundation.

The Table E.2 and E.3 of EC7-2 are given below;

Table%2BE2
tABLE%2BE3

Solved Example

A 2m x 2m square footing is carrying a quasi-permanent SLS combination load of 1040 kN. The Menard Pressuremeter moduli values are as given in the sketch below. Calculate the settlement of the foundation, if the unit weight of the first layer of soil is 18.0 kN/m3.

Question

Solution

σv0 = 18 kN/m3 × 1.2m = 21.6  kN/m2
q = 1040 kN / (2m × 2m) = 260  kN/m2
B = 2.0m (width of the foundation)
λd = 1.12 (square foundation)
λc = 1.1 (square foundation)
α = 0.67 (normally consolidated clay)
Ec = 16.8 MPa
Ed = 3/[(1/16.8) + (1/27) + (1/33)] =  23.447 MPa
Substituting these values into the settlement equation;

equation

Therefore, the settlement of the foundation is 5.358mm.

What do you think about this approach?
Thank you for visiting Structville today.


Design of Timber Roof Truss

By
Ubani Obinna
-

The use of timber as trussed rafters for roofs of buildings is a very popular alternative all over the world. Timber roof trusses are made up of a series of interconnected timber members that are arranged in a triangular pattern. This pattern allows the truss to distribute the weight of the roof evenly across its supports, making it a strong and efficient structural system.

roof truss design
Figure 1: Members of a timber roof truss

The aim of this article is to show the design example of a timber roof truss (trussed rafter). As a direct product of nature, timber has so many variable properties that are more complex than that of concrete, steel, bricks, or aluminium. Some of the characteristics which influence the structural behaviour of timber are;

  • moisture content
  • the direction of applied load (perpendicular or parallel to the grain)
  • duration of loading
  • strength grading of the timber
commercial timber log
Figure 2: Unprocessed timber logs

Steps in the Design of Timber Roof Trusses

Here’s a step-by-step overview of the design process for timber roof trusses:

1. Preliminary Design: In this step, the designer is to determine the roof span, pitch, and overall architectural design of the building. It is important to choose a truss configuration based on structural performance, aesthetic preferences, functional requirements, and available space.

2. Load Analysis: In this case, it is important to identify and analyze all relevant loads, including dead loads (roofing materials, insulation), live loads (snow, occupants), and potential wind or seismic loads. Calculate the total design load that the truss system must support.

3. Material Selection: Choose an appropriate timber species based on its mechanical properties, availability, and durability. It is important to consider factors such as bending strength, compression and tension properties, and resistance to decay.

4. Truss Geometry and Configuration: Based on the chosen truss configuration (e.g., king post, queen post, Howe truss), determine the angles, lengths, and dimensions of each timber member. It is important to ensure that the truss geometry is consistent with both structural and architectural requirements.

5. Member Sizing and Design: Calculate the axial forces, bending moments, and shear forces in each timber member using static equilibrium equations. Size each timber member to withstand the calculated forces while considering factors such as serviceability and durability.

6. Connection Design: Design secure and efficient connections between timber members using appropriate fasteners (nails, screws, bolts) and connectors (metal plates, brackets). Ensure that connections provide load transfer, prevent excessive movement, and account for potential wood shrinkage.

7. Lateral Stability and Bracing: Incorporate diagonal bracing or other lateral stability measures to prevent buckling or twisting of the truss members under lateral loads (wind or seismic forces).

8. Serviceability and Deflection Control: Assess the truss deflection under the designed loads to ensure that it meets acceptable serviceability criteria. Incorporate additional members or adjust member sizes if necessary to control deflection.

9. Fire Resistance and Protective Treatments: Consider fire resistance requirements by choosing fire-rated timber or applying fire-retardant treatments if needed. Apply appropriate protective coatings or treatments to enhance the truss’s durability and resistance to decay.

10. Detailed Drawings and Documentation: Create detailed construction drawings that specify the geometry, dimensions, connections, and material details of the truss members. Prepare design calculations and documentation that outline the design methodology, load assumptions, and member sizing.

