Unit 4 | unit 4 design of flexural members

DSS

Unit - 4

Design of Flexural Members

Q1) What is Design bending strength?

A1)

The design of beam consists 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

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

Q2) Explain laterally restrained and unrestrained beams?

A2)

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

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

Q3) Explain Design of laterally restrained beams using single rolled steel section with and without flange plate?

A3)

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

Q4) 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.

A4)

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

Q5) What is curtailment of flange plates?

A5)

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.

Q6) Explain Low and high shear?

A6)

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

Q7) Explain Check for web buckling, web crippling and deflection?

A7)

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 an 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:

Q8) Explain Design of laterally unrestrained beams using single rolled steel section, check for and deflection?

A8)

Design steps:

1) Maximum shear force

Maximum bending force

2) Selecting of section

Assuming fbd = 130 N/mm^2

Properties:

Bf, tf, tw

Rmin

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

Q9) 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

A9)

1] Maximum shear force

V= WL/2