STRENGTH OF MATERIALS 140AU0402 UNIT 3: BEAMS - LOADS AND STRESSES

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1 STRENGTH OF MATERALS 140AU040 UNT 3: BEAMS - LOADS AND STRESSES Tpes of beams: Suppots and Loads Shea foce and Bending Moment in beams Cantileve, Simpl suppoted and Ovehanging beams Stesses in beams Theo of simple bending Stess vaiation along the length and in the beam section Effect of shape of beam section on stess induced Shea stesses in beams Tpes of beams Thee ae 5 most impotant beams. The ae Simple suppoted beam Cantileve beam Ovehanging beam Fied beam Continuous beam Simple suppoted beam: A beam suppoted o esting feel on the suppots at its both ends, is known as simpl suppoted beam. Cantileve beam: A beam which is fied at one end and fee at the othe end is known as cantive beam. Ove hanging beam: f the end potion of a beam is etended beond the suppot such beam is known as Ovehanging beam Fied beam: A beam whose both ends ae fied o built in walls is known as fied beam.

2 Continuous beam: A beam which is povided moe than two suppots is known as continuous beam. Tpes of suppots Thee ae 5 most impotant suppots. The ae Simple suppots o knife edged suppots Rolle suppot Pin-joint o hinged suppot Smooth suface suppot Fied o built-in suppot Simple suppots o knife edged suppot: in this case suppot will be nomal to the suface of the beam. f AB is a beam with knife edges A and B, then R A and R B will be the eaction. Rolle suppot: hee beam AB is suppoted on the olles. The eaction will be nomal to the suface on which olles ae placed. Pin joint (o hinged) suppot: hee the beam AB is hinged at point A. the eaction at the hinged end ma be eithe vetical o inclined depending upon the tpe of loading. f load is vetical, then the eaction will also be vetical. But if the load is inclined, then the eaction at the hinged end will also be inclined.

3 Fied o built-in suppot: in this tpe of suppot the beam should be fied. The eaction will be inclined. Also the fied suppot will povide a couple. Tpes of loading Thee ae 3 most impotant tpe of loading: Concentated o point load Unifoml distibuted load Unifoml vaing load Concentated o point load: A concentated load is one which is consideed to act at a point. Unifoml distibuted load: A unifoml distibuted load is one which is spead ove a beam in such a manne that ate of loading is unifom along the length. Unifoml vaing load: A unifoml vaing load is one which is spead ove a beam in such a manne that ate of loading vaies fom point to point along the beam. CONCEPT AND SGNFCANCE OF SHEAR FORCE AND BENDNG MOMENT SGN CONVENTONS FOR SHEAR FORCE AND BENDNG MOMENT (i) Shea foce: Fig. 1 shows a simpl suppoted beam AB. caing a load of 1000 N at its middle point. The eactions at the suppots will be equal to 500 N. Hence R A = R B = 500 N. Now imagine the beam to be divided into two potions b the section X-X. The esultant of the load and eaction to the left of X-X is 500 N veticall upwads. And the esultant of the load and eaction to the ight of X-X is ( = 500 N) 500 N downwads. The esultant foce acting on an one of the pats nomal to the ais of the beam is called the shea foce at the section X-X is 500N. The shea foce at a section will be consideed positive when the esultant of the foces to the left to the section is upwads, o to the ight of the section is downwads. Similal the shea foce at a the section will be consideed negative if the esultant of the foces to the left of the section is downwad, o to the ight of the section is upwads. Hee the esultant foce to the left of the section is upwads and hence the shea foce will be positive.