Analysis of Timber Roof Trusses

Ideal trusses are theoretical structures in which the members meet at points called nodes. These nodes are idealized as hinges or pins, which means that they cannot transmit bending moments. Loads are applied to ideal trusses only at the nodes. This keeps the truss members free from shear and bending stresses and makes the analysis of the truss much simpler.

However, practical construction does not allow roof trusses to behave exactly as ideal trusses. The members of real-world trusses are not pinned at the nodes, and loads are often applied along the length of the chords. This means that practical trusses must resist bending moments and shear in addition to axial stress.

As a result, the classical methods of analyzing trusses are only valid for ideal trusses. These methods do not account for the bending moments and shear in practical trusses. As a result, the results of classical methods of analysis can be inaccurate for real-world trusses, even though they are usually employed.

Quickly in this post, I am going to carry out a very simple design example of timber roof truss using BS 5268. A lot of information regarding timber as a structural material can be obtained from specialist textbooks. It is worth knowing that the most current design code for timber structures is Eurocode 5.

Timber%2Broof%2Btruss%2Bdesign

Note:
BS 5268 is based on permissible stress design. When using permissible stress design, the margin of safety is introduced by considering structural behaviour under working/service load conditions and comparing the stresses thereby induced with permissible values. The permissible values are obtained by dividing the failure stresses by an appropriate factor of safety. The applied stresses are determined using elastic analysis techniques, i.e.

Stress induced by working loads ≤  (failure stress/factor of safety)

Since BS 5268 is a permissible stress design code, mathematical modelling of the behaviour of timber elements and structures is based on assumed elastic behaviour.

Solved Example

Let us design the roof truss of a building subjected to the following medium-term loads. The configuration of the roof truss is as shown above.

Data
Span of roof truss = 4.8m
Spacing of the truss = 2.0m
Nodal spacing of the trusses = 1.2m
Service class of roof truss: Service class 2

Load Analysis

(i) Dead Loads
On rafter (top chord)
Self-weight of long-span aluminium roofing sheet (0.55mm gauge thickness) = 0.019 kN/m2
Weight of purlin (assume 50mm x 50mm African Mahogany hardwood timber)
Density of African Mahogany = 530 kg/m3 = 0.013  kN/m = (0.013 × 2m)/(2m × 1.2m) = 0.0108 kN/m2
Self-weight of rafter (assume) = 0.05 kN/m2
Total = 0.0885 kN/m2
Weight on plan = 0.0885 × cos 17.35 = 0.08 kN/m2

On Ceiling Tie Member (bottom chord)
Weight of ceiling (10mm insulation fibre board) = 0.077 kN/m2
Weight of services = 0.1 kN/m2
Self weight of ceiling tie = 0.05 kN/m2
Total = 0.227 kN/m2
Therefore the nodal permanent load on rafter (Gk) = 0.08 kN/m2 × 2m × 1.2m = 0.192 kN
Therefore the nodal permanent load on ceiling tie (Gk) = 0.227 kN/m2 × 2m × 1.2m = 0.5448 kN

(ii) Live Load
Imposed load on top and bottom chord (qk) = 0.75 KN/m2 (treat as medium-term load on plan)
Therefore the nodal permanent load on rafter (Gk) = 0.75 kN/m2 × 2m × 1.2m = 1.8 kN

Analysis of the rafter (top chord)
Span Length = 1.257m
Load = (0.0885 + 0.75) × 2m = 1.667 kN/m

Timber%2BTruss%2BRafter

Results
Analysis of the structure for the loads gave the following results;

on%2Bstaad


All load values are medium-term loads;

Medium-term load is defined in this case by:
Dead load + temporary imposed load

Top Chord Result
Axial force = 10.1 kN (Compression)
Bending Moment = 0.2 kNm
Length of member = 1.26 m

