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MECHANICS OF MATERIALS

MECHANICS OF MATERIALS. SHEAR FORCE AND BENDING MOMENT DIAGRAM. IT IS NOW CLEAR THAT SHEAR FORCE “V” AND BENDING MOMENT “M” IN A BEAM ARE FUNCTIONS OF THE DISTANCE “x” MEASURED ALONG THE LONGITUDINAL AXIS OF THE BEAM.

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MECHANICS OF MATERIALS

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  1. MECHANICS OF MATERIALS

  2. SHEAR FORCE AND BENDING MOMENT DIAGRAM

  3. IT IS NOW CLEAR THAT SHEAR FORCE “V” AND BENDING MOMENT “M” IN A BEAM ARE FUNCTIONS OF THE DISTANCE “x” MEASURED ALONG THE LONGITUDINAL AXIS OF THE BEAM. IN CASE OF DESIGNING A BEAM IT WOULD BE DESIRABLE TO KNOW THE EXACT VALUES OF SHEAR FORCE “V” AND BENDING MOMENT “M” AT EACH POINT OF APPLICATION OF BEAM. AN EASY METHOD TO GET ALL THIS INFORMATION IS DEVELOP GRAPH OF SHEAR-FORCE AND BENDING-MOMENT AS A FUNCTION OF AXIAL DISTANCE ON THE BEAM. SHEAR-FOCE AND BENDING-MOMENT DIAGRAMS

  4. FOR SUCH GRAPHS ABSCISSA IS TAKEN AS A POSITION OF THE CROSS SECTIONS OF THE BEAM AND ORDINATE IS TAKEN AS THE CORRESPONDING VALUE OF EITHER THE SHEAR FORCE OR THE BENDING MOMENT. SUCH DIAGRAMS OF SHEAR FORCE “V” AND BENDING MOMENT “M” AS A FUNCTION OF HORIZONTAL DISTANCE “x” ARE CALLED SHEAR-FORCE AND BENDING-MOMENT DIAGRAMS. TO EXPLAIN THE CONSTRUCTION OF DIAGRAM, LET US CONSIDER A SIMPLE BEAM “AB” WITH A CONCENTRATED LOAD “P”. THE REACTIONS AT BOTH ENDS “A” AND “B”, SIMILAR TO AS WERE DISCUSSED DURING EQUILIBRIUM OF AXIAL LOADING, ARE GIVEN BY SIMPLE RELATIONSHIPS AS FOLLOWS: Ra = Pb/L AND Rb = Pa/L

  5. NOW SUPPOSE THE BEAM IS CUT THROUGH THE LEFT OF THE LOAD “P” AND AT A DISTANCE “x” FROM SUPPORT “A”. BY CONSTRUCTING FREE-BODY DIAGRAM OF THE LEFT-HAND PART OF THE BEAM, FROM EQUILIBRIUM WE GET THE FOLLOWINGS: V = Ra = Pb/L …………… (7) AND M = RaX = PbX/L …….…. (8) THESE EQUATIONS SHOW THAT THE SHEAR FORCE IS CONSTANT FROM SUPPORT “A” TO THE POINT OF APPLICATION OF THE LOAD “P” AND THE BENDING MOMENT VARIES LINEARLY WITH “X”. NOW THE BEAM IS CUT THROUGH THE RIGHT OF THE LOAD SUCH THAT a < x < L. NOW FROM THE EQUILIBRIUM OF THE LEFT-HAND PART OF THE BEAM, FOLLOWING RELATIONSHIPS ARE OBTAIANED:

  6. V = Ra - P V = Pb/L - P = - Pa/L ………. (9) AND M = RaX - P(x – a) M = PbX/L - P(x - a) M = x(Pb/L – P) + Pa M = x(-Pa/L) +Pa M = Pa (1 - x/L) M = Pa/L (L – x) ………………… (10) IT CAN BE OBSERVED FROM ABOVE EQUATIONS THAT SHEAR FORCE IS CONSTANT BUT BENDING MOMENT IS A LINEAR FUNCTION OF “x”.

  7. IT CAN BE OBSERVED FROM ABOVE EQUATIONS THAT SHEAR FORCE IS CONSTANT BUT BENDING MOMENT IS A LINEAR FUNCTION OF “x”. IN DERIVING EQUATIONS (9) & (10) FOR THE SHEAR FORCE AND BENDING MOMENT TO THE RIGHT OF THE LOAD “P”, LEFT-HAND PART OF THE BEAM WAS CONSIDERED THAT WAS ACTED UPON BY TWO FORCES “Ra” AND “P”. HOWEVER, IF THE RIGHT SIDE OF THE BEAM IS CONSIDERED AS FREE BODY, EQUATIONS CAN MORE SIMPLY BE DERIVED.

