MECHANICS OF MATERIALS

1. MECHANICS OF MATERIALS

2. BEAMS - SHEAR FORCE AND BENDING MOMENT

3. IN LAST LECTURES WE DISCUSSED THE STRESS DISTRIBUTION UNDER AXIAL AND TORSIONAL LOADINGS. IN THIS AND IN THE COMING LECTURES STRESS DISTRIBUTION AND DEFLECTION OF BEAMS WOULD BE DISCUSSED. A BEAM IS A STRUCTURAL MEMBER THAT IS DESIGNED TO SUPPORT TRANSVERSE LOADS, LOADS ACTING PERPENDICULAR TO THE LONGITUDINAL AXIS OF THE BEAM. BEAMS DIFFER FROM BAR IN TENSION AND BAR IN TORSION PRIMARILY BECAUSE OF THE DIRECTIONS OF THE LOADS THAT ACT UPON THEM. BEAMS AND ITS TYPES

4. A BEAM RESISTS APPLIED LOADS BY A COMBINATION OF INTERNAL TRANSVERSE SHEAR FORCE AND BENDING MOMENT. BEAMS ARE PLANAR STRUCTURES BECAUSE ALL OF THE LOADS ACT IN THE PLANE OF THE FIGURE AND ALL DEFLECTIONS OCCUR IN THE SAME PLANE WHICH IS CALLED THE PLANE OF BENDING. ASSUMPTION THAT THE LOADS APPLIED ON THE BEAM ACT IN THE PLANE OF THE FIGURE MEANS THAT ALL FORCES HAVE MOMENT VECTORS IN THE PLANE OF THE FIGURE. SIMILARLY THE ALL COUPLES HAVE THEIR MOMENT VECTORS PERPENDICULAR TO THAT PLANE.

5. SIMPLY SUPPORTED BEAM A BEAM HAVING A PIN SUPPORT AT ONE AND A ROLLER SUPPORT AT THE OTHER END IS KNOWN AS SIMPLY SUPPORTED BEAM. SUCH BEAMS ARE STATICALLY DETERMINAGE BEAMS. CANTILEVER BEAM A BEAM WITH A FIXED END AT ONE SIDE AND FREE END AT THE OTHER SIDE IS KNOWN AS CANTILEVER BEAM. SUCH BEAMS ARE ALSO STATICALLY DETERMINATE BEAMS. CONTINUOUS BEAM A BEAM WITH A PIN SUPPORT AT ONE END, A ROLLER SUPPORT AT THE OTHER END, AND ONE OR MORE INTERMEDIATE ROLLER SUPPORTS IS CALLED CONTINUOUS BEAM. SUCH BEAMS ARE STATICALLY INDETERMINATE. TYPES OF BEAMS

6. PROPPED CANTILEVER BEAM A BEAM HAVING A FIXED END AT ONE SIDE AND A ROLLER AT THE OTHER END IS CALLED A PROPPED CANTILEVER BEAM. THESE ARE ALSO STATICALLY INDETERMINAGE BEAMS. OVERHANGING BEAM A BEAM THAT EXTENDS BEYOND THE SUPPORT AT ONE END OR AT BOTH ENDS IS KNOWN AS OVERHANGING BEAM. SUCH BEAMS ARE ALSO STATICALLY DETERMINATE. STATICALLY DERERMINATE MEANS THAT REACTIONS AT BOTH SUPPORTS CAN BE DETERMINED FROM EQUILIBRIUM EQUATIONS. TYPES OF BEAMS

7. WITH REFERENCE TO TYPES OF BEAMS SUPPORTS MIGHT BE ROLLER SUPPORT PIN SUPPORT CENTILEVER SUPPORT (FIXED END) LOADS ON BEAMS MIGHT BE OF FOLLOWING TYPES: CONCENTRATED LOAD DISTRIBUTED LOAD UNIFORM LOAD COUPLE TYPES OF SUPPORTS AND LOADS

8. NOW THE QUESTION IS HOW DOES A BEAM RESPOND TO THE EXTERNAL LOADS, AND HOW DOES IT TRANSMIT THESE LOADS TO ITS SUPPORTS? THIS IS QUITE EASY! LET US INVESTIGATE IT. WHENEVER A BEAM IS LOADED BY LOADS OR COUPLES, INTERNAL STRESSES AND STRAINS ARE CREATED. FOR THIS PURPOSE WE NEED FIND INTERNAL FORCES AND INTERNAL COUPLES THAT ACT ON X-SECTIONS OF BEAMS. SUPPOSE A CENTILEVER BEAM IS ACTED UPON BY A VERTICAL FORCE “P” AT ITS FREE END. IMAGINE THIS BEAM IS CUT AT A X-SECTION “mn” LOCATED AT DISTANCE “X” FROM THE FREE END. LEFT PART OF THE BEAM IS ISOLATED AS A FREE BODY. SHEAR FORCE AND BENDING MOMENT