Design of the Top Chord
Let us try 38mm x 100mm timber
Strength class C18

Compression parallel to grain (σc,g,||) = 7.1 N/mm2
σc,adm,|| = σc,g,|| ×  k2 × k3 × k8 × k12

Bending parallel to grain (σm,g,||) = 5.8 N/mm2
σm,adm,|| = σm,g,|| ×  k2 × k3 × k6 × k7 × k8

k2 = wet exposure (does not apply in this case)
k3 = duration factor = 1.25 (medium-term loading)
k6 = shape factor = 1.0 (rectangular section)
k7 = Depth of section 72mm < h < 300mm
k7 = (300/h)0.11 = (300/100)0.11 = 1.128
k8 = Load sharing factor (does not apply since the spacing of the rafters exceed 610 mm).

Section Properties
Area = 3.8 × 103 mm2
Zxx = 63.3 × 103 mm3
Zyy = 24.1 × 103 mm3
Ixx = 3.17 × 106 mm4
Iyy = 0.457 × 106 mm4
rxx = 28.9 mm
ryy = 11 mm

Applied bending stress
σm,a,|| = M/Z = (0.2 × 106)/(63.6 × 103) = 3.144 N/mm2

Axial compressive stress
σc,a,|| = P/A = (10.1 × 103)/(3.8 × 103) = 2.657 N/mm2

Check for slenderness
Effective length (Le) = 1260 mm (assuming pin end connection)

λ = Le/r = 1260/28.9 = 43.598 < 52 Ok  (clause 2.11.4)

Medium-term load
Compression parallel to grain (σc,g,||) = 7.1 N/mm2

Emin = 6000 N/mm2

k3 = 1.25 (Table 17)

σc,|| = 7.1 × 1.25 = 8.88 N/mm2

E/σc,|| = 6000/8.88 = 675.67
Slenderness λ = 43.598

We can obtain the value of k12 by interpolating from Table 22 of the code
We are interpolating for E/σc,|| = 675.67 and  λ = 43.598

E/σc,||           40         50
600           0.774       0.692
700           0.784       0.711

On interpolating (bivariate interpolation);
k12 = 0.7545

σc,adm,|| = σc,g,|| ×  k2 × k3 × k8 × k12
σc,adm,|| = 7.1 ×  1.0× 1.25 × 1.0 × 0.7545 = 6.699 N/mm2

σm,adm,|| = σm,g,|| ×  k2 × k3 × k6 × k7 × k8
σm,adm,|| = 5.8 ×  1.0× 1.25 × 1.0 × 1.128 × 1.0 = 8.178 N/mm2

Euler critical stress σe =  π2Emin/λ2

σe =  π2(6000)/(43.598)2 = 31.154 N/mm2

For combined bending and compression

For%2Bcombined%2Bbending%2Band%2Bcompression

σm,a,|| =  3.144 N/mm2
σc,a,|| = 2.657 N/mm2
σc,adm,|| =  6.699 N/mm2
σm,adm,|| =  8.178 N/mm2
σe =  31.154 N/mm2

[3.144/(6.699 × 0.9034)] + [2.657/6.699] = 0.919 < 1.0
Therefore, 38mm x 100mm GS C18 Timber is adequate for the rafter

Consider portion over nodes (at supports)
Bending moment = 0.28 kN.m
Axial load (taking the average at that joint) = (10.81 + 8.43)/2 = 9.62 kN

Applied bending stress
σm,a,|| = M/Z = (0.28 × 106)/(63.6 × 103) = 4.40 N/mm2

Axial compressive stress
σc,a,|| = P/A = (9.62 × 103)/(3.8 × 103) = 2.531 N/mm2

At node point, λ < 5.0, and the rafter is designed as a short column at this point;

σc,adm,|| = σc,g,|| ×  k2 × k3 × k8
σc,adm,|| = 7.1 ×  1.0× 1.25 × 1.0  = 8.875 N/mm2

The interaction formula for this scenario is given below;

[σm,a,|| / σm,adm,||] + [σc,ma,|| / σc,adm,||] ≤ 1.0

[4.40 / 8.178] + [2.531 / 8.875] = 0.8232 < 1.0

This shows that the section is satisfactory for rafter.

