Lecture-6 (Flexural Formula).pptx

Flexural Formula
Mechanics of solid (MOS)
Stree strain
Civil engineering

1. Flexural Stresses and Flexural Formula By Dr. Salah Uddin

2. Subject : Mechanics of Solids-I Understand the analysis of bending stresses in beams

3. After the end of this lesson students will able to; State the theory of simple bending. Explain bending stresses. Identify position of neutral axis. Define moment of resistance. Define section modulus. Solve problems related to bending stresses in beam.

4. If a specific length of a beam is subjected to a c on s t a n t bend i n g mo m e n t & shea r f o r ce is z e r o , then the s t r esses s e t u p in th a t l e n g th o f the be a m a r e kn o wn as b end in g s t r esse s . Th e l en g th o f the be a m un d er c on st a n t be n d i ng mom e n t is sa id t o b e in p ur e bending. St r esses in B eams – Bendi n g and Shear

5. Internal bending moment causes beam to deform Top fibers in compression, bottom in tension. N e ut r al sur f ace – n o c h an g e in len g t h . N e ut r al Axis – L i n e of i n t e r se c ti o n o f n e ut r al sur f a c e with the t r an s v e r se se c ti on. All c r os s - sect i on s r em a in p l ane and p e rp e nd i cu la r t o l ong itud i nal a x i s. St r esses in B eams – Bendi n g and Shear

6. A s su m p t i on s in s i mp l e (pu r e) bend i n g theory: M at erial o f bea m is h om o g enous and i s o t r o p ic (same c omp o s iti o n & c on st a n t E in all d i r e c ti on s ). Y oun g ’ s modulus is c on s t a n t in c omp r ession and t ension. T r an s v e r se se c ti o n which a r e p l ane b e f o r e bend i n g r em a in p lain a f t er bend i n g ( Elimi n at e e f f e c ts o f s t r ai n s in o ther d i r e c ti on). B e am is i n itiall y straight, and all l on g itud ina l f il am e n ts bend in ci r cula r a r cs. Ra d i u s o f c u r v a tu r e is l a rg e c ompa r ed with d imens io n of c r o s s sect i ons. E ach l a y er o f the beam is f r ee t o e xp a n d o r c o n t r ac t . St r esses in B eams – Bendi n g and Shear

7. Der i v a ti o n o f R el a ti onsh ip B e t w een Bendi n g St r ess a n d Ra d i u s o f C u r v a tu r e Du e t o a c ti o n of bend i n g , the l e n g t h dx wi l l b e d e f ormed as s h o wn i n t h e f ig u r e (b ) Consider a small length dx of a beam subjected to a simple bending as shown in the figure (a)

8. Du e t o the dec r ease i n len g th of the l a y e r s abo v e , the s e l a y e r s will b e subjec t ed to c omp r ess i v e s t r esse s . Du e t o the i nc r ease i n len g th of the l a y e r s below , the s e l a y e r s will b e subjec t ed to t ens i le s t r esse s .   The amou n t b y wh i c h a l a y er i nc r eases or dec r eas e s i n len g th, depe n d s u po n the pos i t i on of the l a y er w . r . t . .   Der i v a ti o n o f R el a ti onsh ip B e t w een Bendi n g St r ess a n d Ra d i u s o f C u r v a tu r e

9. L e t, R = Radius of cu r v a tu r e of neu t r al l a y er N ’ -N ’ . θ = Angle subjec t ed a t O b y A ’B’ and C ’D’ p r oduc e d. y = D i s t ance f r om the neut r al l a y e r . Or igi na l l en g th of t h e l a y er F r om figu r e (b), Chan g e (In c r ease) in l en g th of t h e St r ain i n the l a y er E F = I n c r ea s e i n l en g th / origi na l l en g th Ac c o r d i n g t o li nea r ela s ti c i t y , , T h a t i s ,   Der i v a ti o n o f R el a ti onsh ip B e t w een Bendi n g St r ess a n d Ra d i u s o f C u r v a tu r e    

10. Th e s t r e ss es i n du c ed i n t h e l a y e r s of t h e bea m c r e a t e c omp r ess i v e and t en s i le f o r ces . Thes e f o r ces wi l l h a v e mome n t about N A . Th e t o t al mome n t o f these f o r ces about N A f o r a s e cti on i s kn o wn as mome n t of r es i st ance of th a t secti o n . Cons i de r a c r os s sect i o n of a bea m as show n D e rivat i o n o f R e l a tio n sh i p B e twe e n B e n d i n g Stress a n d R a d i us o f C urvatu r e (Mom en t o f R esistanc e o f a S ection)

