chapter 8 _ slope and deflection _ strength of materials - part 1

Strength of Materials - Part 1

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Chapter 1 : Simple Stress and

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Chapter 2 : Principal Stresses

and Strains

Chapter 3 : Bending Moment

and Shear Force Diagrams

Chapter 3 : Part 2

Chapter 4 : Simple Bending of

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Chapter 5 : Torsion

Chapter 6 : Thin Cylinders and

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Chapter 7 : Columns and Struts

Chapter 8 : Slope and

Deflection

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Chapter 8 : Slope and Deflection

8.1 RELATIONSHIP BETWEEN MOMENT, SLOPE ANDDEFLECTIONThe curvature of a curve y = f() of radius R is given by,

8.2 MACAULAY’S METHOD OF INTEGRATIONThis method is suitable for beams subjected to concentratedloads and can be extended to uniformly distributed loads also.It consists of successive integration of expressions for bendingmoment in such a way that same constants of integration arevalid for all portions of the beam even though the law ofbending moment differs from portion to portion. The followingprocedure is adopted to apply this method:1. Write the general expression for the bending momentcovering all the loads just before the right hand support.2. A bracket is disregarded when for a particular value of x thecontents of the bracket becomes negative.3. The bracket terms must be kept intact and not multiplied outuntil a numerical substitution is made for x.4. If there is a uniformly distributed load starting anywhereand extending upto the right and then the general equation forbending moment will hold good for the entire beam. However,

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if the load does not extend upto the right end, the load may beextended upto the right end, and an equal and opposite loadmay be added to counteract the effect due to the additionalload. 8.2.1 Concentrated Load on a Simply Supported BeamConsider a simply supported beam AB of span 1 carryingconcentrated load W at C at a distance ‘a’ from left endsupport.



8.2.2 Uniformly Distributed Load on Simply SupportedBeamConsider beam AB of length 1, simply supported at the endsand carrying a uniformly distributed load of intensity w per unitlength, as shown in Fig. 8.2. The bending moment at a sectionx — x is



8.2.3 Concentrated Load at the Free end of a CantileverConsider a cantilever beam AB of span 1carrying concentratedload W at the free end,a shown in Fig. 8.3. Bending moment at adistance x from thefree end is

Consider a cantilever beam AB of span 1carrying uniformlydistributed loadof intensityto per unit length over the whole span, as shownin Fig. 8.4.The bending moment at a distance x from free end A is



8.2.5 Cantilever Carrying a Concentrated Load not at theFree EndConsider a cantilever AB of span 1carrying a concentrated load W at a distance‘a’ from the freeend A, as shown in Fig. 8.5.The bending moment at a distance x from thefree end is,



Example 8.1. A beam AB, 15 m long is simply supportedat the ends. It carries two concentrated loads of 900 kNand 60 kN at 4m and lOm respectively from the left end.Moment of inertia of the beam about the neutral axis is15 x 10 m4 and E = 200 GPa. Calculate the deflection ofthe beam under the two loads.



Example 8.2 A beam AB, simply supported at the ends,is 5m long. It is loaded as shown in Fig. 8.7 (a). If E =210 GPa and I = 1 x 104m4 calculate the deflection atthe centre. Solution. The udi has been extended to the end B and anequal and opposite udi is applied on the span DB, as shown inFig. 8.7 (h).



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