Analysis of Tie Element
Span Length = 1.2m
Load = (0.227 + 0.75) × 2m = 1.954 kN/m

Results
Axial force = 9.74 kN (tension)
Bending Moment = 0.22 kNm
Length of member = 1.2m

Design of the Bottom Chord (ceiling tie)
Let us still try 38mm x 100mm timber
Strength class C18

Tension parallel to grain (σt,g,||) = 3.5 N/mm2
σt,adm,|| = σt,g,|| ×  k2 × k3 × k8 × k14
(width of section) k14 = (300/h)0.11 = (300/100)0.11 = 1.128
σt,adm,|| = 3.5 × 1.0× 1.25× 1.0 × 1.128 = 4.935 N/mm2

Bending parallel to grain (σm,g,||) = 5.8 N/mm2
σm,adm,|| = 5.8 ×  1.0× 1.25 × 1.0 × 1.128 × 1.0 = 8.178 N/mm2

Applied bending stress
σm,a,|| = M/Z = (0.22 × 106)/(63.6 × 103) = 3.459 N/mm2

Axial tensile stress
σc,a,|| = P/Effective Area = (9.74 × 103)/(3.8 × 103) = 2.563 N/mm2

Note: When the ceiling tie is connected to rafter by the means of a bolt, the projected area of the bolt hole must be subtracted from the gross area of the section.

Combined tension and bending
[σm,a,|| / σm,adm,||] + [σt,ma,|| / σt,adm,||] ≤ 1.0

[3.459 / 8.178] + [2.563 / 4.935] = 0.9422 < 1.0

This is ok.

Consider portion over nodes (at supports)
Bending moment = 0.3 kN.m
Axial load (taking the average at that joint) = (9.5 + 9.74)/2 = 9.62 kN

Applied bending stress
σm,a,|| = M/Z = (0.3 × 106)/(63.6 × 103) = 4.7169 N/mm2

Axial tensile stress
σc,a,|| = P/Effective Area = (9.62 × 103)/(3.8 × 103) = 2.531 N/mm2

Combined tension and bending
[σm,a,|| / σm,adm,||] + [σt,ma,|| / σt,adm,||] ≤ 1.0

[4.7169/ 8.178] + [2.531 / 4.935] = 1.089

In this case, the tie element may be increased to 38mm x 175mm or the grade of the timber could be changed to accommodate the combined flexural and axial stress in the member.

Check for deflection
Deflection of trussed rafter under full load = 6.095 mm (calculated on Staad)
Permissible deflection = 14 mm

Deflection is ok.

That is it for now. Thank you so much for visiting Structville today and God bless you. Remember to share with your folks.

Member Design of a 22m Span Steel Roof Truss

By
Ubani Obinna Uzodimma
-

Introduction

We all need a roof over our head. Nowadays, architects frequently use roof pattern and design to enhance the aesthetics and functionality of a building, and it is very important that engineers follow up and ensure that the structural integrity of such roof systems are guaranteed. On the 10th of December 2016, the roof a church building collapsed at Uyo in Akwa Ibom state Nigeria, leaving about 60 people dead, and many other injured. This is why such designs are to be taken seriously, especially when the span is large.

truss

Large span construction of roof systems is usually found in public or commercial buildings. This comes in handy especially when there is need to have a large uninterrupted floor area. However, the efficiency of trusses in dealing with long span structures has been widely recognised in the structural engineering community, but there is no doubt that the longer the span, the more complex the design.



