Chapter Objectives. Design a beam to resist both bendingand shear loads

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1 Chapter Objectives Design a beam to resist both bendingand shear loads

2 A Bridge Deck under Bending Action

3 Castellated Beams

4 Post-tensioned Concrete Beam

5 Lateral Distortion of a Beam Due to Lateral Load Action

6 T-Beam Roof on a Office Building

7 Modern Steel Frame Building

8 Joint in steel frame building

9 Beams with Different Boundary/Support Conditions

10 Beams with Different Vertical Load Conditions Triangular distributed load and concentrated load on an over-hang beam.

11 Beams with Different Vertical Load Conditions Uniform distributed load a simply supported beam.

12 Internal Actions of a simply supported beam loaded with uniform load

13 Internal Actions of a cantilever beam loaded with triangular distributed load Vmax Mmax

14 Internal Actions of a simply-supported beam loaded with an arbitrary loading Mmax Vmax

15 Deformations and Stresses of a Beam Under Bending Effect

17 Deformation shape of a Cantilever Beam under Bending Action

layers/fibers, Strain and stress profiles vary linearly with

at the neutral axis (neutral axis is the centroid of the

18 Stress and Strain Profile of a Section under Bending Observations: Maximum strain and stress occur at the outer most layers/fibers, Strain and stress profiles vary linearly with the distance y from the centroid, There is no strain and stress at the neutral axis (neutral axis is the centroid of the section), The value c defines the distance from the centroid to the outer most layer.

19 Stress and Strain Profile of a Section under Bending Bending Formula (General) σ x = M Iz y Bending Formula (Maximum Stress at a arbitrary section) σ max = M I z c Bending Formula (Absolute Maximum Stress at a section where max. mom. is observed) σ max = M I max z c

23 PRISMATIC BEAM DESIGN Basis of beam design Strength concern (i.e. provide safety margin to normal/shear stress limit) Serviceability concern (i.e. deflection limit) Section strength requirement S req' d = M σ max allow b S= I c S: Section modulus I: Moment of inertia about the centroidal axis c: half-height of the section C c h

24 PRISMATIC BEAM DESIGN (cont) Choices of section (refer to standard tables): Steel sections e.g. AISC standard W 460 X 68 (height = mm weight = 0.68 kn/m)

25 PRISMATIC BEAM DESIGN (cont) Wood sections Nominal dimensions (in multiple of 25mm) e.g. 50 (mm) x 100 (mm) actual or dressed dimensions are smaller, e.g. 50 x 100 is 38 x 89. Built-up sections

26 PRISMATIC BEAM DESIGN (cont) Procedures: Shear and Moment Diagram Determine the maximum shear and moment in the beam. Often this is done by constructing the beam s shear and moment diagrams. For built-up beams, shear and moment diagrams are useful for identifying regions where the shear and moment are excessively large and may require additional structural reinforcement or fasteners.

27 PRISMATIC BEAM DESIGN (cont) Normal Stress If the beam is relatively long, it is designed by finding its section modulus and using the flexure formula, S req d = M max /σ allow. Once S req d is determined, the cross-sectional dimensions for simple shapes can then be computed, using S req d = I/c. If rolled-steel sections are to be used, several possible values of S may be selected from the tables. Of these, choose the one having the smallest cross-sectional area, since this beam has the least weight and is therefore the most economical.

28 PRISMATIC BEAM DESIGN (cont) Normal Stress (cont) Make sure that the selected section modulus, S, is slightly greaterthan S req d, so that the additional moment created by the beam s weight is considered. Shear Stress Normally beams that are short and carry large loads, especially those made of wood, are first designed to resist shear and then later checked against the allowable-bending-stress requirements. Using the shear formula, check to see that the allowable shear stress is not exceeded; that is, use τ allow (V max /A s ), A s is the shear area and is a function of the cross sectional shape.

29 PRISMATIC BEAM DESIGN (cont) Shear Stress (cont) If the beam has a solid rectangular cross section, the shear formula becomes τ allow 1.5(V max /A), and if the cross section is a wide flange, it is generally appropriate to assume that the shear stress is constant over the cross-sectional area of the beam s webso that τ allow V max /A web, where A web is determined from the product of the beam s depth and the web s thickness.

30 PRISMATIC BEAM DESIGN (cont) Adequacy of Fasteners The adequacy of fasteners used on built-up beams depends upon the shear stress the fasteners can resist. Specifically, the required spacing of nails or bolts of a particular size is determined from the allowable shear flow, q allow = VQ/I, calculated at points on the cross section where the fasteners are located.

31 EXAMPLE 1 A beam is to be made of steel that has an allowable bending stress of σ allow = 170 MPa and an allowable shear stress of τ allow = 100 MPa. Select an appropriate W shape that will carry the loading shown in Fig. a.

EXAMPLE 1 (cont) m m 3 = 706x10 3 mm 3 W460 60 W410

1120 10 = 1200 10 = 1030 10 = 1060 10 S = 98410 ( 3)

32 EXAMPLE 1 (cont) m m 3 = 706x10 3 mm 3 W W W W W W S S S S S = = = = = S = ( 3) 3 mm ( 3) 3 mm ( 3) 3 mm ( 3) 3 mm ( 3) 3 mm ( 3) 3 mm 98710

33 EXAMPLE 1 (cont) Solutions The beam having the least weight per foot is chosen, W460 x 60 The beam s weight is W ( 0.60kN/ m)( 6 ) = 3.60kN = m From Appendix B (tables), for a W460 x 60, d = 455 mm and t w = 8 mm. Thus τ avg = 3 ( ) V max = = 24.7MPa < 100MPa (OK) A web ( 455)( 8) Use a W460 x 60.

34 EXAMPLE 2 The laminated wooden beam shown in Fig. 11 8a supports a uniform distributed loading of 12 kn/m. If the beam is to have a height-to-width ratio of 1.5, determine its smallest width. The allowable bending stress is 9 MPa and the allowable shear stress is 0.6 MPa. Neglect the weight of the beam.

EXAMPLE 2 (cont) Solutions Applying the flexure formula, S req = M σ max allow = ( 3) ( 6) 10.67 10 910 = 0.

35 EXAMPLE 2 (cont) Solutions Applying the flexure formula, S req = M σ max allow = ( 3) ( 6) = m 3 = V max Assuming that the width is a, the height is 1.5a. a 3 a = = ( a)( 1.5a) ( 0.75a) 3 12 = = = 1 M I max = S req c m m 3

36 EXAMPLE 2 (cont) Solutions Applying the shear formula for rectangular sections, Vmax τmax = 1.5 = = 0.929> A ( ) ( ) 1.5 ( 1.5)( 0.147)( 0.147) 0.6 MPa Since the design fails the shear criterion, the beam must be redesigned on the basis of shear. Vmax τ allow = 1.5 A ( ) 600= 1.5 a1.5a ( ) a= m= 183mm (Ans) This larger section will also adequately resist the normal stress.

Use Hooke s Law (as it applies in the uniaxial direction),

0.6 STRSS-STRAIN RLATIONSHIP Use the principle of superposition Use Poisson s ratio, v lateral longitudinal Use Hooke s Law (as it applies in the uniaxial direction), x x v y z, y y vx z, z z vx y Copyright