1. Macaulay’s Method Lecture No-2 J P Supale Mechanical Engineering Department SKN SITS LONAVALA Strength of Materials

2. Step 1 – Calculate the reaction at supports. Ler R A be the reaction at support A and R B be the reaction at support B. Strength of Materials A B W1 W2W3 L1 L2L3L4 L

3. Step 2 – Take a section x-x at a distance X from left support, in last part of beam. Strength of Materials A B W1 W2W3 L1 L2L3L4 X X X

4. Step 3 – Take moment of all forces about section x-x Step 4 – Separate the moment of each force by using compartment line Strength of Materials

5. Step 5 – Use the differential deflection equation Step 6 – Equate the left hand side of it with step 4 equation Strength of Materials

6. Step 7 – Integrate the above equation Note:- 1. while integrating keep constant of integration generally in first compartment only. 2. take as a single term only and integrate it. Strength of Materials

7. Step 8 – Again integrate the above equation. Step 9 – Calculate the values of constants of integration C 1 and C 2, by keeping the boundary condition values. Put values of C 1 and C 2, in step 7 to get Slope Equation and in step 8 to get Deflection equation. Strength of Materials

8. Example-1 A Simply supported beam subjected to central point load W. Determine maximum slope and deflection. Strength of Materials W L A B L/2

9. Example-1 Step 1 – Calculate the reaction at supports. R A = R B =W/2 Step 2- Take a section x-x at a distance X from left support, in last part of beam. Strength of Materials W L A B L/2 X X X

10. Example-1 Step 3- Take moment about section X-X. And put compartment line for each force. Step 4 – Equate it with differential deflection equation and put value of R A. Step 5 – Integrate above equation. Strength of Materials

11. Example-1 Step 6 – Again integrate above equation Step 7 – Apply boundary conditions, we know that at A X=0 and y=0 put these values in step 6, we get Step 8 – Put C2 =0, and at B, X=L and y=0 in step 6 again, we get Strength of Materials

12. Example-1 Step 9 – Put values of C 1 and C 2 in step 5 and 6 to get following equations: Slope Equation Deflection Equation Strength of Materials

13. Example-1 Step 10 – Calculate slope at A, put X=0,in slope eq. Calculate slope at B, put X=L in slope eq. Strength of Materials

14. Example-1 Step 11 – Calculate deflection under load, put X = L/2 Strength of Materials

15. x y P B L A M xz Q xy x P P.L Example:

16. P To find C 1 and C 2 : Boundary conditions: (i) @ x=0 (ii) @ x=0 Equation of the deflected shape is: v Max occurs at x=L

17. ab L Macaulay’s Notation y x Example: Q xy M xz P x P

18. Boundary conditions: (i) @ x=0 (ii) @ x=L From (i): From (ii): Since (L-a)=b Equation of the deflected shape is:

19. This value of x is then substituted into the above equation of the deflected shape in order to obtain v Max. To find v Max : v Max occurs where (i.e. slope=0) Assuming y Max will be at x<a, when P v Max Note: if

20. Workout Numerical A cast iron beam 40 mm wide and 80 mm deep is simply supported on a span of 1.2 m. The beam carries a point load of 15 kN at the centre. Find the deflection at the centre. Take E = 108000 N/mm 2.  b=40mm, d=80mm, L=1.2m  W=15kN, E=108000 N/mm 2 Strength of Materials