MECHANICS OF MATERIALS - II

1. MECHANICS OF MATERIALS - II

2. SHEAR STRESSES IN BEAMS -II

3. IN CASE A BEAM IS CIRCULAR WE CAN NOT ASSUME THAT ALL OF THE SHEAR STRESSES ACT PARALLEL TO THE Y-AXIS. HOWEVER, IT CAN BE SHOWN THAT AT A POINT ON THE BOUNDARY OF THE X-SECTION, THE SHEAR STRESS MUST BE TANGENT TO THE BOUNDARY. LET US NOW INVESTIGATE THE SHEAR STRESSES ACTING ALONG A CHORD “pq” AT A DISTANCE “y1” FROM THE NEUTRAL AXIS. THE SHEAR STRESSES AT THE ENDS OF THE CHORD MUST BE TANGENT TO THE BOUNDARY OF THE X-SECTION. AT MID POINT “n” OF THE CHORD, SYMMETRY REQUIRES THAT THE SHEAR STRESS BE PARALLEL TO THE Y AXIS. SHEAR STRESESS IN CIRCULAR BEAMS

5. IF IT FOLLOWS THAT THE LINES OF ACTION OF THE SHEAR STRESS AT POINTS “P” AND “n” WILL INTERSECT AT A POINT ON THE Y AXIS. NOW IN ORDER TO CALCULATE THE SHEAR STRESSES ALONG THE LINE “pq”, WE MUST OBTAIN THE FIRST MOMENT OF AREA (Q) OF THE AREA BELOW THE LINE “pq” WITH RESPECT TO THE Z-AXIS. THE ELEMENT OF AREA AT DISTANCE “y” FROM THE Z AXIS (SHADED AREA) HAS THE THICKNESS “dy” AND LENGTH “2√(r² - y²)” IN WHICH “r” IS THE RADIUS OF THE X-SECTION. THE FIRST MOMENT OF THIS ELEMENT OF AREA IS FOUND BY MULTIPLYING THE AREA BY “y”, AND THUS THE VALUE OF “Q” OF ENTIRE SEGMENT BELOW THE LINE “pq” IS GIVEN AS

6. Q = ∫ 2y √(r² - y²) dy = 2/3 (r² - y1²)³∕² AS THE WIDTH “b” AND MOMENT OF INERTIA “I” ARE GIVEN AS b = 2√(r² - y1²) AND I = πr⁴ / 4 HENCE THE VERTICAL COMPONENT OF THE SHEAR STRESS IS GIVEN AS τy = VQ/Ib = 4V / 3πr⁴ (r² - y1²) NOW AT POINT “P” ON THE BOUNDARY OF THE X-SECTION, TOTAL SHEAR STRESS “τ” IS RELATED TO ITS VERTICAL COMPONENT BY THE EQUATION τ = τy / cos θ

7. τ = τy / cos θ = rτy / √(r² - y1²) BY SUBSTITUTING THIS VALUE IN THE PREVIOUS EQUATION WE GET τ = (4V / 3πr³) √(r² - y1²) THIS EQUATION GIVES THE SHEAR STRESS AT ANY POINT “p” ON THE BOUNDARY AT A DISTANCE “y1” FROM THE Z AXIS. AS WE MOVE ALONG THE LINE “pn”, THE SHEAR STRESS DIMINISHES AND REACHES AT MINIMUM VALUE AT POINT “n” WHERE τ = τy.

8. IN THIS CASE TOO MAXIMUM SHEAR STRESS OCCURS AT THE NEUTRAL AXIS, HENCE BY SUSTITUTING y1 = 0 IN THE ABOVE EQUATION, WE GET τmax = (4V / 3πr²) = 4V/3A THIS EQUATION SHOWS THAT THE MAXIMUM SHEAR STRESS IN A CIRCULAR BEAM IS 4/3 TIMES THE AVERAGE STRESS V/A. HENCE SHEAR STRESSES AT NEUTRAL AXIS MAY DIRECTLY BE CALCULATED BY THE EQUATION (VQ / Ib) FOR ANY TYPE OF GEOMETY. FOR EXAMPLE FOR CIRCULAR X-SECTION I = πr⁴/4, AND b = 2r, Q = πr²/2 (4r/3π) = 2r³/ 3

9. QUESTIONS AND QUERIES IF ANY! IF NOT THEN GOOD BYE SEE ALL OF YOU IN NEXT LECTURE ON-------------------------