Static equilibrium requires a balance of forces and a balance of moments.

Transcription

1 Static Equilibium Static equilibium equies a balance of foces and a balance of moments. ΣF 0 ΣF 0 ΣF 0 ΣM 0 ΣM 0 ΣM 0 Eample 1: painte stands on a ladde that leans against the wall of a house at an angle of 0 o. ssume that the painte is a midheight of the ladde. Calculate the minimum coefficient of fiction fo static equilibium.

2 0 ) Σ Σ P m m P F P P P P F peson ladde μ μ μ μ μ μ 1 ) 1 ) 1 g m m P g m m P p l p l 0 0) ) 0) 0) 1 Sin L g m m P LCos LSin P p l μ Static equilibium: Sum of vetical and hoiontal foces must be eo. Taking the moment about point O and setting it equals to eo:

3 Load nalsis 3-D Newton Fist Law: bod at est tend to emain at est and a bod in motion at constant velocit will tend to maintain that velocit unless acted upon b an etenal foce. Newton Second Law: The time ate of change of momentum of a bod is equal to the magnitude of applied foce and acts in the diection of the foce. Newton Thid Law: when two paticles inteact, a pai of equal and opposite eaction foces will eist at thei contact point. This foce pai will have the same magnitude and act along the same diection line but have opposite sense.

4 Newton s second law can be witten fo a igid bod in two foms, one fo linea foces and one fo moments o toques M G H G ma F & Σ Σ FFoce; mmass, a acceleation M G moment about the cente of gavit G H & time ate of change of the moment o the angula momentum about the CG ma F ma F ma F Σ Σ Σ k j i H G ˆ ˆ ˆ ω ω ω M M M ω ω α ω ω α ω ω α ) ) ) Σ Σ Σ f the, and aes ae chosen to coincide with the pincipal aes of inetia of the bod. Whee, and ae the pincipal centoidal mass moments of inetia second moments of mass). n 3-D Eule s Equations

5 Fo -D: Whee α is the angula acceleation ΣF ma ΣF ma ΣM Moment of a Foce about an ais α

6 3-D Equilibium Eample: Two tansmission belts pass ove sheaves welded to an ale suppoted b beaings at B and D. The adius at.5 and at C. The ale otates at constant speed. Find T and the eaction foces at B, D. ssume that beaing at D eets no aial thust and neglect weights of sheaves and ale.

7 8in 6in 6in

8 B C D 33.75lb 67.5lb 33.75lb 33.75lb 33.75lb 0.5lb-in

9 B C D 4lb 70lb 8lb 4lb 8lb 336lb-in

10 Definitions Centoid of ea Cente of Gavit of an aea): Point, ) that defines the geometic cente of the aea Fist Moment of an ea with espect to the - ais Fist Moment of an ea with espect to the - ais Q Q τ VQ b

11 Eample : Calculate the cente of gavit of the ectangle: a)without a hole b)with a hole of dimensions c and d a) Without a hole 1 1 a b a b a a b b a b a b n i i n i i

12 b) With a hole d c b a d c c e b a a d c b a d c d f b a b n i i n i i ) ) 1 1 Second Moment o Moment of netia of an ea O J ρ Rectangula moments of inetia Pola moments of inetia

13 Radius of Gation O O O J Eample 3: Find the moment of inetia of the cicula aea about the and aes, the pola moment of inetia and the adius of gation about the and aes. φ φ Cos Sin Cos ea φ φ φ Cos J Cos Sin π π π φ φ φ π π

14 Paallel-is Theoem ' d Eample 4: ) 4 ' 4 ' π. π π d π

15 σ M

16

17 Mass Moment of netia t is the poduct of the element s mass and the squae of the element s distance fom the ais. m ) m a m ) m a m ) m a

18 Load Classification and Sign Convention Classification with espect to the method of application: a)nomal tensile b)nomal compessive c)shea d)bending e)tosion f) Combined

19 υ > 0 υ deflection The sign convention that will be used hee is as follows:

20

21 Distibuted loads, Shea Foce and Bending Moment in Beams

22 E -q q V E V V M E M E M υ υ υ υ

23 Successive ntegation Method V q V ) V ) q ) 600 C V ) 600 C Fo 0 < < 1 V ) 600 C1 Fo 0 C V ) Fo 1m then V 1) 350N < < 1 650N /3m 1050N

24 N V then m Fo V C V Fo C V Fo 1050 ) ) ) ) 1 < < 550 1) ) ) ) ) ) < < M then Fo C M then Fo C M V M V Fo

25 0 ) ) ) ) ) ) < < M then Fo C M then Fo C M V M V Fo M Ma 550N-m

26 Nomal Stess and Stain: Whee F: foce, nomal to σ F the coss-sectional aea, 0 : oiginal coss-sectional aea 0 1Pa 1 N.m - ; 1MPa 10 6 Pa; 1GPa10 9 Pa Nomal stain o tensile stain o ial stain ε l l l 0 0 Δl l 0

27 Hooke s Law When stains ae small, most of mateials ae linea elastic. Young s modulus Nomal: Δl F o lo E σ Εε σ Spings: the sping ate k F Δl o E l o E ε

28 Tosion Loading esulting fom the twist of a shaft. l Z θ θ γ θ, l G G G θ θ γ τ θ θ,, G Shea Modulus of Elasticit Shea stain Twist Moment o Toque l G T θ τ θ θ,, J ea Pola Moment of netia J G l T l J G T θ θ o J T τ θ, Thus: J T o Ma τ ngula sping ate: l G J T k a θ

29 Maimum Stesses in Beams We can obtain the nomal and shea stesses fom fleue and shea fomulas σ M τ VQ b Whee: σ is the nomal stess acting on the coss section, M is the bending moment, is the distance fom the neutal ais and is the moment of inetia of the coss sectional aea with espect to the neutal ais. τ is the shea stess at an point in the coss section, V is the shea foce, Q is the fist moment of the coss sectional aea outside of the point in the coss section whee the stess is being found, and b is the width of the coss section.