4 ` Fig 1 Fig (ii) Bending moment. The bending moment at a section is consideed positive if the bending moment at that section is such that it tends to bend the beam to a cuvatue having concavit at the top as shown in Fig.. Similal the bending moment at a section is consideed negative if the bending moment at that section is such that it tends to bend the beam to a cuvatue haling conveit at the top. The positive B.M. is often called sagging moment and negative B.M. as hogging Moment. MPORTANT PONTS FOR DRAWNG SHEAR FORCE AND BENDNG MOMENT DAGRAMS The shea foce diagam is one which shows the vaiation of the shea foce along the length of the beam. And a bending moment diagam is one which show the vaiation of the bending moment along the length of beam. n these diagams, the shea foce o bending moment ae epesented b odinates wheeas the length of the beam epesents abscissa. The following ae the impotant points fo dawing shea foce and bending moment diagams 1. Conside the left o the ight potion of the section.. Add the foces (including eaction) nomal to the beam on one of the potion. f ight potion of the section is chosen, a foce on the ight potion acting downwads is positive while foce acting upwads is negative. f the left potion of the section is chosen, a foce on the left potion acting upwads is positive while foce acting downwads is negative.

5 3. The positive values of shea foce and bending moments ae plotted above the base line, and negative values below the base line. 4. The shea foce diagam will incease o decease suddenl i.e., b a vetical staight line at a section whee thee is a vetical point load. 5. The shea foce between an two vetical loads will be constant and hence the shea foce diagam between two vetical loads will be hoizontal. 6. The bending moment at the two suppots of a simpl suppoted beam and at the fee end of a cantileve will be zeo. SHEAR FORCE AND BENDNG MOMENT DAGRAMS FOR A CANTLEVER BEAM WTH A PONT LOAD A cantileve beam of length m caies the point loads as shown in fig. daw the shea foce and B.M diagams fo the cantileve beam. Shea foce diagam: The shea foce at D is +800N. this shea foce emains constant between D and C. At C, due to point load the foce becomes 1300N. between C and D, the shea foce emains 1300N. At B again, the shea foce becomes 1600N. the shea foce between B and A emains constant and equal to 1600N. hence the shea foce at diffeent points will be as follows: S.F. at D, FD= N S.F. at C. Fe.= = 1300N S.F. at B, Fa= =1600N S.F. at A, FA = N. The shea foce, diagam is shown in Fig. which is dawn as: Daw a hoizontal line AD as base line. On the base line mak the points B and C below the point loads. Take the odinate DE = 800 N in the upwad diection. Daw a line EF paallel to AD. The point F is veticall above C. Take vetical line FG is 500 N. Though G, daw a hoizontal line GH in which

6 point H is veticall above B. Daw vetical line H = 300 N. Fom, daw a hoizontal line J. The point J is veticall above A. This completes the shea foce diagam. Bending Moment Diagam The bending moment at D is zeo: (i) The bending moment at an section between C and Data distance: and D is given b, M = X which follows a staight line law. At C, the value of = 0.8 m. B.M. at C, = X 0.8 = Nm. (ii) The B.M. at an section between B and C at a distance fom D is given b (At C, = 0.8 and at B, = = 1.5 m. Hence hee vaies fom 0.8 to 1.5). M = (- 0.8) Bending moment between B and C also vaies b a staight line law. B.M. at B is obtained b substituting = 1.5 m in equation (i). M B = -800 X ( ) = = 1550 Nm.

7 (iii) The B.M. at an section between A and B at a distance fom D is given b (At B, = 1.5 and at A, =.0 m. Hence hee vaies fom 1.5m to.0 m M = ( - 0.8) 300 (- 1.5) Bending moment between A and B vaies b a staight line law. B.M. at A is obtained b substituting =.0 m in equation (ii), M A = X ( - 0.8) ( - 1.5) = X X X 0.5 = = Nm. Hence the bending moments at diffeent points will be as given below : MD = 0 Mc = Nm M B = Nm, M A= Nm SHEAR FORCE AND BENDNG MOMENT DAGRAMS FOR A CANTLEVER BEAM WTH A UNFORMLY DSTRBUTED LOAD A cantileve beam of length m caies a unifoml distibuted load of kn/m length ove the whole length and a point load of 3kN at the fee end. daw the shea foce and B.M diagams fo the cantileve beam. Shea Foce diagam

