Unit - 4
Design of Flexural Members
The design of beam consist of selecting a section based on plastic sectional modulus and checking for its shear capacity deflection, web crippling and web buckling. Most of equation available in IS 800-2007
Design bending strength
1) If V> 0.6Vd
Or
V< 0.6Vd
Then
For simply supported beams
Where,
2) If V>0.6Vd
Then Md= Mdv
Where
Md= Design of bending strength under high shear
For plastic and compact sections
=
1. Laterally Restrained Beams
Beams subjected to BM develop compressive and tensile forces and the flange subjected to compressive forces has the tendency to deflect laterally. This out of plane bending is called lateral bending or buckling of beams. The lateral bending of beams depends on the effective span between the restraints, minimum moment of inertia (Iyy) and its presence reduces the plastic moment capacity of the section.
Beams where lateral buckling of the compression flange are prevented are called laterally restrained beams. Such continuous lateral supports are provided in two ways:
i) The compression flange is connected to an RC slab throughout by shear connectors.
Ii) External lateral supports are provided at closer intervals to the compression flange so that it is as good continuous lateral support. Cl 8.2.1 (pg - 52 and 53) gives the specifications in this regard.
Design of such laterally supported beams are carried out using Clauses 8.2.1.2, 8.2.1.3, 8.2.1.5, 8.4, 8.4.1, 8.4.1.1, 8.4.2.1 and 5.6.1 (Deflection) In addition, the beams shall be checked for vertical buckling of web and web crippling. The design is simple, but lengthy and does not involve trial and error procedure.
Design steps:
- Calculate the factored load and the maximum bending moment and shear force
- Obtain the plastic section modulus required
Zp = (M*Vmo /Fy)
A suitable section for the beam-ISLB, ISMB, ISWB or suitable built-up sections (doubly symmetric only). (Doubly symmetric, singly symmetric and asymmetric- procedures are different)
3. Check for section classification such as plastic, compact, semi-compact or slender. Most of the sections are either plastic or compact. Flange and web criteria.
4. Calculate the design shear for the web and is given by
5. Calculate the design bending moment or moment resisted by the section (for plastic and compact)
6. Check for buckling
7. Check for crippling or bearing
8. Check for deflection
2. Laterally Unrestrained Beams
Beams subjected to BM develop compressive and tensile forces and the flange subjected to compressive forces has the tendency to deflect laterally. This out of plane bending is called lateral bending or buckling of beams. Lateral buckling of beams involves three kinds of deformations namely lateral bending, twisting and warping. The lateral bending of beams depends on the effective span between the restraints, minimum moment of inertia (Iyy) and can reduce the plastic moment capacity of the section.
The value of Mer can be calculated using the equations given in cl. 8.2.2.1 pg-54 for doubly symmetric c/s and annex E (pg 128-129) for c/s symmetrical about the minor axis. The design bending compressive strength can be calculated using a set of equations as specified in cl.8.2.2 (Table 13a and 13 pg-54 to 57.
Design steps:
1) Maximum shear force
Maximum bending force
2) Selecting of section
Assuming fbd = 130 N/mm^2
Properties:
Bf , tf , tw
Rmin
Zp
3) Classification of cross section
Flange
Web
Section is plastic or semi plastic
4) Design shear strength
Strength in bending Md need not be reduced due to shear
5) Design bending strength
6) Check Deflection limit
Key takeaways:
- Beams subjected to BM develop compressive and tensile forces and the flange subjected to compressive forces has the tendency to deflect laterally.
- This out of plane bending is called lateral bending or buckling of beams. Lateral buckling of beams involves three kinds of deformations namely lateral bending, twisting and warping.
1. Laterally restrained beams Design steps:
- Calculate the factored load and the maximum bending moment and shear force
- Obtain the plastic section modulus required
Zp = (M*Vmo /Fy)
A suitable section for the beam-ISLB, ISMB, ISWB or suitable built-up sections (doubly symmetric only). (Doubly symmetric, singly symmetric and asymmetric- procedures are different)
3. Check for section classification such as plastic, compact, semi-compact or slender. Most of the sections are either plastic or compact. Flange and web criteria.
4. Calculate the design shear for the web and is given by
5. Calculate the design bending moment or moment resisted by the section (for plastic and compact)
6. Check for buckling
7. Check for crippling or bearing
8. Check for deflection
Examples:
1) A simply supported beam of effective span 4 m carries a factored point load of 350 kN at mid span. The section is laterally supported throughout the span. Design the cross section using I section.