Some basic considerations in the design of trusses
(1)The connection design is as important as the member design. The connections could be welded or bolted or both.
(2) 2D idealisation of the structure is usually very sufficient for analysis and design.
(3) Joints are hardly pinned in reality, but the use of pinned joint is usually encouraged for the purpose of analysis and design.
(4) Trusses can be analysed as having continuous top and bottom chords, with the internal members pinned. In this case, loads can be applied at locations other than the joints with the resulting bending moment appropriately accounted for in the design.
(5)  It is usually important to consider out of plane buckling for compression members in trusses. However, purlins usually take care of the top chord, while bracing can be used to take care of the bottom chord.

Read Also…
Design of Roof Purlins

(6) For efficient structural performance, it is recommended that the truss span to depth ratio be kept between 10 to 15.
(7) In the layout of truss systems, it is more preferable (in terms of economy and efficiency) to have the shorter members in compression and the longer members in tension.
(8) Welded connections offer more advantages in terms of  deflection behaviour of trusses. Slack displacements are possible in bolted connections especially when non-preloaded bolts are used.
(9) In the design of compression members, buckling is the most critical.
(10) When member lines do not intersect at a node, it is important that the moment that arises due to the eccentricity be included in the design.


Design Example
The structural layout of a roof system is given below;

roof%2Blayout

We are required to carry out the member design of the trusses for this 23m span open hall. The design data is given below;

Data
Design code: BS EN 1993-1-1:2005
Design wind pressure: 1.5 kN/m2
Variable load = 0.75 kN/m2

To make this post simpler and shorter, a 29 page fully referenced design paper for all the trusses in the roof system has been attached.

Click HERE to download the calculation sheet in PDF format (premium).

Thank you for visiting Structville today, and remember to tell your colleagues about us.

truss33

Read Also...
Practical Analysis and Design of Roof Trusses

Structural Analysis of Free-Standing Staircase

By
Ubani Obinna Uzodimma
-

Free-standing staircases offer a very pleasing solution for vertical circulation in residential and commercial buildings. They are usually constructed in such a way that the landing is freely supported, while the supports at the flight are fully fixed. It is possible to have the flights supported in other ways, but that makes the analysis more complicated. Freestanding staircase structures are complex in analysis and design, but with finite element analysis packages, simple solutions can be easily obtained as shown in this post.

The most widely available manual solution for the analysis of free-standing staircases was developed by Cusens and Kuang using the strain-energy principles. Expressions were developed relating the horizontal restraint force H and the bending moment Mo at the mid-point of the free-standing stair. By solving both equations simultaneously, the values of Mo and H can be substituted into the general expressions to obtain the bending moments and forces at any point in the structure.

In this article, we are going to compare the results obtained with Staad Pro software with results from manual analysis using the method proposed in Table 175, Reynolds and Steedman, 2005.

Dimensions

FREE%2BSTANDING%2BPLAN
SECTION


Solved Example
The geometry of a free-standing staircase is given below. We are expected to analyse the staircase for the ultimate moment using the formula given in Reynolds and Steedman (2005) and compare the answer with the result from Staad Pro.

QUESTION

Data
Thickness of waist of flight and landing = 250 mm
Depth of riser = 150mm
Unit weight of concrete = 25 kN/m3

Actions on the stairs
Concrete self weight (waist area) = 0.25 × 25 = 6.25 kN/m2 (normal to the inclination)
Stepped area = 1⁄2 × 0.15 × 25= 1.875 KN/m2 (global vertical direction)
Finishes (say) = 1.2 kN/m2

We intend to apply all gravity loads purely in the global y-direction, therefore we convert the load at the waist of the stair from local to global direction by considering the angle of inclination of the flight area to the horizontal;

γ = tan-1⁡(1.5/3) = 26.565°

Therefore UDL from waist of the stair in the global direction is given by = (6.25 × cos 26.565) = 5.59 kN/m2

Total permanent action on flight area (gk) = 5.59 + 1.875 + 1.2 = 8.665 kN/m2
Total permanent action on landing; (gk) = 6.25 + 1.2 = 7.45 kN/m2