8 The shea foce at B = 3 kn Conside an section at a distance fom the fee end B. The shea foce at the section is given b. F = w. ( +ve sign is due to downwad foce on ight potion of the section) = X The above equation shows that shea foce follows a staight line law. At B, = 0 hence F B = 3.0 kn At A. = m hence FA = 3 + = 7 kn. The shea foce diagam is shown in Fig (b), in which F B = BC = 3 kn and FA = AD = 7 kn. The points C and D ae joined b a staight line. Bending Moment Diagam The bending moment at an section at a distance fom the fee end B is given b. M = - ( 3 + w. /) = - ( 3 + /) = - (3 + ) ( The bending moment will be negative as fo the ight potion of the section. the moment of loads at is clockwise) Equation (i) shows that the B. M. vaies accoding to the paabolic law. Fom equation (i) we have At B. = 0 hence M B = -( ) = 0 At A, = m hence M A = - ( 3 + ) = - 10 kn/m Now the bending moment diagam is dawn n this diagam. AA' = 10 knm and points A' and B ae joined b a paabolic cuve. SHEAR FORCE AND BENDNG MOMENT DAGRAMS FOR A CANTLEVER CARRYNG A GRADUALLY VARYNG LOAD A cantileve of length 4 m caies a gaduall vaing load, zeo at the fee end to Kn/m. at the fied end. Daw the S.F. and B.M. diagams fo the cantileve.

9 Shea Foce Diagam The shea foce is zeo at B. The shea foce at C will be equal to the aea of load diagam ABC. Shea foce at C = (4 ) / = 4 kn The shea foce between A and B vaies accoding to paabolic law. Bending Moment Diagam The B.M. at B is zeo. The bending moment at A is equal to M A = w. l / 6 = - 4 / 6 = knm. The B.M. between A and B vaies accoding to cubic law. SHEAR FORCE AND BENDNG MOMENT DAGRAMS FOR A SMPLY SUPPORTED BEAM WTH PONT LOAD A simpl suppoted beam of length 6 m, caies point load of 3 kn and 6 kn at distances of m and 4 m fom the left end. Daw the shea foce and bending moment diagams fo the beam.

10 Sol. Fist calculate the eactions RA and RB. Taking moments of the foce about A, we get RB X 6 = 3 X + 6 X 4 = 30 RB = 30/ 6 = 5 kn RA = Total load on beam - RB = (3 + 6) 5 = 4 kn Shea Foce Diagam Shea foce at A, FA= + RA= + 4 kn Shea foce between A and C is constant and equal to + 4 kn Shea foce at C, Fc = = + 1 kn Shea foce between C and D is constant and equal to + 1 kn. Shea foce at D, FD= = - 5 kn The shea foce between D and B is constant and equal to - 5 kn. Shea foce at B, FB= - 5 kn Bending Moment Diagam B.M. at A, MA = 0

11 B.M. at C, MC = RA X = 4 X = +8kNm B.M. at D, MD = RA X 4-3 = =+ 10 knm B.M. at B, MB= 0 SHEAR FORCE AND BENDNG MOMENT DAGRAMS FOR A SMPLY SUPPORTED BEAM WTH A UNFORMLY DSTRBUTED LOAD Daw the S.F. and B.M. diagams of a simpl suppoted beam of length 7 m caing unifoml distibuted load Sol. Fist calculate the eactions RA and RB, Taking moments of all foces about A, we get RB X 7 = 10 X 3 X (3/) + 5 X X ( 3++(/) = = 105 RB = 105 /7 = 15 kn and RA =Total load on beam - RB = ( 10X 3 +5 X ) - 15 = = 5kN S.F. Diagam The shea foce at A is + 5 kn The shea foce at C = RA - 3 X 10 = = - 5 kn The shea foce vaies between A and C b a staight line law. The shea foce between C and D is constant and equal to - 5 kn. The shea foce at B is - 15 kn The shea foce between D and B vaies b a staight line law. The shea foce is zeo at point E between A and C. Let us find the location of E fom A. Let the point E be at a distance fom A. The shea foce at E = RA - 10 = 5-10 But shea foce at E = = 0 o 10 = 5