9. THIS FREE BODY WOULD BE IN EQUILBRIUM IF THE FORCE “P” IS BALANCED BY THE STRESSES THAT ACT OVER THE X-SECTION “mn”. IN FACT EXACT DISTRIBUTION OF STRESSES AT X-SECTION IS NOT KNOWN BUT IT MUST BE SUCH THAT IT CAN MAINTAIN EQUILBRIUM OF THE FREE BODY. NOW THE RESULTANT IS CONSIDERED TO BE A SHEAR FORCE “V” ACTING PARALLEL TO THE X-SECTION AND A BENDING COUPLE OF MAGNITUDE OF MOMENT “M”. BOTH THE SHEAR FORCE AND BENDING COUPLE ACT IN IN THE PLANE OF THE BEAM AND THE MOMENT OF THE BEDNIG COUPLE IS CALLED THE BENDING MOMENT (M).

10. BOTH THE SHEAR FORCES AND BENDING MOMENTS ARE CALLED STRESS RESULTANTS BECAUSE THESE ARE THE RESULTANT OF STRESSES DISTRIBUTED OVER THE X-SECTION SIMILAR TO AXIAL FORCES IN IN BARS AND TWISTING COUPLES IN SHAFTS. THE STRESS RESULTANTS IN STATICALLY DETERMINATE BEAMS CAN BE CALCULATED FROM EQUATIONS OF STATIC EQUILIBRIUM. THERFORE, FOR ABOVE-MENTIONED FREE BODY DIAGRAM WE HAVE FOLLOWING EQUATIONS: V = P AND M = Px A POSITIVE SHEAR FORCE TENDS TO DEFORM THE ELEMENT BY CAUSING THE RIGHT-HAND TO MOVE DOWNWARD WITH RESPECT THE LEFT-HAND FACE.

11. SIMILARYL A POSITIVE BENDING MOMENT ELONGATES THE LOWER PART OF THE BEAM AND COMPRESSES THE UPPER PART. AS THE SIGNS FOR V AND M ARE RELATED TO THE DEFORMATION OF THE MATERIAL, THESE SIGN CONVENTIONS ARE CALLED DEFORMATION SIGN CONVENTIONS. IN AXIAL DEFORMATION WE USED A SIGN CONVENTION SUCH AS THE TENSION IS ALWAYS POSITIVE AND COMPRESSION IS NEGATIVE. IN THIS WAY A DIFFERENT SIGN CONVENTION, KNOWN AS STATIC SIGN CONVENTIONS, WAS USED IN STATIC EQUILIBRIUM.

12. IN STATIC SIGN CONVENTION FORCES ARE ALWAYS TAKEN AS POSITIVE AND NEGATIVE WHEN ACTING IN THE POSITIVE AND NEGATIVE DIRECTIONS OF A COORDINATE SYSTEM RESPECTIVELY. IN ORDER TO EXPLAIN THESE TWO TYPES OF SIGN CONVENTIONS, EQUATION OF EQUILIBRIUM FOR THE TWO PARTS OF BEAM UNDER CONSIDERATION IS WRITTEN DOWN. ACCORDING TO DEFORMATION SIGN CONVENTION FOR STRESS RESULTANTS SHEAR FORCE AND BENDING MOMENT ARE POSITIVE. HOWEVER, ACCORDING TO STATIC SIGN CONVENTION, LOAD P IS GIVEN POSITIVE SIGN AND V IS GIVEN NEGATIVE SIGN. HENCE ΣFy = 0, P - V = 0

13. IT SHOULD BE KEPT IN MIND THAT A POSITIVE SHEAR FORCE MAY APPEAR IN A FORCE EQUILIBRIUM EQUATION WITH EITHER A POSITIVE OR A NEGATIVE SIGN DEPENDING UPON THE FREE-BODY DIAGRAM TO BE CONSIDERED. EXACTLY THE SAME SITUATION EXISTS FOR BENDING MOMENTS WHEN MOMENT EQUILIBRIUM EQUATIONS ARE USED. HOWEVER, SUCH DIFFICULTIES WITH SIGNS MAY BE AVOIDED BY KEEPING IN MIND THAT TWO TYPES OF SIGN CONVENTIONS ARE USED IN MECHANICS.