  8. IN THIS CASE FROM EQUILIBRIUM OF THE RIGHT-HAND PART, FOLLOWING TWO EQUATIONS CAN BE OBTAINED: V = - Rb = - Pa/L AND M = Rb(L – x) = Pa/L (L – x) WHICH ARE THE SAME RESULTS AS OBTAINED EARLIER. BY LOOKING AT THE SHEAR FORCE AND BENDING DIAGRAM, FOLLOWING CONCLUSIONS CAN BE DRAWN: SLOPE OF THE SHEAR-FORCE DIAGRAM dV/dx IS ZERO IN THE REGIONS 0 < x < a AND a < x < L.

  9. ALSO IN THE SAME REGIONS THE SLOPE OF BENDING MOMENT dM / dx IS EQUAL TO SHEAR FORCE “V”. AT THE POINT OF APPLICATION OF THE LOAD, THERE IS ABRUPT CHANGE IN THE SHEAR-FORCE DIAGRAM AND A CORRESPONDING CHANGE IN THE SLOPE OF THE BENDING-MOMENT DIAGRAM. SLOPE OF THE BENDING MOMENT DIAGRAM IS POSITIVE AND EQUAL TO “Pb/L” TO THE LEFT OF THE LOAD AND NEGATIVE AND EQUAL TO” -Pa/L” TO THE RIGHT OF THE LOAD. THE AREA OF THE SHEAR-FORCE DIAGRAM IS EQUAL TO “Pab/L” AS IS MOVED FROM x = 0 TO x = a ALONG THE BEAM.

  10. THE AREA OF THE SHEAR-FORCE DIAGRAM IS EQUAL TO “-Pab/L” AS IS MOVED FROM x = a TO x = L ALONG THE BEAM. POSITIVE AREA REPRESENTS THE INCREASE IN BENDING MOMENT BETWEEN POINTS x = 0 TO x = a. NEGATIVE AREA REPRESENTS THE DECREASE IN BENDING MOMENT BETWEE POINTS x = a TO x = L. BENDING MOMENT WOULD BE ZERO A BOTH ENDS OF THE BEAM, A AND B.

  11. IN ORDER TO DESIGN BEAMS THE MAXIMUM AND MINIMUM VALUES OF SHEAR FORCES AND BENDING MOMENTS ARE ALWAYS REQUIRED. THUS FOR A SIMPLE BEAM WITH A SINGLE CONCENTRATED LOAD, THE MAXIMUM SHEAR FORCE OCCURS AT THE END OF THE BEAM NEAREST TO THE CONCENTRATED LOAD, AND MAXIMUM BENDING MOMENT OCCURS UNDER THE LOAD ITSELF. NOW LET US CONSIDER THE CASE OF A SIMPLE BEAM WITH A UNIFORMLY DISTRIBUTED LOAD. IN THIS CASE EACH OF THE REACTIONS “Ra” AND “Rb” IS EQUAL TO “qL/2”.

  12. THE SHEAR FORCE “V” AND BENDING MOMENT “M” AT A DISTANCE “x” FROM THE LEFT-HAND END “A” CAN BE CALCULATED BY THE FOLLOWING RELTIONSHIPS: V = qL/2 - qx …………. (11) M = qLx/2 - qx² ………. (12) EQUATION (11) SHOWS THAT THE SHEAR-FORCE DIAGRAM CONSISTS OF AN INCLINED STRAIGHT LINE HAVING ORDINAGES qL/2 and –qL/2 at x=0 and x=L RESPECTIVELY. THE SLOPE OF THIS LINE IS EQUAL TO –q.

  13. THE BENDING MOMENT DIAGRAM IS A PARABOLIC CURVE THAT IS SYMMETRIC ABOUT THE MIDDLE OF THE BEAM. AT EACH X-SECTION, THE SLOPE OF THE BENDING MOMENT IS EQUAL TO THE SHEAR FORCE “V”. dM /dx = d/dx (qLx/2 - qx²/2) qL/2 - qx = V THE MAXMIMUM VALUE OF BENDING MOMENT OCCURS AT THE POINT WHERE dM/dx = 0, THAT THE POINT WHERE VALUE OF SHEAR FORCE IS ZERO. THIS IS THE MIDDLE POINT OF THE BEAM.