12 =.5m B.M. Diagam B.M. at A is zeo B.M. at B is zeo B.M. at C, M C = R A X 3-10 X 3 3/ = 5 X 3 45 = = 30 knm At E, =.5 and hence B.M. at E, M E = R A X.5-10 X.5 X (.5 /) = 5 X.5-5 X 6.5 = = 31.5 knm B.M. at D. M D = 5(3 + ) - 10 X 3 X ((3/) + ) = = 0 knm The B.M. between AC and between BD vaies accoding to paabolic law. But B.M. between C and D vaies accoding to staight line law. SHEAR FORCE AND BENDNG MOMENT DAGRAMS FOR OVER HANGNG BEAM

13 A beam of length 1 m is simpl suppoted at two suppots which ae 8m apat, with an ovehang of m on each side as shown in Fig. The beam caies a concentated load of 1000 N at each end. Daw S.F. and B.M. diagams. As the loading on the beam is smmetical. Hence eactions R A and R B will be equal and thei magnitude will be half of the total load. R A = R B = ( )/ = 1000N S.F. at C = N S.F. emains constant (i.e., = N) between C and A S.F. at A = 1000+R A = = 0 S.F. emains constant (i.e., = 0) between A and B S.F. at B = =+ 1000N S.F. emains constant (i.e., 1000 N) between B and D B.M. Diagam B.M. at C = 0 B.M. at A = = Nm

14 B.M. between C and A vaies accoding to staight line law. The B.M. at an section in AB at a distance of fom C is given b, M X = X + R A ( - ) = X ( - ) = Nm Hence B.M. between A and B is constant and equal to Nm. B.M. at D = 0 STRESSES N BEAMS When some etenal load acts on a beam, the shea foce and bending moments ae set up at all sections of the beam. Due to the shea foce and bending moment, the beam undegoes cetain defomation. The mateial of the beam will offe esistance o stesses against these defomations. These stesses with cetain assumptions can be calculated. The stesses intoduced b bending moment ae known as bending stesses. f a length of a beam is subjected to a constant bending moment and no shea foce (i.e., zeo shea foce), then the stesses will be set up in that length of the beam due to B.M. onl and that length of the beam is said to be in pue bending o simple bending. The stesses set up in that length of beam ae known as bending stesses. A beam simpl suppoted at A and B and ovehanging b same length at each suppot is shown in Fig A point load W is applied at each end of the ovehanging potion. The S.F. and B.M. fo the beam ae dawn as shown in Fig. 7.1 (b) and Fig. 7.1 (c) espectivel. Fom these diagams, it is clea that thee is no shea foce between A and B but the B.M. between A and B is constant. This means that between A and B, the beam is subjected to a constant bending moment onl. This condition of the beam between A and B is known as pue bending o simple bending. THEORY OF SMPLE BENDNG

