1.  2005 Pearson Education South Asia Pte Ltd TUTORIAL-3 : BENDING 1 Using graphical method, draw the shear and bending moment diagrams for the beam shown in the figure. Determine the absolute maximum bending stress. PROBLEM-1 FREE BODY DIAGRAM and SUPPORT REACTIONS  M A = 0; D y (6) – P 2 (4) – P 1 (2) = 0 D y = (32 + 10)/6 = 7 kN  F y = 0; A y – P 1 – P 2 + D y = 0 A y = P 1 + P 2 – D y = 6 kN DyDy AyAy B 2 m P 1 = 5 kN P 2 = 8 kN C D A P 1 = 5 kN C A B P 2 = 8 kN D 2 m 6 m 2 m 5 cm5 cm 8 cm8 cm

2.  2005 Pearson Education South Asia Pte Ltd TUTORIAL-3 : BENDING 2 PROBLEM-1 DyDy AyAy B 2 m P 1 = 5 kN P 2 = 8 kN C D A SHEAR FORCE DIAGRAM V (kN) 6 1 –7 A1A1 A2A2 A3A3 SHEAR FORCE AREA: A 1 = (6)(2) = 12 kN.m A 2 = (1)(2) = 2 kN.m A 3 = (–7)(2) = –14 kN.m

3.  2005 Pearson Education South Asia Pte Ltd TUTORIAL-3 : BENDING 3 PROBLEM-1 DyDy AyAy B 2 m P 1 = 5 kN P 2 = 8 kN C D A M (kN.m) 12 14 A B C D BENDING MOMENT DIAGRAM M 0 = 0 M 2 = M 0 + A 1 = 0 + 12 = 12 kN.m M 4 = M 2 + A 2 = 12 + 2 = 14 kN.m M 6 = M 4 + A 3 = 14 – 14 = 0 Absolute maximum bending stress: 5 cm5 cm 8 cm8 cm

4.  2005 Pearson Education South Asia Pte Ltd TUTORIAL-3 : BENDING 4 PROBLEM-2 FREE BODY DIAGRAM and SUPPORT REACTIONS  M A = 0;  F y = 0; P 1 (2) – P 2 (2) + C y (4) = 0 C y = (30 – 20)/4 = 2.5 kN A y – P 1 – P 2 + C y = 0 A y = 22.5 kN 6 m P 1 =10 kN 2 m P 2 =15 kN B C A P 1 =10 kN P 2 =15 kN 2 m AyAy CyCy B C A If the rod bar has a diameter of 100 mm, Determine the absolute maximum bending stress in the shaft.

5.  2005 Pearson Education South Asia Pte Ltd TUTORIAL-3 : BENDING 5 PROBLEM-2 SHEAR FORCE DIAGRAM P 1 =10 kN P 2 =15 kN 2 m C y = 2.5 kN A y = 22.5 kN V (kN) - 10 12.5 - 2.5 A1A1 A2A2 A3A3 SHEAR FORCE AREA: A 1 = (–10)(2) = –20 kN.m A 2 = (12.5)(2) = 25 kN.m A 3 = (–2.5)(2) = –5 kN.m

6.  2005 Pearson Education South Asia Pte Ltd TUTORIAL-3 : BENDING 6 PROBLEM-2 BENDING MOMENT DIAGRAM P 1 =10 kN P 2 =15 kN 2 m C y = 2.5 kN A y = 22.5 kN M (kN.m) 5 M 0 = 0 M 2 = M 0 + A 1 = 0 – 20 = –20 kN.m M 4 = M 2 + A 2 = –20 + 25 = 5 kN.m M 6 = M 4 + A 3 = 5 – 5 = 0 - 20 Absolute maximum bending stress:

7.  2005 Pearson Education South Asia Pte Ltd TUTORIAL-3 : BENDING 7 EXAMPLE-3 If the shaft has a diameter of 50 mm, Determine the absolute maximum Bending stress in the shaft.

8.  2005 Pearson Education South Asia Pte Ltd TUTORIAL-3 : BENDING 8 EXAMPLE-3

9.  2005 Pearson Education South Asia Pte Ltd TUTORIAL-3 : BENDING 9 PROBLEM-4 Shaft is supported by smooth journal bearings at A and B. Due to transmission of power to and from the shaft, the belts on the pulleys are subjected to the tensions shown. Determine the smallest diameter of the shaft using the maximum-shear-stress theory, with  allow = 50 MPa.

10.  2005 Pearson Education South Asia Pte Ltd TUTORIAL-3 : BENDING 10 PROBLEM-4 Support reactions are calculated and shown on the free-body diagram of the shaft.

11.  2005 Pearson Education South Asia Pte Ltd TUTORIAL-3 : BENDING 11 PROBLEM-4 Bending-moment diagrams for M x and M z are shown above. y z y x M A1 = 0 M C1 = (475)(0.250) = 118.75 N.m M A2 = 0 M C2 = (150)(0.250) = 37.5 N.m M B2 = (150)(0.50) = 75 N.m

12.  2005 Pearson Education South Asia Pte Ltd TUTORIAL-3 : BENDING 12 PROBLEM-4 Torque diagram is also shown.

13.  2005 Pearson Education South Asia Pte Ltd TUTORIAL-3 : BENDING 13 EXAMPLE 11.6 (SOLN) By inspection, critical pts for bending moment occur either at C or B. Also, just to the right of C and at B the torsional moment is 7.5 N · m. Moment at B is At C, resultant moment is

14.  2005 Pearson Education South Asia Pte Ltd TUTORIAL-3 : BENDING 14 PROBLEM-4 Since the design is based on the maximum-shear-stress theory, Eqn 11-2 applies. The radical √(M 2 + T 2 ) will be the largest at section just to the right of C. We have Thus the smallest allowable diameter is

15.  2005 Pearson Education South Asia Pte Ltd TUTORIAL-3 : BENDING 15 PROBLEM-5 Determine the length of the length of the center portion of the bar so that the maximum bending stress at section A, B, and C is the same. The bar has a thickness of 10 mm. Reaction forces: Since the total weight is located in the middle of beam, the support reaction has the same value; W/2 = (350L)/2 = 175L W W/2 = 175L Free-body diagram

16.  2005 Pearson Education South Asia Pte Ltd TUTORIAL-3 : BENDING 16 PROBLEM-5 Stress concentration factor: Referring to the graph: K = 1.45

17.  2005 Pearson Education South Asia Pte Ltd TUTORIAL-3 : BENDING 17 PROBLEM-5

18.  2005 Pearson Education South Asia Pte Ltd TUTORIAL-3 : BENDING 18 PROBLEM-5