Solution:
- Maximum shear force
V= reaction = W/2 = 1575 kN
2. maximum bending moment
M= WL/4= 350 kn-m
3. selection of cross section (using yielding criteria)
Zp required =
Select an ISLB 500@75kg/m (using annex H Is 800-2007)
Sectional properties d=500 mm, Izz= 38549 x 1064 mm^4
Zp= 1773.7x 10^3 mm^3 > Zp required
Ze= 1545.2 X 10^3 mm^3
4. classification of section
5. Design shear strength
6. Design bending strength
7. Checks
- Web buckling
At a concentrated load W
Buckling class C
From table 9© IS 800-2007 and fy= 250 Mpa
Kl/r | Fcd |
130 | 74.3 |
131.78 | ? |
140 | 66.2 |
Interpolation
Fcd= 72.85 N/mm^2
At concentrated point load
At reaction R
B. Web crippling
C. Deflection limits
Working load = 350/1.5 = 233.33kN
Provide an ISLB 500@ 75 kg/m beam
Key takeaways:
- Maximum shear force
- Maximum bending moment
M= WL/4
3. selection of cross section (using yielding criteria)
Zp required =
4. classification of section
5. Design shear strength
6. Design bending strength
7. Checks
Web buckling
Web crippling
Deflection limits
In beam B.M is very less near to the simple support therefore, no need to provide same size of flange plate. To achieve economy flange thickness are curtailed near to the support called curtailment of flange.
Its position from support is calculating by equating the B.M at that section to the design bending strength at the section.
The shear verification affects bending and axial resistances according to EN 1993-1-1. The "low shear" is defined by the following condition:
Where,
Vsd= The shear force
Vpl,rd = The shear resistance
Otherwise, "High shear" appears. If there are shear forces because of torsion within side the cross-section, the fee of resistance Vpl, T ,Rd (decreased fee because of torsion, defined within side the chapter "Shear pressure because of torsion") is used as opposed to the resistance Vpl,Rd. The layout values of the axial and bending resistances are decreased for "High shear". The discount is primarily based totally at the discount aspect for yield power consistent with the subsequent formula
Where
= the reduction factor mixing an allowance for the presence of shear force
The reduction factor is given by formula
1. Web buckling:
The web of the beam is thin and can buckle under reactions and concentrated loads with the web behaving like a short column fixed at the flanges. The unsupported length between the fillet lines for sections and the vertical distance between the flanges or flange angles in built up sections can buckle due to reactions or concentrated loads. This is called web buckling.
For safety against web buckling. The resisting e shall be greater than the reaction or the concentrated load. It will be assumed that the reaction concentrated load is dispersed into the web at 45" as shown in the Fig.
Let Resisting force F
Thickness of web = t
Design compressive stress in web =fcd
Width of bearing plate= b1
Width of dispersion = n1
For concentrated loads, the dispersion is on both sides and the resisting force can be expressed as, Fwb = [(b₁ +2 n₁) tw fcd] ≥ Concentrated load, P
The design compressive stress fd is calculated based on a effective slenderness ratio of
Where,
d = clear depth of web between the flanges
Ry = radius of gyration about y-y axis and is expressed as,
Design compressive stress in web, fcd for the above slenderness ratio is obtained from curve, C (Buckling class C) (Table 9c, pg 42)
2. Web crippling:
Web crippling causes local crushing failure of web due to large bearing stresses under reactions at supports or concentrated loads. This occurs due to stress concentration because of the bottle neck condition at the junction between flanges and web. It is due to the large localized bearing stress caused by the transfer of compression from relatively wide flange to narrow and thin web. Web crippling is the crushing failure of the metal at the junction of flange and web. Web crippling causes local buckling of web at the junction of web and flange.
For safety against web crippling, the resisting force shall be greater than the reaction or the concentrated load. It will be assumed that the reaction or concentrated load is dispersed into the web with a slope of 1 in 2.5 as shown in the Fig.
Let Resisting force = Fwc
Thickness of web = tw
Yield stress in web = fyw
Width of bearing plate = b₁
Width of dispersion = n₂
For concentrated loads, the dispersion is on both sides and the resisting force can be expressed as,
3. Deflection
A beam may have adequate strength in flexure and shear and can be unsuitable if it deflects excessively under the service loads. Excessive deflection causes problems in the functioning of the structure. It can harm floor finishes, cause cracks in partitions and excessive vibrations in industrial buildings and pending of water in roofs. Cl.5.6.1, 5.6.1.1 and Table 6 gives relevant specifications with respect to deflection.
The beam size may have to be taken based on deflection, if the spans and loadings are large. Typical maximum deflection formulae for simple loadings are given below:
Key takeaways
- A beam may have adequate strength in flexure and shear and can be unsuitable if it deflects excessively under the service loads.
- Excessive deflection causes problems in the functioning of the structure.
- It can harm floor finishes, cause cracks in partitions and excessive vibrations in industrial buildings and pending of water in roofs.