15 THEORY OF SMPLE BENDNG WTH ASSUMPTONS MADE Befoe discussing the theo of simple bending, let us see the assumptions made in the theo of simple bending. The following ae the impotant assumptions: 1. The mateial of the beam is homogeneous* and isotopic**.. The value of Young's modulus of elasticit is the same in tension and compession. 3. The tansvese sections which wee plane befoe bending, emain plane afte bending 4. The beam is initiall staight and all longitudinal filaments bend into cicula acs with a common cente of cuvatue. 5. The adius of cuvatue is lage compaed with the dimensions of the coss-section. 6. Each lae of the beam is fee to epand o contact, independentl of the lae, above o below it. A beam subjected to simple bending. Conside a small length fit of this pat of beam. Conside two sections AB and CD which ae nomal to the ais of the beam N - N. Due to the action of the bending moment, the pat of length & will be defomed as shown in Fig.(b). Fom this figue, it is clea that all the laes of the beam, which wee oiginall of the same length, do not emain of the same length an moe. The top lae such as AC has defomed to the shape NC. This lae has been shotened in its length. The bottom lae BD has defomed to the shape B'D'. This lae has been elongated. Fom the Fig. 7. (b), it is clea that some of the laes have been shotened while some of them ae elongated. At a level between the top and bottom of the beam, thee will be a lae which is neithe shotened no elongated. This lae is known as neutal lae o neutal suface. This lae in Fig.(b) is shown b N' N' and in Fig.(a) b N N. The line of intesection of the neutal lae on a coss-section of a beam is known as neutal ais (witten as N.A.). The laes above N N (o N' N') have been shotened and those below, have been elongated. Due to the decease in lengths of the laes above N N, these laes will be subjected to compessive stesses. Due to the incease in the lengths of laes below N N, these laes will be subjected to tensile stesses. We also see that the top lae has been shotened maimum. As we poceed towads the lae N N, the decease in length of the laes deceases. At the lae N N, thee is no change in length. This means the compessive stess will be maimum at the top lae. Similal the incease in length will be maimum at the bottom lae. As we poceed fom bottom lae towads the lae N N, the incease in length of laes deceases. Hence the amount b which a lae inceases o deceases in length, depends upon the position of the lae with espect to N - N. This theo of bending is known as theo of simple bending.

16 Simple Bending Theo OR Theo of Fleue fo nitiall Staight Beams (The nomal stess due to bending ae called fleue stesses) Peamble: When a beam having an abita coss section is subjected to a tansvese loads the beam will bend. n addition to bending the othe effects such as twisting and buckling ma occu, and to investigate a poblem that includes all the combined effects of bending, twisting and buckling could become a complicated one. Thus we ae inteested to investigate the bending effects alone, in ode to do so; we have to put cetain constaints on the geomet of the beam and the manne of loading. Assumptions: The constaints put on the geomet would fom the assumptions: 1. Beam is initiall staight, and has a constant coss-section.. Beam is made of homogeneous mateial and the beam has a longitudinal plane of smmet. 3. Resultant of the applied loads lies in the plane of smmet. 4. The geomet of the oveall membe is such that bending not buckling is the pima cause of failue. 5. Elastic limit is nowhee eceeded and E' is same in tension and compession. 6. Plane coss - sections emains plane befoe and afte bending. Let us conside a beam initiall unstessed as shown in fig 1(a). Now the beam is subjected to a constant bending moment (i.e. Zeo Sheaing Foce') along its length as would be obtained b appling equal couples at each end. The beam will bend to the adius R as shown in Fig 1(b) As a esult of this bending, the top fibes of the beam will be subjected to tension and the bottom to compession it is easonable to suppose, theefoe, that somewhee between the two thee ae points at which the stess is zeo. The locus of all such points is known as neutal ais. The adius of cuvatue R is then measued to this ais. Fo smmetical sections the N. A. is the ais of smmet but whateve the section N. A. will alwas pass though the cente of the aea o centoid.

17 The above estictions have been taken so as to eliminate the possibilit of 'twisting' of the beam. Concept of pue bending: Loading estictions: As we ae awae of the fact intenal eactions developed on an coss-section of a beam ma consists of a esultant nomal foce, a esultant shea foce and a esultant couple. n ode to ensue that the bending effects alone ae investigated, we shall put a constaint on the loading such that the esultant nomal and the esultant shea foces ae zeo on an coss-section pependicula to the longitudinal ais of the membe, That means F = 0 Since o M = constant. Thus, the zeo shea foce means that the bending moment is constant o the bending is same at eve coss-section of the beam. Such a situation ma be visualized o envisaged when the beam When a membe is loaded in such a fashion it is said to be in pue bending. The eamples of pue bending have been indicated in EX 1and EX as shown below:

