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☻. 1.0 Axial Forces. 2 .0 Bending of Beams. Now we consider the elastic deformation of beams (bars) under bending loads. M. M. www.engineering.auckland.ac.nz/mechanical/MechEng242. Normal Force:. F n. F n. S.B. Shear Force:. F t. F t. Bending Moment:. M t. M t. K . J .
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☻ 1.0 Axial Forces 2.0 Bending of Beams Now we consider the elastic deformation of beams (bars) under bending loads. M M www.engineering.auckland.ac.nz/mechanical/MechEng242
Normal Force: Fn Fn S.B. Shear Force: Ft Ft Bending Moment: Mt Mt K.J. Torque or Twisting Moment: Mn Mn Application to a Bar
Excavator Atrium Structure Yacht Car Chassis Examples of Devices under Bending Loading:
(Refer: B,C & A –Sec’s 6.3-6.6) 2.2 Stresses in Beams sx sx P x Mxz Mxz 2.3 Combined Bending and Axial Loading (Refer: B,C & A –Sec’s 6.11, 6.12) P1 P2 2.4Deflections in Beams (Refer: B,C & A –Sec’s 7.1-7.4) 2.5Buckling (Refer: B,C & A –Sec’s 10.1, 10.2) 2.0 Bending of Beams 2.1 Revision – Bending Moments (Refer: B,C & A – Sec’s 6.1,6.2)
12 kN Q (SFD) 3m 3m 0 (BMD) M 0 2.1Revision – Bending Moments RECALL… (Refer: B, C & A – Chapter 6) Last year Jason Ingham introduced Shear Force and Bending Moment Diagrams.
y x Deflected Shape B A Mxz Mxz RAy RBy Mxz Mxz What stresses are generated within, due to bending? Consider the simply supported beam below: (Refer: B, C&A – Sections 1.14, 1.15, 1.16, 6.1) Radius of Curvature, R P
P A B W u Mxz Mxz RAy RBy Recall: Axial Deformation Bending Load (W) Bending Moment (Mxz) Axial Stiffness Flexural Stiffness Extension (u) Curvature (1/R)
sx (Compression) Mxz Mxz sx=0 sx (Tension) sx is NOT UNIFORM through the section depth sxDEPENDS ON: Axial Stress Due to Bending: Mxz=Bending Moment y x Beam Unlike stress generated by axial loads, due to bending: (i) Bending Moment, Mxz (ii) Geometry of Cross-section
Mxz Mxz -vesx Qxy Qxy +vesx +VE (POSITIVE) “Happy” Beam is +VE “Sad” Beam is -VE Qxy=Shear Force Sign Conventions: y Mxz=Bending Moment x
Mxz=P.L RAy=P P.L Mxz P Qxy P x Mxz Mxz Mxz Qxy Qxy Qxy Q & M are POSITIVE Example 1: Bending Moment Diagrams P y x A B L
P.L Mxz Qxy x P P +ve Qxy 0 Mxz 0 -ve -P.L To find sx and deflections, need to know Mxz. P y L x A B Shear Force Diagram (SFD) Bending Moment Diagram (BMD)
a b x P a Mxz A Qxy Where can only be +VE or ZERO. y P Example 2: Macaulay’s Notation x A C B
(i) When 0 (ii) When Mxz BMD: Eq. 1 Eq. 2 +ve 0 A C B y P a b x A C B x 1 2
Distributed Load w per unit length x RAy=wL wL2 Mxz= 2 Mxz wx Qxy wL wL2 Mxz 2 Qxy Example 3: Distributed Load y A B x L
Mxz L BMD: 0 -ve x -wL2 2
Apart from the revision problems on Sheet 4, you might try these sources: • B, C & A Worked Examples, pg 126-132 Problems, 6.1 to 6.8, pg 173 • Jason Ingham’s problem sheets: www.engineering.auckland.ac.nz/mechanical/EngGen121 Summary – Is anything Necessary for Revision Generating Bending Moment Diagrams is a key skill you must revise. From these we will determine: • Stress Distributions within beams, • and the resulting Deflections