14. ONE DEFORMATION SIGN CONVENTION IS USED FOR STRESS RESULTANTS AND THE OTHER STATIC SIGN CONVENTION IS USED FOR EQUATIONS OF EQUILIBRIUM. DEFORMATION SIGN CONVENTIONS ARE BASED UPON HOW THE MATERIAL IS DEFORMED, AND THE STATIC SIGN CONVENTIONS IS BASED UPON DIRECTIONS IN SPACE.

15. NOW SOME IMPORTANT RELATIONSHIPS BETWEEN LOADS ON A BEAM “P”, SHEAR FORCE “V” AND BENDING MOMENT “M” WOULD BE ESTABLISHED. THESE RELATIONSHIPS WOULD BE VERY HELPFUL IN CONSTRUCTING SHEAR-FORCE AND BENDING-MOMENT DIAGRAMS. LET US CONSIDER AN ELEMENT OF A BEAM CUT OUT BETWEEN TWO X-SECTIONS THAT ARE A DISTANCE “dx” APART FROM EACH OTHER. THERE ARE A SHEAR FORCE “V” AND BENDING MOMENT “M” ACTING ON THE LEFT-HAND FACE OF THE ELEMENT IN THEIR POSITIVE DIRECTIONS. RELATIONSHIP BETWEEN LOAD, SHEAR FORCE AND BENDING MOMENT – DISTRIBUTED LOAD

16. AS “V” AND “M” ARE FUNCTIONS OF THE DISTANCE “x” MEASURED ALONG THE AXIS OF THE BEAM, “V” AND “M” ON THE RIGHT-HAND FACE OF THE BEAM WOULD BE WITH INCREMENTS “dV” and “dM”. HENCE THE CORRESPONDING VALUES OF SHEAR FORCE AND BENDING MOMENT (STRESS RESULTANTS) ON THE RIGHT-HAND FACE WOULD BE V+dV and M+dM. THE LOADING ACTING ON THE TOP SURFACE OF THE ELEMENT MAY BE A DISTRIBUTED LOAD, A CONCENTRATED LOAD, OR A COUPLE. NOW LET US ASSUME THAT THE LOAD IS DISTRIBUTED ON THE TOP SURFACE AND ITS INTENSITY IS “q”.

17. CONSIDERING THE EQUILIBRIUM OF FORCES IN THE VERTICAL DIRECTION, FOLLOWING RELATIONSHIP IS OBTAINED: V – (V+dV) - qdx = 0 => dV / dx = -q …………… (1) THIS RELATIONSHIP SHOWS THAT SHEAR FORCE “V” VARIES WITH THE DISTANCE “x”, ITS RATE OF CHANGE WITH RESPECT TO “x” IS EQUAL TO “-q”. THE SHEAR FORCE WOULD BE CONSTANT IF THERE IS NO LOAD ON BEAM. IN DERIVING THE ABOVE EQUATION IT WAS ASSUMED THAT DISTRIBUTED LOADS ARE POSITIVE WHEN ACTING DOWNWARD AND NEGATIVE ACTING IN THE UPWARD DIRECTION.

18. EQUATION (1) CAN BE INTEGRATED BETWEEN TWO POINTS ON A BEAM IN ORDER TO OBTAIN AN RELATIONSHIP TO CALCULATE SHEAR FORCES ACTING AT TWO DIFFERENT X-SECTIONS. THUS INTEGRATING THE EQUATION BETWEEN TWO POINTS “A” AND “B” ON THE AXIS OF BEAM PROVIDES Vb - Va = - = - (AREA OF LOAD-INTENSITY DIAGRAM BETWEEN A & B) . (2) THE LEFT-HAND SIDE OF THIS EQUATION EQUALS THE DIFFERENCE BETWEEN SHEAR FORCES AT POINTS “B” & “A” ON THE AXIS OF THE BEAM.

19. THE INTEGRAL ON THE RIGHT-SIDE REPRESENTS THE AREA OF THE LOAD INTENSITY DIAGRAM BETWEEM “A” AND “B”, WHICH IS ALSO EQUAL TO THE RESULTANT OF THE DISTRIBUTED LOAD BETWEEN “A” AND “B”. IT SHOULD BE REMEMBERED THAT EQ. (1) WAS DERIVED FOR AN ELEMENT OF THE BEAM SUBJECTED ONLY TO A CONTINUOUSLY DISTRIBUTED LOAD OR TO NO LOAD. EQ. (1) CAN NOT BE USED AT A POINT WHERE A CONCENTRATED LOAD IS APPLIED. SIMILAR IS THE CASE FOR EQUATION (2). AREA OF THE LOAD-INTENSITY DIAGRAM MAY BE POSITIVE OR NEGATIVE.