11. D e rivat i o n o f R e l a tio n sh i p B e twe e n B e n d i n g Stress a n d R a d i us o f C urvatu r e (Mom en t o f R esistanc e o f a S ection) E u l e r – B e rn o u l l i B e n d i n g E q u a tion         Total moment of the forces on the section of the beam (or moment of resistance)     or    

12. Sect i o n Modu lu s (Z) It is the r a tio o f mome n t o f i nert i a o f a sect i on abo u t the n e ut r al a x is t o the d i st ance o f t he ou t ermo s t l a y er f r o m the n e ut r al a x i s. T hus , mome n t o f r esi s t an c e o f f e r ed b y the se c ti o n i s m a ximum when Z i s m a x imum. Hen c e, Z r ep r ese n ts the s t r en g th o f the se c ti on.   I = Mome n t o f Inerti a about neut r al a x i s. = D i s t ance o f the o u t ermo s t l a y er f r o m the neut r al a x i s.   Hence,  

13. Sect i o n Modu lu s (Z) 1 . Recta n g u la r Secti o n      

14. Sect i o n Modu lu s (Z) 2 . Recta n g u la r Holl o w Secti o n      

15. Sect i o n Modu lu s (Z) 3 . Circula r Section      

16. Sect i o n Modu lu s (Z) 4 . Circula r Hollo w Section      

17. P r ob lems A c a n t i l e v er of l e n g th 2 m f ai l s w he n a l oa d of 2 kN i s a pp l i ed a t the f r ee en d . I f the sect i o n of t h e beam i s 4 m m X 6 mm , f i n d the s t r ess a t the f ail u r e. So l ut i on : P r ob le m S k et c h:

18. L e t, σ m a x = St r ess a t f ail u r e (M a x imum s t r ess) S i nc e R i s no t gi v en, w e c annot use In s t ead, use OR   , wh e re   a n d,     He n c e ,    

19. P r ob lem A squa r e bea m i n sec t i o n a n d 2 m l on g i s s i mp l y s u ppor t ed a t the ends. Th e beam f ails when a p o i n t l oa d of 40 N i s app l ied a t the ce n t r e of the bea m . W h a t un i f orm l y d i s tr i bu t ed l oa d pe r m e t er len g th will b r eak a c a n til e v er of the sam e m a t erial 4 m m wi de , 60m m d e ep and 3 m l ong? So l ut i o n: Squa r e c / s. : 2 m m x 2 mm ; L = 2m ; W = 40 N R ec t angular c / s. : 4 m m x 6 mm ; L = 3m ; w = ? E qua t e the maxim u m s tress i n the two c ase s .  

20. M a x imum s t r ess i n bea m of squa r e c / s.: , whe r e  

21. M a x imum s t r ess i n bea m of r ec t angular c / s.: M = σ m a x x Z , whe r e M a x imum bend i n g mome n t f o r a c a n til e v er beam l oade d wi t h UDL f o r the e n t i r e spa n i s gi v en by M m ax = w L 2 / 2 He n ce, M m ax = M m ax = 450 w N-mm

22. P r ob lem A ti m ber beam of rectan g u l ar secti o n of l e ngth 8 m i s sim p l y supporte d . The beam c a r r i es a U. D . L . of 1 2 kN/m run over the en t ire le n gth and a p o i nt load o f 1 kN at 3m f r om t h e l eft support . If t h e depth i s two t i mes the wi d th and the stres s i n t he timber i s not t o exc e ed 8 N/mm 2 , find t he s uitable dime n sion s of the sectio n . Solution : Probl e m sk e tch

23. Gi v e n D a t a: F i n d the m a x imum bend i n g mome n t i n the beam and us e M = σ m a x x Z t o find b an d d .

24. Fin d t he react ion forces at the supports Fin d t he shea r force at all the points of in t ere st Fin d t he m axi m um bending m o m ent wher e S F is z ero . Shea r Fo r ces : F B = - R B = - 51750 N ; F C = +18250 N ; F A = +54250 N Shea r force cha nges i t s sign between B and C . Let D be the point wher e S F = 0. Let x be the dis t ance ( i n m ete rs) of t his point on the beam from B. By calculat i on, x = 4.3125 m

25. Fin d t he BM at D . Fin d t he Mo m ent of Iner t ia ( I ) or s ec t ion m odulus (Z) of t he bea m ’ s c r oss se ct i on Us e the equat i on, M = σ ma x x Z Hence, b = 275.5 m m d = 551 m m

26. Any Questions ?

27. Thank you 28