# Computer Aided Design - Universiti Teknologi Malaysia

Bending Shear and Moment Diagram, Graphical method to construct shear and moment diagram, Bending deformation of a straight member, The flexure formula 1 2 Shear and moment diagram Axial load diagram Torque diagram Both of these diagrams show the internal forces acting on the members. Similarly, the shear and moment diagrams show the internal shear and moment

acting on the members 3 Type of Beams Statically Determinate Simply Supported Beam Overhanging Beam Cantilever Beam 4 Type of Beams Statically Indeterminate Continuous Beam

Propped Cantilever Beam Fixed Beam 5 6 Example 1 F A V B

M x Equilibrium equation for 0 x 3m: F y 0 F V 0 V 9kN M 0 Vx M 0

M 9 x(kNm) * internal V and M should be assumed +ve Shear Diagram F V M x Sign convention: V= -9kN Lecture 1 8 Shear Diagram

F V = -9 kN M = -9x kN.m x M=-9x Sign convention: M= -9x kNm X=0: M= 0 X=3: M=-27kNm 9 At cross section A-A At section A-A

V=9kN M=X X 10 Example 2 1) Find all the external forces M A 0; 10(1) C y ( 2) 0 C y 5 kN Fy 0

A y 5 10 0 A y 5 kN 11 Force equilibrium Moment equilibrium 0 x 1m 0 x 1m F M

y 0 A 0 M Vx 0 F V 0 M Vx 5kNm(ccw) V 5kN (down) Force equilibrium

Moment equilibrium 1 x 2m 1 x 2m F M y 0 F 10 V 0 V 5kN (up)

A 0 Vx 10(1) M 0 M 10 Vx M (10 5 x)kNm(ccw) 12 Boundary cond for V and M 0 x 1m V 5kN (down) M 5 x(ccw) M=5x

1 x 2m V 5kN (up ) M 10 5 x(ccw) M=10-5x Solve it Draw the shear and moment diagrams for simply supported beam. 14 15 Distributed Load For calculation purposes, distributed load can be represented as a single load

acting on the center point of the distributed area. Total force = area of distributed load (W : height and L: length) Point of action: center point of the area 16 Example 17 Example 18 Solve it Draw the shear and moment diagrams the beam:

19 Solving all the external loads Distributed load will be F Wl 8(6) 48kN Solving the FBD M A 0

Bx 4 F (3) 0 48(3) 4 Ay 12kN By 36kN Ax 0 Boundary Condition 0 x 4 Equilibrium eq F

Y 0 12 8 x V 0 V 12 8 x M A 0 M Vx 4 x 2 0 M (12 8 x) x 4 x 0 M 12 x 4 x 2 0

M 12 x 4 x 2 21 Boundary Condition 4 x 6 Equilibrium eq F Y 0 12 36 8 x V 0

V 48 8 x M A 0 M Vx 8 x( x / 2) 0 M (48 8 x) x 36( 4) 4 x 2 0 M 48 x 144 4 x 2 0 M 4 x 2 48 x 144 22 0 x 4

V 12 8 x x=0 V= 12 kN x=4 V=-20 kN M 12 x 4 x 2 x=0 M= 0 kN x=4 V=-16 kN 4 x 6 V 48 8 x x=4 V= 16kN x=6 V= 0 kN M 4 x 2 48 x 144 x=4 V= -16kN x=6 V= 0 kN

23 Graph based on equations Straight horizontal line y=c y = mx + c y=3x + 3 y=-3x + 3 y = ax2 + bx +c y=3x2 + 3 y=-3x2 + 3

24 25 Graphical method Relationship between load and shear: Fy 0 : V w x x (V V ) 0 V w x x Relationship between shear and bending moment: M o 0 : Vx M w x x k x M M 0 M Vx w x k x

2 26 26 Dividing by x and taking the limit as x0, the above two equations become: Regions of distributed load: dV w(x) dx Slope of shear = distributed diagram at load intensity

each point at each point dM V dx Slope of moment diagram at each point = shear at each point 27

