Transverse shear stress

Transverse shear stress

1. PRESENTATION ON TRANSVERSE SHEAR STRESS AMIT KUMAR SINGH 15ME63R30 MECHANICAL SYSTEMS DESIGN BY-

2. Table of content Concept Transverse Shear Load Transverse Shear stress Difference between Bending and Shear stress Assumptions Derivation Analysis in rectangular cross section Other cross section example Points to remember References

3. TRANSVERSE SHEAR LOAD

4. Beams are often subjected to transverse loads which generate both bending moments ‘M’ and shear forces ‘V’ along the beam . The bending moments cause bending normal stresses σ to arise through the depth of the beam, and the shear forces cause transverse shear-stress distribution through the beam cross section as shown TRANVERSE SHEAR STRESS

5. Difference between Bending and Shear stress Bending stress It acts perpendicular to plane of cross section. Bending stress varies linearly over the depth of beam. At extreme fibre bending stress is maximum. At neutral axis bending stress is zero. Shear stress It acts parallel to the plane of cross section of beam. It varies parabolically over the depth of beam. At extreme fibre it is zero. At neutral axis shear stress have some value.

6. ASSUMPTIONS Shear stresses are uniformly distributed across the width of the beam . The shear formula is applicable for prismatic beams. Accuracy of shear formula for rectangular beam is directly proportional to depth to width (d/b) ratio . Beam should be of homogeneous material.

7. DERIVATION To determine the shear stress distribution equation, look at a loaded beam as Fig.

8. Consider now a segment of this element at distance y above the N.A. up to the top of the element , Look at a FBD of the element dx with the bending moment stress distribution only,

9. stresses due to the bending moments only form a couple, therefore the force resultant is equal to zero horizontally. =0 - dA+ Substituting the values of the stresses : ( ) =   Let the width of section at a distance y from the N.A. be a function of y and call It ‘b’. Applying the horizontal equilibrium equation , gives: I n order for it to be in equilibrium, a shear stress τ must be present.

10. Where, P= The shear force carried by the section I= Moment of inertia b= The sectional width at the distance y from the N.A. Q= The First moment of Area M= Moment Y= Distance of centroid of hatched area from N.A. =   solving for , = ( )( ) ( ) = Load ‘P ’ =Q  

11. TRANVERSE SHEAR STRESS ANALYSIS OF RECTANGULAR CROSS SECTION-

12. A= b[d/2-y] = y+1/2[d/2 -y] =1/2[d/2+y] Q=A =b/2[( - ] = = At external fibres y=d/2 Therefore =0 = = = WHERE     [d/2 - ]   d b

13. Triangular cross section   =4/3  

14.       Rectangular cross section with vertical and horizontal diagonals

15. I section

16. POINTS TO REMEMBER In case of rectangular , square and circular cross section shear stress is maximum at N.A. For square and rectangle A (because =constant) For circular , triangular , and square of vertical and horizontal diagonals ( because =constant ) For I-section (because =constant) at junction of flange and web.  

17. REFERENCES Strength of Materials lecture by Prof : S . K. Bhattacharya Department of Civil Engineering, IIT , Kharagpur . Mechanics of material by Timosenko .

18. Thank you!