chapter 4- beams design(5)

8/18/2019 CHAPTER 4- Beams Design(5)

1/31

CHAPTER 4: BEAM DESIGN

0 Steel Beam Design

0 Timber Beam Design

Jan - May 2016 1


2/31

Beam

The most frequently used and possibly the earliest used,

structural element is beam.

Loading can be imposed on a beam from one or several of a

number of sources , e.g. other secondary beams, columns,walls, floor systems or directly from installed plant or

equipments.

Jan - May 2016 2


3/31

Steel Beam

Jan - May 2016 3


4/31


5/31

Jan - May 2016 5


6/31


7/31


8/31

Figure 5-1 Flow chart of a beam designJan - May 2016 8


9/31


10/31

The d/t ratio also needs to be checked due to thestrength of the web will be governed by itsresistance to shear buckling if it is too high.

Provided in cl. 4.2.3. for a rolled section, the webshould be checked for shear buckling when d/t exceeds 70√(275/py) where the d is the web depthclear of the fillets.

Average shear stress on web = shear force/webarea = V/ (D x t)

Jan - May 2016 10


11/31

Shear failure in concrete beam Failure in steel beam

Jan - May 2016 11


12/31

Moment Capacity

Mult = y x S

where Mult = ultimate moment

y = ultimate yield stress

S = plastic section modulus

Jan - May 2016 12


13/31

Table 1: Universal Steel Beams (Dimension and Properties)

Jan - May 2016 13


14/31

Example 1

0 A 254 x 146 x 31 kg/m universal beam is used as a 3 m long

cantilever. What is the maximum ultimate load that can beapplied at the end of the cantilever if the steel yields at a

stress of 275 N/mm2?

Jan - May 2016 14


15/31

Solution

Mult = y x S

= 275 x 396/103

= 108.9 kNm

Mult = 3 x F 3F = 108.9

F = 36.3 kN

Jan - May 2016 15


16/31

A single span beam is simply supported between two columns and carries

a reinforced concrete slab as shown in Figure below. Using the design load indicated, determine the dimensions of suitable standard universal beam.

Assume S275 steel.

Example 2

13.7 kN/m

6m

a) Universal Beam SectionJan - May 2016 16


17/31

Solution :

a) Design a suitable universal beam section

Solution

13.72kN/m

A B

6 m

Step 1: Determine reaction at support

Reaction R A = R B= (13.72 x 6)/2 = 41.2 kNMmax = wL

2 /8 = 13.72 x 62 /8 = 61.74 kNm

Jan - May 2016 17


18/31


19/31

Solution :

Step 4 : CalculationA typical yield stress for mild steel is 275 N/mm2

S = Mult / y = 61.74 x 106/ 275 x 103 = 224.5 cm3

Try 254 x 102 x 22 kg/m universal beam

( S = 2 6 2 c m3, D = 254 mm, t = 5.8 mm)

Average shear stress = 41.2 x 103/ 254 x 5.8 = 28 N/mm2 ( < 165

N/mm2)

Use 254 x 102 x 22 kg/m universal beam

Example 2 : Full Lateral Restraint Beam

Jan - May 2016 19


20/31

A simply supported 406 x 178 x 74 UB S275 is required tospan 4.5 and to carry an ultimate design load of 40 kN/m.Check the suitability of the section with respect the shear.

Solution :Section properties : 406 x 178 x 74 UB S275

t= 9.7 mm, D= 412.8 mm, d= 360.4 mm, d/t = 37.2

Design shear force at the end of beam Fv = (40 x 4.5)/2= 90kN

Clause 4.2.3: Pv = 0.6 yAvFor a rolled UB section Av = tD = 9.7 x 412.8 = 4 004 mm

2

Example 3 : Shear check of a Simply SupportedBeam

Jan - May 2016 20


21/31


22/31


23/31


24/31

Timber Beam

Space Beam

• Blocked and securely nailed at

frequent intervals to enable

individual member to act as an

integral unit.

Glue Laminated Timber

Jan - May 2016 24


25/31

Timber Joists

Small timber beams are often used in the floor construction ofhouses, where they are known as joists.

Jan - May 2016 25


26/31

Timber Joists

Jan - May 2016 26


27/31

TIMBER JOISTS

Jan - May 2016 27


28/31

Timber Joists

Jan - May 2016 28


29/31


30/31

Solution:

Total load = 0.45 + 1.5 = 1.95 kN/m2

Load/m, w on each joist = 1.95 x 0.4 = 0.78 kN/m

For a simply supported floor joist with a uniformly distributed load:

M = wL2/8

= 0.78 L2/8

For a beam with a rectangular cross-section:

Elastic section modulus Z = BD2/6= 50 x 1502 /6

= 187 500 mm3

From above M = max x Z = 7.5 x 187 500

= 1 406 000 Nmm

= 1.406 kNmEquating the two values of M:

1.406 = 0.78 L2/8

L = 3.79 m

Allowable span = 3.79 mJan - May 2016 30


31/31

A single span beam is simply supported between two columns andcarries a reinforced concrete slab in addition to the column and loading

shown below. Using the design load indicated, select a suitable section

considering section classification, shear and bending only. Assume S275

steel and that dead loads are inclusive of self-weights.

Exercise

41.2 kN

3.17 kN/m

3 m 3 m

Universal Beam Section

Jan - May 2016 31

chapter 4- beams design(5)

Documents