20. NOW AGAIN CONSIDER THE PREVIOUS EXAMPLE USED TO CALCULATE SHEAR FORCE. BY CONSIDERING THE EQUILIBRIUM BY SUMMING BENDING MOMENTS “M” ABOUT AN AXIS THROUGH THE LEFT-HAND FACE OF THE ELEMENT, WE GET -M - qdx(dx/2) - (V+dV)dx + (M+dM) = 0 IN THIS WAY FOLLOWING EQUATION IS OBTAINED IF PRODUCTS OF DIFFERENTIALS ARE DISCARDED AS THESE ARE NEGLIGIBLE AS COMPARED TO OTHER TERMS: dM/dx = V ……………….. (3) THIS EQUATION SHOWS THAT RATE OF CHANGE OF BENDING MOMENT WITH RESPECT TO “X” IS EQUAL TO THE SHEAR FORCE. “M” WOULD BE CONSTANT IF SHEAR FORCE IS ZERO.

21. EQUATION (3) APPLIES ONLY IN THE REGIONS WHERE DISTRIBUTED LOADS ACT ON THE BEAM. AT A POINT WHERE A CONCENTRATED LOAD ACTS, A SUDDEN CHANGE OR DISCONTINUITY IN THE SHEAR FORCE RESULTS AND THE DERIVATIVE dM/dx UNDEFINED. IN A SIMILAR WAY EQUATION (3) CAN ALSO BE INTEGRATED BETWEEN TWO POINT “A” & “B”ON THE BEAM, WHICH GIVES THE INTEGRAL ON THE LEFT SIDE IS EQUAL TO THE DIFFERENCE OF BENDING MOMENTS AT POINTS “A” AND “B” ON THE BEAM.

22. THE INEGRAL ON THE RIGHT-HAND SIDE SHOWS SHEAR FORCE AS A FUNCTION OF “x” WHICH VARIES WITH “x”. HENCE THIS INEGRAL REPRESENTS THE AREA OF SHEAR FORCE DIAGRAM BETWEEN POINTS “A” & “B”. = AREA OF SHEAR-FORCE DIAGRAM BETWEEN A AND B -- (4) THIS EQUATION CAN ALSO BE USED WHEN CONCENTRATED LOADS ARE ACTING ON THE BEAM BETWEEN SUPPORTS A AND B. HOWEVER, THIS EQUATION IS NOT VALID IN CASE A COUPLE IS APPLIED AS A COUPLE PRODUCE A SUDDEN CHANGE IN THE BENDING MOMENT AND THE LEFT-HAND SIDE OF THIS EQUATION CAN NOT BE INTEGRATED ACROSS SUCH A DISCONTINUITY.

23. NOW CONSIDER A CONCENTRATED LOAD “P” ACTING ON THE BEAM ELEMENT AS CONSIDERED IN THE PREVIOUS CASE. IN A SIMILAR WAY THE STRESS RESULTANTS ON THE LEFT-HAND FACE ARE DENOTED BY “V” AND “M”. ON THE RIGHT-HAND FACE, SHEAR FORCE AND BENDING MOMENT ARE DENOTED BY “V+V1” AND BENDING MOMENT “M+M1”, WHERE “V1” AND “M1” REPRESENT THE POSSIBLE INCREMENTS IN SHEAR FORCE AND BENDING MOMENT. IN CASE OF CONCENTRATED LOAD, FOR SIGN CONVENTION IT IS ASSUMED HERE THAT A DOWNWARD LOAD IS POSITIVE. THEREFORE FOR EQUILIBRIUM OF FORCES IN THE VERTICAL DIRECTION, WE HAVE RELATIONSHIP - IN CASE OF CONCENTRATED LOAD

24. V - P - (V+V1) = 0 OR V1 = - P ………………. (3) IT MEANS THAT AN ABRUPT CHANGE OCCURS IN THE SHEAR FORCE AT A POINT WHERE A CONCENTRATED LOAD ACTS. THEREFORE, AS WE PASS FROM LEFT TO RIGHT THROUGH THE POINT OF LOAD APPLICATION, THE SHEAR FORCE DECREASES BY AN AMOUNT EQUAL TO THE MAGNITUDE OF THE DOWNWARD LOAD.

25. SIMILARLY FROM EQUILIBRIUM OF MOMENTS, WE GET AS FOLLOWS - M – P(dx/2) – (V+V1)dx + M + M1 = 0 M1 – P(dx/2) – Vdx – V1dx = 0 M1 = P(dx/2) + Vdx + V1dx …… (4) AS THE LENGTH “dx” OF THE ELEMENT IS INFINITESIMALLY SMALL, HENCE THE INCREMENT “M1” IN THE BENDING MOMENT IS ALSO INFINETESIMALLY SMALL AS IS EVIDENT FROM THE EQUATION.