18 When a beam is subjected to pue bending ae loaded b the couples at the ends, cetain coss-section gets defomed and we shall have to make out the conclusion that, 1. Plane sections oiginall pependicula to longitudinal ais of the beam emain plane and pependicula to the longitudinal ais even afte bending, i.e. the coss-section A'E', B'F' ( efe Fig 1(a) ) do not get waped o cuved.. n the defomed section, the planes of this coss-section have a common intesection i.e. an time oiginall paallel to the longitudinal ais of the beam becomes an ac of cicle. We know that when a beam is unde bending the fibes at the top will be lengthened while at the bottom will be shotened povided the bending moment M acts at the ends. n between these thee ae some fibes which emain unchanged in length that is the ae not stained, that is the do not ca an stess. The plane containing such fibes is called neutal

19 suface.the line of intesection between the neutal suface and the tansvese eploato section is called the neutal ais Neutal ais (N A). Bending Stesses in Beams o Deivation of Elastic Fleual fomula : n ode to compute the value of bending stesses developed in a loaded beam, let us conside the two coss-sections of a beam HE and GF, oiginall paallel as shown in fig 1(a).when the beam is to bend it is assumed that these sections emain paallel i.e. H'E' and G'F', the final position of the sections, ae still staight lines, the then subtend some angle q. Conside now fibe AB in the mateial, at a distance fom the N.A, when the beam bends this will stetch to A'B' Since CD and C'D' ae on the neutal ais and it is assumed that the Stess on the neutal ais zeo. Theefoe, thee won't be an stain on the neutal ais Conside an abita a coss-section of beam, as shown above now the stain on a fibe at a distance ' fom the N.A, is given b the epession

20 Now the tem is the popet of the mateial and is called as a second moment of aea of the coss-section and is denoted b a smbol. Theefoe This equation is known as the Bending Theo Equation. The above poof has involved the assumption of pue bending without an shea foce being pesent. Theefoe this temed as the pue bending equation. This equation gives distibution of stesses which ae nomal to coss-section i.e. in -diection. Section Modulus: Fom simple bending theo equation, the maimum stess obtained in an coss-section is given as Fo an given allowable stess the maimum moment which can be accepted b a paticula shape of coss-section is theefoe Fo ead compaison of the stength of vaious beam coss-section this elationship is sometimes witten in the fom s temed as section modulus STRESSES N BEAMS n pevious chapte concen was with shea foces and bending moment in beams. Focus in this chapte is on the stesses and stains associated with those shea foces and bending moments.

21 Loads on a beam will cause it to bend o fle. P A B plane = plane of bending ν z A P B Deflection Cuve Coss Section assumed smmetic about - plane PURE BENDNG AND NONUNFORM BENDNG Pue Bending = fleue of a beam unde constant bending moment shea foce = 0 ( V = 0 = dm / d ); no change in moment. Nonunifom Bending = fleue of a beam in the pesence of shea foces bending moment is no longe constant Moment Diagam eample: Pa M a a CURVATURE OF A BEAM A O O = cente of cuvatue d ds PURE BENDNG d P B NONUNFORM BENDNG A beam in NONUNFORM BENDNG (V 0) will have a vaing cuvatue. ds = cuve length A beam in PURE BENDNG ( V = 0) will have will have constant cuvatue = cicle ds = ac length of cicula segment

22 = adius of cuvatue = cuvatue = -1 1 = d = ds Fo small deflections: ds d d 1 (1) d Sign Convention: O NORMAL STRANS a + cuvatue - cuvatue b opposite fom tet because we use + O z M o c d M o a M o c d b M o Lengthened in tension Somewhee between the top and bottom of the beam is a place whee the fibes shotened ae neithe in compession in tension o compession. neutal ais of the coss section dashed line = neutal suface of the beam when bent: a b lengthens c d shotens causes nomal stains,

