Physics, asked by dikshadubey9684, 1 year ago

) design a simply supported beam of 10 m effective span carrying a total factored load of 60 kn/m. the depth of beam should not exceed 500 mm. the compression flange of the beam is laterally supported by floor construction. assume stiff end bearing is 75 mm?

Answers

Answered by ashwani381
6
INTRODUCTION

The following points should be considered in the design of a beam.

 Bending moment consideration: The section of the beam must be able to resist the maximum bending moment to which it is subjected.

 Shear force consideration: The section of the beam must be able to resist the maximum shear force to which it is subjected.

 Deflection consideration: The maximum deflection of a loaded beam should be within a certain limit so that the strength and efficiency of the beam should not be affected. Limiting the deflection within a safe limit will also prevent any possible damage  to finishing. As per the I.S. code, generally the maximum deflection should not exceed 1/325 of the span.

 Bearing stress consideration: The beam should have enough bearing area at the supports to avoid excessive bearing stress which may lead to crushing of the beam or the support itself.

Buckling consideration: The compression flange should be prevented from buckling. Similarly the web, the beam should also be prevented from crippling. Usually these failures do not take place under normal loading due to proportioning of thickness of flange and web. But under considerably heavy loads, such failures are possible and hence in such cases the member must be designed to remain safe against such failures

14.2 SHEAR AND BEARING STRESSES

When the beams are subjected to loads, then, these are also required to transmit large shear forces either at supports or at concentrated loads. For simply supported beams, the shear force is maximum at the supports. The values of shear force at the concentrated loads also remain large. Due to shear force, the shear stresses are setup along with the bending stresses at all sections of the beams. The shear stress at any point of the cross-section is given by



Where  is the shear stress, 

F = the shear force at cross-section,

Q = Static moment about the neutral axis of the portion of cross-sectional area beyond the location at which the stress is being determined.

I = Moment of inertia of the section about the neutral axis

t = Thickness of web (width of section at which the stress is being determined)



The distribution of shear stresses for rectangular section of beam and I-beam section are shown in Fig. 14.1. The maximum shear stress occurs at the neutral axis of the section. The maximum shear stress in a member having regard to the distribution of stresses in conformity with the elastic behavior of the member in flexure (bending) should not exceed the value of maximum permissible shear stress, τvm found as follows.

τvm=0.45fy

Where fy is the yield stress of structural steel to be used. It is to note that in the case of rolled beams and channels, the design shear is to be found as the average shear. The average shear stress for rolled beams or channels calculated by dividing the shear force at the cross-section by the gross-section of the web. The gross-cross-section of the web is defined as the depth of the beam or channel multiplied by its web thickness.

Average shear stress for rectangular beam is given by 

Average shear stress for rectangular beam is given by 

For rolled steel beams and channels, it is assumed that shear force is resisted by web only. The portion of shear resisted by the flanges is neglected. The average shear stress τva.cal, in a member calculated on the gross cross-section of web (when web buckling is not a factor) should not exceed in case of unstiffened web of the beam,

τva= 0.4 fy

The allowable shear stress as per AISC, AASHTO and AREA specifications are as follows:



When the beams are subjected to co-existent bending stresses (tension or compression) and shear stress, then the equivalent stress, σe.cal is obtained from the following formula



The equivalent stress  due to co-existent bending (tension or compression) and shear stresses should not exceed the maximum permissible equivalent stress σe found as under



When the bearing stress σp is combined with tensile or compressive bending and shear stresses under the most unfavourable conditions of loading, the equivalent stress σe.cal obtained as below should not exceed



σbc.cal, σbt.cal, τvm.cal and σp.cal are the numerical values of the co-existent bending (compression or tension), shear and bending stresses. When bending occurs about both the axes of the member, σbt.cal and σbc.cal should be taken as the sum of the two calculated fibre stresses, σe is the maximum permissible equivalent stress.

The bearing stress in any part of a beam when calculated on the net area of contact should not exceed the value of σpcalculated as below

Where σp is the maximum permissible bearing stress and fy is the yield stress.

14.3 EFFECTIVE SPAN AND DEFLECTION LIMITATION


The large deflections of beams are undesirable for the following reasons:


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Answered by sanjumccullum42
0

Answer:

3. Design a simply supported (laterally supported) of effective span 12m to carry a factored load of 70kN/m. The depth of the beam is restricted to 500mm.

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