Stress Analysis Lecture 4 ME 276 Spring Dr./ Ahmed Mohamed Nagib Elmekawy

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1 Stress Analysis Lecture 4 ME 76 Spring Dr./ Ahmed Mohamed Nagib Elmekawy

2 Shear and Moment Diagrams Beam Sign Convention The positive directions are as follows: The internal shear force causes a clockwise rotation of the beam segment on which it acts. The internal moment causes compression in the top fibers of the segment such it bends the segment so that it hold water.

3 Example 1 Draw the shear and moment diagrams for the beam shown in the attached figure. (1) 0 0 x L w V V wx wl F y Shear and Moment Diagrams () 0 0 x Lx w M M x wx x wl M

4 Shear and Moment Diagrams M max w L L L wl 8 Note V dm dx

5 Shear and Moment Diagrams Example Draw the shear and moment diagrams for the beam shown in the attached figure. 0 x 1 5 F y V 0 V 5.75 kn (1) M M x M x 80 kn.m () 1 1

6 Shear and Moment Diagrams 5 x V F y x 5V x kn (3) x M M 5 0 x x 15.75x 9.5 kn.m (4) x 5 x M 0

7 The Bending Formula

8 The Bending Formula

9 The Bending Formula

10 The Bending Formula M I y Where max M I c M I c max y the resultant internal moment acting at the cross section. the moment of inertia of the cross-sectional area about the neutral axis. perpendicular distance from the neutral axis to a point farthest away from the neutral axis. This is where max acts. the maximum bending stress in the shaft, which occurs at the outer surface. perpendicular distance from the neutral axis to a point where is to be calculated

11 Bending a Cantilever Beam under a Concentrated Load 11

12 Bending a Cantilever Beam under a Concentrated Load 1

13 Bending Stress 13

14 Bending Stress 14

15 Bending Stress 15

16 Bending Stress 16

17 Bending Stress 17

18 Bending Stress 18

19 Bending Stress 19

20 Bending of Curved beam Displacement Stress in x direction 0

21 The Bending Formula Example 1 A beam has a rectangular cross section and is subjected to the stress distribution shown in the figure. Determine the internal moment M at the section caused by the stress distribution. M max 3 I bh 1 M c I 88* N.mm 840*10 mm 0 M *10 4

22 The Bending Formula Example The simply supported beam has the cross-sectional area shown in the figure. Determine the absolute maximum bending stress in the beam and draw the stress distribution over the cross section at this location.

23 I The Bending Formula I 1 1 max 1 1 M c I 1.7 Ad * 50* *10 mm MPa 6.5*10 * *10 ; max 6

24 The Bending Formula Example 3 The beam shown in the figure has a cross-sectional area in the shape of a channel as shown in the attached figure. Determine the maximum bending stress that occurs in the beam at section a-a. y ~ ya A *100*15*00 10* 0*50 *15*00 0* mm 0 400* *59.09 M 0 M M NA 4859*10 3 N.mm

25 The Bending Formula I I Ad * 00* * 0* *10 6 mm 4 max M c I 16. MPa *10 * *10 ; max 6

26 The Shear Formula VQ It Where V The shear stress in the member at the point located a distance y from the neutral axis. This stress is assumed to be constant and therefore averaged across the width t of the member. The internal resultant shear force, determined from the method of sections and the equations of equilibrium.

27 The Shear Formula I The moment of inertia of the entire cross-sectional area calculated about the neutral axis. t The width of the member s cross-sectional area, measured at the point where is to be calculated. Q =, where A is the area of the top portion of the member s cross-sectional area, ya above the section plane where t is measured, and neutral axis to the centroid of A. y is the distance from the Example 1 Determine the distribution of the shear stress over the cross section of the beam shown in the attached figure.

28 The Shear Formula 4 1 y h b y h y h y y A Q Applying the shear formula, we have y h bh V b bh b y h V It VQ This indicates that the shear stress distribution over cross section is parabolic with maximum value of (1.5V/A) at the neutral axis and zero at top and bottom as shown in the attached figure.

29 The Shear Formula It is important to realize that max also acts in the longitudinal direction of the beam as shown in the attached figure. It is this stress that can cause a timber beam to fail as shown in the figure. Here horizontal splitting of the wood starts to occur through the neutral axis at the beam s ends, since the vertical reactions subject the beam to large shear stress and wood has a low resistance to shear along its grains, which are oriented in the longitudinal direction.

30 Shear Stress due to bending 30

31 The Shear Formula Example A steel wide-flange beam has the dimensions shown in the figure. If is subjected to a shear of V = 80 kn, plot the shear-stress distribution acting over the beam s cross sectional area. I *10 mm For point B t B Q B 300 mm ya *0 66*10 4 mm

32 The Shear Formula B VQ It B For point B B 80*1000*66* *10 * MPa t B Q B 15 Q mm B 66*10 4 mm 3 B VQ It B B 80*1000*66* *10 * MPa For ppint C t C Q C 15 mm ya 110 0* * *10 3 mm 3

33 The Shear Formula C VQ It C C 80*1000*735* *10 * MPa From the attached figure, note that the shear stress occurs in the web and is almost uniform throughout its depth, varying from.6 MPa to 5. MPa. It is for this reason that for design, some codes permit the use of calculating the average shear stress on the cross section of the web rather than using the shear formula; that is, avg A V 80* * MPa

34 The Shear Formula

35 State of Stress Caused by Combined Loadings In previous chapters we developed methods for determining the stress distribution in a member subjected to either an internal axial force, a shear force, a bending moment, or a torsional moment. Most often, however, the cross section of a member is subjected to several of these loadings simultaneously. When this occurs, the method of superposition can be used to determine the resultant stress distribution.

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43 Summary Principal Stresses 43

44 Example 1 For the state of plane stress shown in Fig., determine (a ) the principal planes, (b ) the principal stresses, (c ) the maximum shearing stress. 44

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47 Sample Problem 7.1 in Beer s Book 47

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52 State of Stress Caused by Combined Loadings Example 1 A force of 15 kn is applied to the edge of the member shown in the figure. Neglect the weight of the member and determine the state of stress at points B and C.

53 State of Stress Caused by Combined Loadings Normal Force A P * MPa Bending Moment max Mc I 15000* MPa Superposition

54 State of Stress Caused by Combined Loadings x 100 x x 33.3 mm Example B C MPa MPa The member shown in the figure has a rectangular cross section. Determine the state of stress that the loading produces at point C.

55 State of Stress Caused by Combined Loadings Normal Force c A P 16.45* * MPa Shear Force c 0

56 State of Stress Caused by Combined Loadings Bending Moment C Mc I Superposition 1.93*1000*1500* MPa C MPa

57 State of Stress Caused by Combined Loadings Example 4 The solid rod shown in the figure has a radius of 7.5 mm. If it is subjected to the force 500 N, determine the state of stress at point A. Normal Force A y P A MPa

58 State of Stress Caused by Combined Loadings Bending Moment A y Mc I Superposition 500*140* MPa MPa A y

59 State of Stress Caused by Combined Loadings Example 5 The solid rod shown in the figure has a radius of 7.5 mm. If it is subjected to the force 800 N, determine the state of stress at point A. Shear Force Q ya 4* mm

60 State of Stress Caused by Combined Loadings yz A VQ It 800* * Bending Moment A MPa Torque yz A Tc J 11*1000* MPa Superposition yz A MPa

61 State of Stress Caused by Combined Loadings yz A VQ It 800* * Bending Moment A MPa Torque yz A Tc J 11*1000* MPa Superposition yz A MPa