23 The nomal stain is: ( ) Whee, = distance fom neutal ais Fom Eqn ( ): + - = + ε (elongation) + = - ε (shotening) fo + Tansvese Stains: NORMAL STRESSES N BEAMS z Whee = Poisson s Ratio f mateial is elastic with linea stess-stain diagam, THEN: = E (Hooke s Law) = E = - E ( 3 ) vaies lineal with Whee is longitudinal ais of beam and is the nomal stesses in this diection acting on the coss section. These stesses vaies lineal with the distance fom the neutal suface. REF: + CURVATURE = + STRESSES above neutal suface Mo c c 1 z da da E da 0 must equal ZERO because thee is NO esultant nomal foce that acts on the ENTRE coss section da 0 ( 4 ) Eqn ( 4 ) is the 1 st Moment of the Aea of the coss section w..t. z-ais and it is zeo z-ais must pass thu the centoid of the coss section. z-ais is also the neutal ais neutal ais passes thu the centoid of the coss section Limited to beams whee -ais is the ais of smmet.

24 , z aes ae the PRNCPAL CENTRODAL AXES. Conside the Moment Resultant of : RECALL Eqn ( 3 ): E dm M O O da da E M E da M = -M 0 da whee, = Moment of netia of coss sectional aea w..t. z-ais ( neutal ais ) 1 M E E = FLEXURAL RGDTY E M substitute into Eqn ( 3) M M Fleue Fomula = Bending Stess c c 1 + M 1 MAXMUM STRESSES: Mc 1 1 Mc Tet defines Section Moduli as: S 1 c 1 S c

25 1 M S 1 M S Section Modulus is hand to use when evaluating bending stess w..t. to moment which vaies along length of a beam. f coss section is smmetical w..t. z-ais, then: c 1 = c = c 1 Mc Moments of netia to know: b O h h d O z 3 bh 1 Poblems fo Pactice d 64 4 A high-stength steel wie of diamete d = 4 mm, modulus of elasticit E = 00 GPa, popotional limit pl = 100 MPa is bent aound a clindical dum of adius R 0 = 0.5 m. FND: a. bending moment, M b. maimum bending stess, ma R 0 C d Poblems fo Pactice The beam shown which is constucted of glued laminated wood. The unifom load includes the weight of the beam. 9 ft P = 1 k q = 1.5 k / ft A B h = 7 in. L = ft FND: a. Maimum Tensile Stess in the beam due to bending. b. Maimum compessive stess in the beam due to bending. b = 8.75 in.

26 DESGN of BEAMS fo BENDNG STRESSES Afte all factos have been consideed (i.e., mateials, envionmental conditions) it usuall boils down to Allow > Beam Allow Hee is whee the section modulus is useful. M ma c RECALL: M S thus, S M ma allow Appendi E and F give popeties of beams. i.e., W8 15 Weight pe foot ( lbs / ft ) Nominal Depth ( in ) DEPTH W Shape = wide flange WDE FLANGE Wood Beams - 4 eall is: net dimensions (should alwas use net dims.) h A A h A h h A Ah 4 Ah 1 S 4 Ah c h Fo W Shapes; S 0.35 Ah You want as much mateial as possible, as fa fom the neutal ais as possible because this is whee the geatest stess is occuing. Howeve, ou have to be caeful because if the web is too thin, it could fail b: 1.) being ovestessed in shea.) buckling SHEAR STRESSES N BEAMS (RECTANGULAR CROSS-SECTONS) Assumptions: 1. acts paallel to V (shea foce, also, ais)

27 Conside a sstem of beams. is unifom acoss coss-section P h REF: τ P h τ τ Assuming thee is no ( o ve little ) fiction, the top beam can slide w..t. to the bottom beam. Thus, thee must be shea stesses pesent that pevent sliding. Looking at the dotted ectangle: b h h M 1 M + dm Nomal Foce: ( left side ) Total Foce: ( left side ) da M da M F 1 da ( 1 ) similal, M dm Total Foce: F da ( ) ( ight side ) Shea Foce: F 3 = b d ( 3 ) 0 F F F 1 F 1 m n z TOTAL FORCE F 1 F1 F3 0 d F3 F F1 Sub Eqns (1), (), and (3) into (4): ( 4 ) da CROSS SECTON OF BEAM AT SUBELEMENT = distance at which shea stess acts. REF: d m b n z

