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# STRUC STEEL, TIMB, CONCR ARCH 3126

OK State

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This 0 page Class Notes was uploaded by Katheryn Stamm on Sunday November 1, 2015. The Class Notes belongs to ARCH 3126 at Oklahoma State University taught by Staff in Fall. Since its upload, it has received 30 views. For similar materials see /class/232813/arch-3126-oklahoma-state-university in Architecture at Oklahoma State University.

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Date Created: 11/01/15

ARCH 3126 Statics Strength Review El Chapter O H1 Equilibrium of Beams 0 O H11 Introduction In Statics we have looked at equilibrium of beams This is the most important topic covered We will repeat this subject along with the appropriate topics from Strength of Materials Let us classify some of the different types of beams as follows show these The first three are statically determinate while the last two are statically indeterminate Simply supported beam Cantilever beam Overhanging beam Propped end cantilever beam Continuous beam 959 Loading types that beams must resist are as follows Concentrated force Concentrated moment Uniformly distributed load Linearly varying distributed load General distributed load 959 The positive sign convention for shear is down on a right cut section or up on a left cut section Show a positive shear deformation and how a ball would roll in the plus x direction The positive moment sign convention is thumb out on right cut section or thumb in on left cut section Show a positive bending deformation and how smiling is the positive sense 0 O H12 Equilibrium of Beams Using Finite Freebody Diagrams We can solve for internal forces in three ways as follows ARCH 3126 Statics Strength Review Simply cut a freebody diagram at a specific location to find the internal forces at that section not really that useful Cutting a freebody diagram within a general region of the members to derive the equation for the internal forces within that region If equations are derived for all section a diagram can be drawn that represents the internal forces use for finding deflection equations or if loading is more than 1St degree lntuitively applying the relations between load force and moment to graphically describe the forces and moments The basis is the differential calculus relationships best if only shear and moment values are needed The normal steps of procedure for internal forces are as follows These apply to both values at a specific point and within a region of a beam 1 2 3 4 Draw freebody diagram of the entire structures showing all known and unknown forces Apply the equations of equilibrium to the total freebody diagram to find the reactive forces Draw the freebody diagram at the section of the member was the internal forces are to be found Apply the equations of equilibrium to the sectioned freebody diagrams to find the internal forces The general location of the cut is normally given the coordinate of x in general measured from the end of the member Normally we use a right cut section called a lefthand freebody diagram This means that x is measured from the left end to the cut section At the cut the unknown shear force V down and bending moment M by RHR are drawn sometimes normal force N ARCH 3126 Statics Strength Review Problem O H11 For the simply supported beam AD in Figure O H11 determine the following a V1x and M1x the internal resultants at an arbitrary section between A and B that is 0ltxlt6in and b V2x and M2x the internal resultants at an arbitrary section between B and C that is 6inltxlt8in Just for fun let find c V3x and M3x the internal resultants at an arbitrary section between C and D that is 8inltxlt10in The total beam freebody is drawn first and the reactions solved for Then freebody diagrams are cut at the three unique regions of the beam Last the shear and moment equations are derived from equilibrium of the cut freebody diagrams Total FBD EMA O 0 2 201b in6in6in 501b8m Dy 10m 2 Fy 0 Ay 201M in6in SOlb Dy Ay 460011 46011 FBD OltXlt6in ZFy o Ay x201bmj Vlx ARCH 3126 Statics Strength Review V1x 46 gx2jlb 2M0 O Ayxx20binj x M1x M1x 46x Sx lbin FBD 6inltxlt8in Z Fy 0 Ay 6in201b m V2x V2 x 14lb 2 MO O Ayx 6in201binx 6inj M2 x M2x 240 14xlbm FBD 8inltxlt10in Z Fy 0 Ay 6in201b m SOlb V3 x V3x 64lb 2 M0 0 Ayx 6in201b inx 6171 501bx 8m M3x ARCH 3126 Statics Strength Review M3 x 640 64xlb m ARCH 3126 Statics Strength Review Figure O H1 1 501b 20 lblin E I I AL 1 BL MCL 2an A A A A J 6m 2m L 4 L 4 3 sx2 3 201bin Mlx Ay quot v x 60 1b 50 1b ARCH 3126 Statics Strength Review 0 O H13 Equilibrium Relationships Among Load Shear Force and Bending Moment The general relationship between distributed load w shear force V and bending moment M was derived in Statics could draw FBD of beam segment Ax if time allows to derive The sign convention used in this book for load is up therefore down is negative The primarily concepts and relationships are given below dV E w x Shear Slope Load Value x2 AVIZ V2 V1 Iwxdx Shear Change Load Area x1 dM V dx Moment SlopeShearValue x2 AMIZ M2 M1 Ide Moment Change Shear Area x1 Talk about effect of concentrated force on shear diagram changes shear in the force direction and concentrated moment on moment diagram internal moment is reactive to applied moment and is therefore opposite in sign 0 O H14 Shear Force and Bending Moment Diagrams The normal steps of procedure for drawing diagrams are as follows 1 Draw freebody diagram of the entire structures showing all known and unknown forces 2 Apply the equations of equilibrium to the