26. BY CONSIDERING EQUATIONS (3) AND (4), IT CAN BE CONCLUDED THAT AN ABRUPT CHANGE OCCURS IN THE SHEAR FORCE AT A POINT WHERE A CONCENTRATED LOAD ACTS. AS WE PASS FROM LEFT TO RIGHT THROUGH THE POINT OF LOAD APPLICATION, THE SHEAR FORCE DECREASES BY AN AMOUNT EQUAL TO THE MAGNITUDE OF THE DOWNWARD LOAD. BENDING MOMENT DOES NOT CHANGE AS WE PASS THROUGH THE POINT OF APPLICATION OF A CONCENTRATED LOAD. HOWEVER, THE RATE OF CHANGE OF BENDING MOMENT (dM/dx) UNDERGOES AN ABRUPT CHANGE. THIS ABRUPT CHANGE IS TO LOAD P AS AT THE LEFT-HAND SIDE dM/dx = V AND AT THE RIGHT-HAND SIDE dM/dx = V + V1.

27. NOW CONSIDER A COUPLE Mo IS APPLIED ON THE BEAM AS SHOWN IN THE FIGURE. FROM THE EQUILIBRIUM OF FORCES, FOLLOWING IS GET: V - (V + V1) = 0 V1 = 0 …………………….. (5) FROM THE EQUILIBRIUM OF BENDING MOMENMTS, WE GET - M + Mo – (V +V1) dx + M + M1 = 0 NEGLECTING VALUES WITH DIFFERENTIALS M1 = -Mo …………………. (6) RELATIONSHIPS - IN CASE OF A COUPLE

28. IT CAN BE CONCLUDED FROM EQUATIONS (5) & (6) THAT: SHEAR FORCE DOES NOT CHANGE AT THE POINT OF APPLICATION OF A COUPLE. THERE IS AN ABRUPT DECREASE IN THE BENDING MOMENT IN THE BEAM DUE TO THE APPLIED COUPLE Mo AS WE MOVE FROM LEFT TO RIGHT THROUGH THE POINT OF APPLICATION OF LOAD. EQUATIONS FROM (1) TO (6) SO FAR DERIVED WOULD BE USEFUL WHILE COMPLETE INVESTIGATION OF THE SHEAR FORCES AND BENDING MOMENTS IN THE BEAM WOULD BE REQUIRED.

29. A BEAM “ABC” WITH AN OVERHANG SUPPORTS A UNIFORM LOAD OF INTENSITY “q = 6kN /m AND A CONCENTRED LOAD “P = 30 kN. CALCULATE THE SHEAR FORCE “V” AND BENDING MOMENT “M” AT A CROSS SECTION “D” LOCATED 5m FROM THE LEFT-HAND SUPPORT. THIS PROBLEM CAN BE SOLVED FIRST BY CALCULATING THE REACTIONS FROM EQUATIONS OF EQUILIBRIUM OF THE ENTIRE BEAM. THUS TAKING MOMENTS ABOUT THE SUPPORT B, FOLLOWING IS OBTAINED: - Ra (8) + (28) (5) + (6) (10) (3) = 0 Ra = 40 kN “V” PROBLEM

30. SIMILARLY TAKING MOMENTS ABOUT SUPPORT A, FOLLWING VAUES ARE OBTAINED: Rb (8) - (28) (3) - (6) (10) (5) = 0 Rb = 48 kN IN ORDER TO CALCULATE THE SHEAR FORCE AND BENDING MOMENT , BEAM IS CUT AT A SECTION D AND CONSTRUCT A FREE-BODY DIAGRAM OF LEFT HAND PART OF THE BEAM. AT THE TIME OF DRAWING THE DIAGRAM, IT IS ASSUMED THAT THE UNKNOWN STRESS RESULTANTS “V” AND “M” ARE POSITIVE. “V” PROBLEM

31. NOW THE EQUATIONS OF EQUILIBRIUM FOR THE FREE BODY DIAGRAM ARE: ΣFy = 0 Ra - P - qx - V = 0 V = - 18 Kn ΣM = 0 - Ra (5) + P (2) + q (5) (2.5) + M = 0 M = 69 KN.m IN A SIMILAR WAY FOR RIGHT HAND PART, BOTH EQUATIONS WOULD BE ΣFy = 0 Rb - qx - V = 0 ΣM = 0 Rb (3) - q (5) (2.5) - M = 0 “V”

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