Industrial Roof Trusses: Understanding and Designing

The industrial buildings such as godowns and factory floors are often low rise structures with few or none internal walls. In such buildings, special Care needs to be taken while designing industrial roof trusses, since large spans need to support the entire roofing system without intermittent support. Trusses with roof covering materials make up of the entire roofing assembly here.

What are Trusses?
Trusses are triangular formation of metal sections, usually used to span large lengths in space instead of solid girders. The external load apply mostly axial forces on the members in a truss. Depending upon how the force is applied, trusses can be designed in the following two ways:

Plane Trusses: where the external load is placed on the plane of the truss.

Space Trusses: where the external load can be applied to any three-dimensional space within.

How are Trusses Built?
Trusses mostly consist of axially loaded members to support loads. The reason for this that when steel members are subjected to axial forces, they perform better in bearing that load, than members that are in flexure. This is because the cross-section of such a system is uniformly stressed under axial forces.

Trusses are very common in most architecture. Mostly used to span long distances, they are well suited to bear the load of single-storey industrial buildings. They can also be designed to bear gravity loads in long span floors. For the same reason they are also mounted to bear loads of long span bridges.

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Published By
Rajib Dey
www.constructioncost.co
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Some useful tips to design a T-beam with the use of limit state method

This construction video briefly describes how to create the design of t-beam with the use of limit state method.

A T-beam (or tee beam) is applied in construction extensively. It is a structural element developed with reinforced concrete, wood or metal. It contains a t-shaped cross section that is formed by a stem and a flange of reinforced concrete or rolled metal.

The top of the t-shaped cross section acts as a flange or compression member in withstanding compressive stresses. The web (vertical section) of the beam underneath the compression flange contributes to withstand shear stress and offer better separation for the coupled forces of bending.

A T-beam has the capacity to tolerate large loads by providing resistance in the beam or by internal reinforcements. It operates similar to singly reinforced beam.

In order to enhance the structural strength of a T-beam, just utilize an inverted T-beam together with a floor slab or bridge deck linking the tops of the beams. If it is accomplished perfectly, the slab performs as the compression flange.

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Published By
Rajib Dey
www.constructioncost.co
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How to measure crackwidth caused by bending

Cracking in reinforced concrete structures normally occurs because of various factors like bending, shear, twisting, axial tension, and restraint from movement. As the tensile strength of concrete is weak, so the cracking is inevitable. Cracking normally stands for a serviceability limit state problem. Besides, damaging concrete, the cracking also leads to stability issues, and leakage problem in water retaining structures.

Cracking is supposed to take place in a concrete section when the restraint strain surpasses the tensile strength ability of the concrete. It signifies that when cracking takes place, some part or the entire of the concrete section should be in tension. Crackwidth in concrete is estimated by multiplying crack inducing strain (strain dispersed by the incidence of cracking) with the crack spacing.

The strain that produce crack because of flexure (bending) is demonstrated in expression 7.9 of EN 1992-1-1 as;

where;
= denote strain in reinforcement.
= denote strain in the concrete among cracks.
= stress in the reinforcement depending on cracked section properties under quasi permanent load combination 
αe = modular ratio, Es/Ecm (normally a value of 7 may be used.)
kt = 0.6 for short term loading and
0.4 for long term loading.
fct,eff = fctm at 3 days and/or 28 days.
ρp,eff = As/Ac,eff (this is computed for each face)

Where; As = area of reinforcement provided, mm2
Ac,eff = Area of concrete in tension whose depth is:
min[0.5h , 2.5(c + 0.5ϕ) , (h – x)/3] for each face of a wall

where;
h = thickness of wall
c = nominal cover
ϕ = bar diameter
x = depth to neutral axis
d = effective depth.
Es = elastic modulus for reinforcement = 200,000 MPa

Solved Example
Workout the crackwidth because of externally applied load on a 200 mm thick slab with the following data;
Design serviceability bending moment = 14.96 kNm/m
Area of tension reinforcement provided at ultimate limit state As1 = H12@200 c/c (Asprov = 565 mm2/m)
Concrete cover = 25 mm
Effective depth d = 200 - 12/2 - 25 = 169 mm

To perform the calculation in excel, download the following excel sheet drive.google.com



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Published By
Rajib Dey
www.constructioncost.co
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