Design steps:
1) Maximum shear force
Maximum bending force
2) Selecting of section
Assuming fbd = 130 N/mm^2
Properties:
Bf , tf , tw
Rmin
Zp
3) Classification of cross section
Flange
Web
Section is plastic or semi plastic
4) Design shear strength
Strength in bending Md need not be reduced due to shear
5) Design bending strength
6) Check Deflection limit
Examples:
Q.1 Design a suitable I section for and simply supported beam of span 5m carrying a dead load of 20 kN/m and imposed load of 40 kN/m. The beam is laterally unsupported throughout the span. Take fy= 250 MPa
Solution:
1] Maximum shear force
V= WL/2
Maximum bending force
Total factored load W= 1.5 x(40+20) = 90 kN/m
2] Selecting of section
Assuming fbd = 130 N/mm^2
Properties:
Bf = 250 mm
Tf = 14.7 mm
Tw= 9.9 mm
Rmin= 49.6 mm
Zp = 2351.35 x 10^3 mm^3
3] Classification of cross section
Flange
Web
Section is plastic
4] Design shear strength
Strength in bending Md need not be reduced due to shear
5] Design bending strength
Effective length KL= 5000mm
As ends are restrained against torsion but compression flange is laterally unsupported
KL/rmin = 5000/49.6 = 100.86
Using table 14 of IS 800-2007
KL/rmin h/tw | 30 | 34.01 | 35 |
100 | 322.9 | A | 309.3 |
100.86 |
| X |
|
110 | 270.9 | B | 257.3 |
By interpolation
A= 311.99 N/mm^2
B= 259.99 N/mm^2
For X,
h/tf = 34.01 and KL/rmin = 100.86
X= fcr,b
Using Table 13 (a ) of IS 800-2007
Fbd ?
Fcr,b | Fbd |
350 | 172.7 |
307.51 | ? |
300 | 163.6 |
By interpolation
6] Deflection limit
Key take ways:
- Maximum shear force
Maximum bending force
2. Selecting of section
Assuming fbd = 130 N/mm^2
Properties:
Bf , tf , tw
Rmin
Zp
3. Classification of cross section
Flange
Web
Section is plastic or semi plastic
4. Design shear strength
Strength in bending Md need not be reduced due to shear
5. Design bending strength
6. Check Deflection limit
References:
- Design of Steel Structure, N Subramanian, Oxford University Press, New Delhi
- Limit State Design in Structural Steel, M. R. Shiyekar, PHI, Delhi
- Fundamentals of structural steel design, M L Gambhir, Tata McGraw Hill Education Private limited, New Delhi.
- Limit State Design of Steel Structure, Ramchandra & Gehlot, Scientific Publishers, Pune
- Analysis and Design: Practice of Steel Structures, Karuna Ghosh, PHI Learning Pvt. Ltd. Delhi
- Structural Design in Steel, Sarwar Alam Raz, New Age International Publisher
- Limit State Design of Steel Structure, V L Shah & Gore, Structures Publication, Pune
- IS Codes :- IS 800-2007: Code of practice for general construction in steel, Bureau of Indian Standards, New Delh
Case Study
Design of Laterally Restrained Beams
Beams are structural individuals that assist transverse masses and are consequently subjected in general to flexure, or bending. Beams can be cantilevered, absolutely supported, constant ended or continuous. If a sizeable quantity of axial load is likewise present, the member is referred to as a beam–column. Although a few diploma of axial load could be found in any structural member, in lots of sensible conditions this impact is negligible and the member may be treated as a beam. The resistance of a metallic beam in bending relies upon at the go section resistance or the incidence of lateral instability.
As with a compression member, instability may be in an ordinary shape or it may be local. Overall buckling happens whilst a beam bends, the compression region (above the neutral axis) has similarities to a column, and in a way much like a column, it's going to buckle if the member is slim enough. Unlike a column, however, the compression part of the go segment is restricted via way of means of the anxiety portion, and the outward deflection (flexural buckling) is followed via way of means of twisting (torsion). This shape of instability is referred to as lateral-torsional buckling (LTB). Lateral torsional buckling may be averted via way of means of bracing the beam in opposition to twisting at sufficiently near intervals. This may be achieved with both of forms of balance bracing: lateral bracing and torsional bracing.
Design of Laterally Unrestrained Beams
A cantilever beam is absolutely constant at one cease and loose on the other. In the case of cantilevers, the help situations within side the transverse aircraft have an effect on the instant pattern. For layout purposes, it's miles handy to apply the idea of notional powerful period, k, which could consist of each loading and help effects. The notional powerful period is defined because the period of the notionally certainly supported (within side the lateral aircraft) beam of similar section, which could have an elastic essential second below uniform second same to the elastic essential second of the real beam below the real loading situations.
Recommended values of ‘k’ for some of instances are given in Table. It may be seen from the values of ‘k’ that it's miles greater powerful to save you twist on the cantilever side rather than the lateral deflection. Generally, in framed structures, non-stop beams are furnished with overhang at their ends. These overhangs have the traits of cantilever beams. In such instances, the form of restraint furnished on the outermost vertical help is maximum significant. Effective prevention of twist at this area is of specific importance.