30 The nomal stesses obtained fom the fleue fomula have thei maimum values at the fathest distance fom the neutal ais. The nomal stesses ae calculated at the coss section of maimum bending moment. The shea stess obtained fom the shea fomula usuall have thei highest value at the neutal ais. The shea stesses ae calculated at the coss section of maimum shea foce. n most cicumstances, these ae the onl stesses that ae needed fo design puposes. Howeve to obtain a moe complete pictue of the stesses, we will need to detemine the pincipal stesses and maimum shea stesses at vaious points in the beam.

31 Beams of Rectangula Coss Section Conside a simple ectangula beam below, and a coss section to the left of the load. Points and E ae at the top and bottom of the beam. Point C is in the midheight of the beam and points B and D ae in between. f Hooke s law applies linea elasticit), the nomal and shea stesses at each of these five points can be eadil calculated fom the fleue and shea fomulas. ll the elements of vetical and hoiontal faces, ae in plane stess, because thee is no stesses acting pependicula to the plane of the figue.

32 Points and E elements ae in uniaial compessive and tensile stesses espectivel. Point C neutal ais) element is in pue shea. Points B and D elements have both nomal and shea stesses. Stesses in a beam of ectangula coss section: a) simple beam with points, B, C, D, and E on the side of the beam; b) nomal and shea stesses acting on stess elements at points, B, C, D, and E; c) pincipal stess; and d) maimum shea stesses.

33 We ma use eithe the tansfomation equations of plane stess o the Moh s cicle to find the stesses at each point along the height of the beam o to descibe how the pincipal stesses changes as we go fom to top to the bottom of the beam. B investigating the stesses at man coss sections of the beam, we can detemine how the pincipal stesses va thoughout the beam. Stess tajecto: Gives the diections of the pincipal stesses. Stess Contous: Cuves connecting points of equal pincipal stess Pincipal-stess tajectoies fo beams of ectangula coss section: a) cantileve beam, and b) simple beam. Solid lines epesent tensile pincipal stesses and dashed lines epesent compessive pincipal stesses.)

34 Eample 8-3: simple beam B with a span length L 6ft suppots a concentated load P 10800lb acting a distance c ft fom the ighthand suppot see figue below). The beam is made of steel and has a ectangula coss section width bin and height h 6in). nvestigate the pincipal stesses and maimum shea stesses at coss section mn located at a distance 9in fom the end of the beam. Conside onl the in-plane stesses) Point in) 3 B C 1 D 0 E -1 F - G -3

35 Solution The eaction of the beam at suppot is R P/3 3600lb, and theefoe the bending moment and shea foce at the section mn ae M R 3600lb)9in) 3400lb-in V R 3600lb Nomal stess on coss section mn bh ) 3400 lb in ) 3 in ) 6 in ) M M 1 σ X Whee has units in inches and σ has units in psi. The stesses calculated ae positive when in tension. Note that a positive value of uppe half of the beam) gives a negative stess, as epected. Shea stesses on coss section mn The shea stesses ae given b the shea fomula τ in which the fist moment Q fo a ectangula coss section is VQ b

36 ) ) the shea fomula becomes thus, 4 h bh V h b b bh V b VQ τ h b h h b Q 4 h b b Q h/ V Ma 3 τ

37 The shea stesses τ acting on the face of the stess element ae positive upwads, wheeas the actual shea stesses τ act downwad. Theefoe VQ 6V h τ 3 b bh 4 Substituting the numeical values into this equation gives τ XY lb ) in ) 6in ) n which has units of inches and τ has units of psi Note: The maimum shea stess that occus at the neutal ais in a ectangula section can be simplified b using the following equation: τ Ma 3V 6in ) 50 ) 4 3 9

38 Calculation of stesses on coss section mn We divide the height of the beam into si equal intevals and label the coesponding points fom to G. Point in) σ psi) τ psi) B C D E F G The nomal stesses va lineal fom a compessive stess of -700psi at the top of the beam point ) to a tensile stess of 700psi at the bottom of the beam point G). The shea stesses have a paabolic distibution with a maimum stess at the neutal ais point D).

39 Pincipal Stesses and Maimum Shea Stesses The pincipal stesses and maimum shea stesses at each of the seven points though G ma be detemined fom the following equations σ σ σ σ σ 1, ± σ σ τ MX τ ) τ )

40 Since thee is no nomal stess in the diection, this equation simplifies to: σ τ MX τ ) σ σ σ 1, ± τ ) Thus, b substituting the values of σ and τ, we can calculate the pincipal stesses σ 1 and σ and the maimum shea stess τ ma. Point in) σ psi) τ psi) σ 1 psi) σ psi) τ ma psi) B C D E F G

41 The lagest tensile stess anwhee in the beam is the nomal stess at the bottom of the beam at the coss section of maimum bending moment σ tens ) ma 14400psi). The lagest shea stess occus to the ight of the load P V R B 700lb). Theefoe, the lagest value that occus at the neutal ais is τ ) ma 900psi The lagest shea stess anwhee in the beam occus at 45 o planes at eithe the top o bottom τ ) ma / 700psi