28 bd bd V b dm d ( M dm ) da dm da 1 da b da M da = distance at which the shea foce acts b = thickness of the coss-sectional aea whee the stess is to be evaluated. da = fist moment of that potion of the coss-sectional aea between the tansvese line (m n) whee the stess is to be evaluated AND the eteme fibe of the beam. LET: Q da RECALL: dm V d VQ SHEAR FORMULA b Fo RECTANGULAR coss-sections, ( see tet pg 339 fo deivation ) V h 1 4 Ma Shea occus at the neutal ais, 1 = 0 τ τ ma V h 3V ma (5) 8 A WHERE: A = bh = TOTAL CROSS-SECTONAL AREA and = NOTE: bh 3 Shea Eqn ( 5 ) is limited to Coss-Sectional shapes that have sides paallel 1 to the -ais. SHEAR STRESSES N BEAMS ( CRCULAR CROSS SECTONS ) Lagest Shea Stesses occu at neutal ais. z

29 We can assume with good accuac that: 1. τ acts paallel to V ( shea foce, also, ais ). τ is unifom acoss coss-sections These assumptions ae the same used when we developed the shea fomula: b VQ Theefoe, we can use this to find τ at the N.A. (Neutal Ais) which is τ ma Q and A A Q b WHERE: TOTAL CROSS-SECTONAL AREA A = π HOLLOW CRCULAR CROSS SECTONS ) ( 3 ) ( ) ( Q b NOTE: f 1 = 0 we get ou pevious equation fo solid cicula coss-section. SHEAR STRESSES N THE WEB OF BEAMS WTH FLANGES The shea fomula b VQ still applies because the same assumptions ae made ma 3 4 A V 4 3 ma V V b VQ A V 3 4 ma 1 z WHERE: 1 A = TOTAL CROSS-SECTONAL AREA h b h 1 t 1 z τ h h τ min τ min h o at, neutal ais, whee, 1 = 0

30 We will use: V Q t whee: t = web thickness b t Q ( h h1 ) ( h1 41 ) bh bh1 th1 1 whee: b = width of flange h = height of beam h 1 = web height (inside flanges) 1 = distance fom N.A. Fo Wide-Flange Beams, we use the AVERAGE Shea Stess in the web: ave V th 1 whee: t = web thickness h 1 = web height (inside flanges) Poblems fo Pactice 1. A beam is to be made of steel that has an allowable bending stess of allow = 4 ksi and an allowable shea stess of τ allow = 14.5 ksi. Select an appopiate W shape that will ca the loading shown A 40 kip 0 kip B 6 6 6

31 . The laminated beam shown suppots a unifom load of 1 kn/m. f the beam is to have a height to width atio of 1.5, detemine the smallest width. allow = 9 MPa, τ allow = 0.6 MPa. Neglect the weight of the beam. 1 kn / m A B 1.5b 1 m 3 m b TEXT/ REFERENCE BOOKS 1. Popov E.P, Engineeing Mechanics of Solids, Pentice-Hall of ndia, New Delhi, Ramamutham.R., Stength of Mateials,16 th Edition, Dhanpat ai Publishing compan, Bansal.R.K., Stength of Mateials,4 th Edition, Lami Publications, Rajput.R.K. Stength of Mateials,4 th Edition, S.Chand & compan,new Delhi Rde G.H, Stength of Mateials, Macmillan ndia Ltd., Thid Edition, Nash W.A, Theo and poblems in Stength of Mateials, Schaum Outline Seies, McGaw-Hill Book Co, New Yok, 1995