27 Example 28 29 The previous equations become: V w( x) dx change in shear = Area under

distributed load M V ( x) dx change in moment = Area under shear diagram 30 +ve area under shear diagram 31

32 33 Bending deformation of a straight member Observation: - bottom line : longer - top line: shorter - Middle line: remain the same but rotate (neutral line) 34 Strain s ' s

s 0 s lim Before deformation s x After deformation, x has a radius of curvature , with center of curvature at point O s x Similarly Therefore s' ( y )

( y ) s 0 y lim 35 Maximum strain will be max max

( c y/ ) c/ y ( ) max c y ( ) max c

-ve: compressive state +ve: tension 36 The Flexure Formula The location of neutral axis is when the resultant force of the tension and compression is equal to zero. FR F 0 Noting dF dA 0 dF dA A

y ( ) max dA c A max c ydA A 37 Since max

0, therefore c ydA 0 A Therefore, the neutral axis should be the centroidal axis 38 ( M R ) Z M Z Maximum normal stress M ydF y dA A

A y y ( ) max dA c A max c max 2 y dA A

Mc I Normal stress at y distance My I 39 Line NA: neutral axis Mc

max Red Line: max normal stress I c = 60 mm Yellow Line: max compressive stress c = 60mm Mc max I Line NA: neutral axis My1 1 Red Line: Compressive stress

I y1 = 30 mm Yellow Line: Normal stress My y2 = 50mm 2 2 I Refer to Example 6.11 pp 289 I: moment of inertial of the cross sectional area I x x I x x bh 3

12 r 4 D4 4 64 Find the stresses at A and B I: moment of inertial of the cross sectional area Locate the centroid (coincide with neutral axis) n

yA i y i i 1 n A i i 1

y1 A1 y2 A2 A1 A2 150(50)(300) 325(50)(300) (50)(300) (50)(300) 237.5mm I: moment of inertial of the cross sectional area Profile I I about Centroidal axis II A

A bh 3 50(300) 3 112.5(10) 6 mm 4 12 12 I about Axis A-A using parallel axis theorem bh3 Ad 2 112.5(10) 6 (50)(300)(87.5) 2 12 227.344(10) 6 mm 4 ( I I ) A A

Profile II bh 3 (300)(50)3 Ad 2 (300)(50)(87.5) 2 12 12 6 4 117 .969(10) mm ( I II ) A A Total I I A A ( I I ) A A ( I II ) A A 227.344(10) 6 117 .969(10) 6 mm 4

345.313(10) 6 mm 4 * Example 6-12 to 6-14 (pp 290-292) Solve it If the moment acting on the cross section of the beam is M = 6 kNm, determine the maximum bending stress on in the beam. Sketch a three dimensional of the stress distribution acting over the cross section If M = 6 kNm, determine the resultant force the bending stress produces on the top board A of the beam Total Moment of Inertia 300( 40) 3 40(300) 3 2 I I 2[

(300)(40)(170) ] 12 12 786.8(10) 6 mm 4 Max Bending Stress at the top and bottom M top 1.45MPa Mc 6000(10)3 (190) 1.45MPa I I

M bottom 1.45MPa Bottom of the flange 1.14MPa 6kNm M f _ top Mc 6000(10) 3 (150) 1.14MPa I I M f _ bottom 1.14MPa Resultant F = volume of the trapezoid

300 mm (1.45 1.14) (40)(300) 15540 N 2 15.54kN FR 40 mm 1.45MPa 1.14MPa Solve it The shaft is supported by a smooth thrust load at A and smooth journal bearing at D. If the shaft has the cross section shown, determine the absolute maximum bending stress on the shaft

External Forces Draw the shear and moment diagram M A 0 FD (3) 3(0.75) 3(2.25) FD 3kN FA 3kN Absolute Bending Stress Mmax = 2.25kNm

Mc 2250(10) 3 (40) max I ( 40 4 254 ) 4 52.8MPa

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