Variable load on staircase (qk) = 4 kN/m2

The load on the flight area at ultimate limit state = 1.35gk + 1.5qk
nf = 1.35(8.665) + 1.5(4) = 17.67 kN/m2

The load on the landing at ultimate limit state = 1.35gk + 1.5qk
nl = 1.35(7.45) + 1.5(4) = 16.06 kN/m2

From Table 175, Reynolds and Steedman (2005), the approximate formula for calculating the critical design moments for free-standing stairs with the flights fully fixed is given below;

H
MO
K

From the given question;
Load on flight nf = 17.67 kN/m2
Load on landing nl  = 16.06 kN/m2
Thickness of flight hf = 250 mm
Thickness of landing hl = 250 mm
a = 3.35m
b = 1.4m
b1 = 2.0m
γ = 26.565°

Plugging these values into the equations above;
K = 0.746
H = 222.637 kN
M0 = 70.541 kNm

Comparing the above answer with ultimate limit state answer from Staad Pro;

Longitudinal Moment

MY

You can see that M0 from Staad Pro is 65.1 kNm. This about 8.3% less than the anser gotten from manual analysis, and further reinforces the fact that finite element analysis approach to this problem yields a more economical result.

Transverse Moment
The moment in the x direction due to ultimate load is given below;

The maximum moment in the x-direction can be found to be 45.5 kNm. Cusens and Kuang (1966) recommends that the transverse reinforcement be concentrated in the vicinity of the flight and the landing. This results offers a good insight.

Torsion
The twisting moment on the staircase due to the load is given below;

Torsion

A little consideration of the above result will show that considerable twisting is occurring at the mid-span section of the flights. This completely agrees with the conclusions made by Cusens and Kuang (1966). In their own words,

“Large torsional moments are present in the flights of free-standing stairs and a proper thickness of concrete must be chosen to resist these moments, due to the difficulty of reinforcing shallow-wide sections against torsion.”

Cusens and Kuang (1966)


Longitudinal Shearing Stresses

SY

As you can see, the maximum longitudinal stress is occurring at point O with a value of 1.64 N/mm2.

We are undertaking further studies on the dynamic behaviour of free standing stairs. We will update in due time. Thank you for visiting, and God bless you.

References
Cusens A.R., Jing Gwo Kuang (1966): Experimental Investigation of Free Standing Stairs. Journal of the American Concrete Institute, Proceedings V. 63, No. 5, May 1966.

Reynolds C.E., Steedman J.C (2005): Reinforced Concrete Designers Handbook. Spon Press, Taylor and Francis Group, London ISBN 0-419-14540-3

Introduction to Structville Research

By
Ubani Obinna Uzodimma
-

We are excited to announce that we have recently added a unique new page to our blog called ‘Structville Research‘. You can visit the page by clicking the link below;

Structville Research is a unique page dedicated to download of research articles, textbooks, design/analysis calculations, spreadsheets, drawings, and other useful educational resources and materials. Currently, we have about 10 articles waiting for your free download! However, some contents are premium, and purchases can be made easily on the spot through Paystack (see www.paystack.com).
Structville Research will be updated on regular basis, and we encourage you to always visit the page for the latest article. From Structville’s home page, you can easily find Structville Research at the menu bar and at the side bar.
Furthermore, if there is any article, drawing, or spreadsheet that you have the copyright permission to distribute, and you will like to have it featured on Structville Research, kindly let us know by sending an e-mail to info@structville.com.
We look forward to seeing you at Structville, where passion for excellence reflects in everything. Thank you, and enjoy the rest of your day. God bless.
1...565758...66Page 57 of 66

EDITOR PICKS

POPULAR POSTS

POPULAR CATEGORY

ABOUT US

Structville is a media channel dedicated to civil engineering designs, tutorials, research, and general development. At Structville, we stop at nothing in giving you new dimensions to the profession of civil engineering.

Contact us: info@structville.com

FOLLOW US

© (2016 - 2024) Structville Integrated Services Limited. All rights reserved