  14. HENCE BY SUBSTITUTING “x = L/2” IN THE EXPRESSION FOR BENDING MOMENT, ITS MAXIMUM VALUE CAN BE CALCULATED BY THE FOLLOWING EXPRESSION: Mmax = qL²/8 ……….. (13) THE DIAGRAM OF LOAD INTENSITY HAS AN AREA OF qL. THE SHEAR FORCE DECREASES BY THIS AMOUNT AS WE MOVE ALONG THE BEAM FROM A TO B. THE AREA OF SHEAR-FORCE DIAGRAM IS EQUAL TO qL²/8 BETWEEN x=0 and x=L/2, AND THIS AREA REPRESENTS THE INCREASE IN THE BENDING MOMENT BETWEEN THESE TWO POINTS.

  15. IN THE SAME WAY BENDING MOMENT DECREASES BY qL²/8 AS IS MOVED FROM x=L/2 to x=L ALONG THE BEAM. IN CASE SEVERAL CONCENTRATED LOADS ACT ON A SIMPLE BEAM, EXPRESSIONS FOR “V” AND “M” CAN EASILY BE DETERMINED FOR EACH REGION OF THE BEAM B/W THE POINTS OF LOAD APPLICATIONS AS FOLLOWS: SUPPOSE THREE DIFFERENT LOADS P1, P2 AND P3 ACT ON A BEAM AB AT DISTANCES a1, a2 AND a3. AGAIN MEASURING DISTANCE x FROM END A OF THE BEAM, FOLLOWING EXPRESSIONS ARE OBTAINED:

  16. V = Ra AND M = Rax FOR FIRST REGION, 0< x < a1 V = Ra - P1 AND M = Rax - P1(x – a1) FOR SECOND REGION, a1 < x < a2 V = -Rb + P3 AND M = Rb(L – x) – P3(L – b3 – x) FOR THIRD REGION, a2 < x < a3 V = -Rb AND M = Rb(L – x) FOR THE FOURTH REGION FROM ABOVE EQUATIONS IT IS EVIDENT THAT IN EACH REGION OF THE BEAM SHEAR FORCE REMAINS CONSTANT AND HAS THE SHAPE AS SHOWN.

  17. SIMILARLY THE BENDING MOMENT IN EACH PART OF THE BEAM IS A LINEAR FUNCTION OF x, HENCE THE CORRESPONDING DIAGRAM WOULD BE CONSISTED OF STRAIGHT LINES AS SHOWN. SHEAR-FORCE AND BENDING-MOMENT DIAGRAMS MAY BE CONSTRUCTED BY PUTTING x = a1, a2, a3 IN ABOVE EQUATIONS TO GET VALUES FOR BENDING MOMENTS. IT IS OT BE NOTED THAT EACH POINT WHERE A CONCENTRATED LOAD ACTS, THE SHEAR-FORCE CHANGES ABRUPTLY BY AN AMOUNT EQUATO TO THE LOAD.

  18. SIMILARLY AT EACH DISCONTINUITY IN SHEAR FORCE, THERE IS A CORRESPONDING CHANGE IN THE SLOPE OF THE BENDING-MOMENT DIAGRAM. BENDING MOMENT HAS A MAXIMUM AND MINIMUM VALUES AT X-SECTIONS WHERE THE SHEAR FORCE CHANGES THE SIGN, THAT OCCURS UNDER A LOAD. IN GENERAL, THE MAXIMUM POSITIVE OR NEGATIVE BENDING MOMENTS IN A BEAM MAY OCCUR AT A CONCENTRATED LOAD, AT A REACTION, AT A X-SECTION WHERE THE SHEAR FORCE EQUALS ZERO, OR AT A SECTION WHERE A COUPLE IS APPLIED.

  19. IN CASE SEVERAL LOADS ACT ON A BEAM, SHEAR-FORCE AND BENDING-MOMENT DIAGRAMS CAN BE OBTAINED BY SUMMATION OF DIAGRAMS OBTAINED FROM EACH OF THE LOAD ACTING SEPARATELY. THE CONCLUSIONS ABOUT SUMMATION OF SHEAR-FORCE AND BENDING-MOMENT DIAGRAMS FOLLOW FROM THE FACT THAT SHEAR FORCE AND BENDING MOMENT ARE LINEAR FUNCTIONS OF THE APPLIED LOAD.

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

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