total freebody diagram to find the reactive forces ARCH 3126 Statics Strength Review 3 Draw the shear force diagram using the loaded beam and loadshear relations 4 Draw the bending moment diagram using the loaded beam shear force diagram and the shearmoment relations Problem O H12 Sketch the shear diagram and moment diagrams for compound beam AE internal pinned at C in Figure O H12 The total beam freebody is drawn first but the reactions cannot all be solved for Therefore separate freebody diagrams are drawn for section AC and CD since the moment is zero at C it is a pin Next the shear diagram is drawn using the loadshear relations Last the moment diagram is drawn using the shearmoment relations FBD section AC EMA 0 6kz p4flCy8fl Cy 3000kip 300kz39p ZFy O Ay 6kz p Cy Ay 3000kz39p 300kz39p FBD section CE ZME 0 Cleft2kft10fz5ftME ME 1300kz39p 130kl39pft ZFy 0 Cy 2k1 pfl10flEy ARCH 3126 Statics Strength Review Ey 2300kz39p 230kzp Shear Diagram VA 3kl39p AVAB 0 VC 2 3 6kl39p 3kip AVBC 0 AVCE 2kip fz10fz 20kl39p VE 3 20kip 23kz39p Moment Diagram MA Okl39pfl AMAB 3kl39p4ft 12kz39p ft MB 2 0 12kl39pfl 12kl39pfl AMBC 3kl39p4fl 12kz39p ft MC 12 12k139pft Okipftm AMCE 3 223fz 10ft 130kz39pft starting shearreaction shear changeload area concentrated shear changearea changearea moment0 for pinned end moment IIshear area change area NO KIDDING change area ARCH 3126 Statics Strength Review ME 0 130k139pfl 130kipft slopevalue Figure O H12 iifms 2k y Flil A B c E 4a 4a 10 4 4 4 jfs 2k y I u A 4 4a 10a L 4 4 6 kips 4 AV 4 4n 7 CY Icy 2kmy 39 ME FE 3 v a u MHp 10 ARCH 3126 Statics Strength Review El Chapter O H2 Area Properties 0 O H21 Centroids We will only cover only centroids of composite bodies but for strange shapes use I instead of Z The equations for Volumes Areas Lines and Points are listed below XZZVJ C yZZIj 22212 ZV 2K ZV 221476 2242 2242 2A 2A1 2A 6 242 y 242 2 gm XL XL 2L 2 2x 2 2 222 71 n 71 Problem O H21 Determine the location of the centroid x and y for the area shown in Figure O H21 Element Aiin2 xi in yi in Aixiin3 Aiyiin3 bh2 1 342 1 3333 6 2o 6 bh2 2 342 11 3333 66 2o 6 bh 3 122 6 1 144 24 24 2 36 216 64 ARCH 3126 Statics Strength Review Figure O H21 A w 7 J all 9 H L3 1 0 O H22 Moment of Inertia but moment of inertia of composite bodies remember the following We will only cover J Z Irsz r2x2y2 A 1y Jx2dA A 511 ARCH 3126 Statics Strength Review The Parallelaxis equation is the basis for composite areas but remember you must go to or from the centroid of each piece LzZpVXJ Adyz 1yZ139yAdf JZJ Ad2 Another useful structural property is the radius of gyration as follows Problem O H22 Determine the centroidal moment of inertia lXX for the shape shown in Figure O H22 Must first find the centroid y Element Aimm2 yi mm Aiyi lx iin4 dyiyyi Aidy12in4 mm3 Top bh 152 bn312 154 152 Flange 15015 75 16875 15050 1455 48290000 2250 12 42190 bh bh312 15490 Web 15150 90 202500 50150 54 9215000 2250 12 4219000 Bottom 1ch nr24 154 215 Flange 35012 215 1588500 Ewan4 51 29220000 78540 4909000 2 12354 1907975 9159000 85725000 y 2A7 1907875mm3 15444mm y 154mm 2A7 12354mm2 1x Z1Adj 9169000 86726000mm 95900000mm4 1x 95900000mm ARCH 3126 Statics Strength Review Figure O H22 15mm Kl 150mm El Chapter O H3 Stresses in Beams 0 O H31 Introduction In this chapter we will determine how bending moment and shear force produce normal stress and shear stress respectively The derivations included all three concepts of deformablebody mechanics geometry of deformation material behavior and equilibrium Some new definitions are in order A beam with a longitudinal plane of symmetry LPS will be used to produce what is known as symmetric bending This is not exactly symmetric bending but rather is about one axis or plane of bending show these ARCH 3126 Statics Strength Review If we consider as loaded beam we see that some of the fibers of the beam are being compressed and some are being stretched In fact some fibers are not being deformed at all These fibers are located at the neutral surface NS The intersection of the neutral surface and the longitudinal plane of symmetry is a line known as the deflection curve 0 O H32 StrainDisplacement Analysis As we derive the equations of bending there are many considerations but one assumption is critical to use This assumption is that a cross section that is plane before bending remains plan after bending some this Let us consider a bending beam of length Ax The bending curve at the NS has a radius that is the center of the curvature This center point is also the intersection of the plane sections at each end of the length Ax The angle of rotation between these plane is A9 See Figure ExO H31 Figure ExO H31 ARCH 3126 Statics Strength Review Now let us look at a line AB ABAB after bending of ld ggth along the NS and another line PQ PQgtPQ after bending of length Ax at a distance y form AB y is towards the center The length of the line is a function of the distance from the center and the angle From this we can find the normal strain as follows ABABPQAxpA6 PQAxp yA6 8 LF LO PQ PQ Ax Ax p yA6 pA6 Z x LO PQ Ax pA6 0 In general form this can be written as follows independent of 2 Skip This SAWpix The reciprocal of the radius of curvature p is the longitudinal curvature orjust curvature K That is K1p If we consider that something that strains in one direction x must have an opposite strain in the other directions y and 2 then that following comes from Poisson s ratio 8y 2 52 V8x Since the crosssection has both tension and compression strains in the xdirection then they must have equal and opposite 2 strains and the beam section also curves but in the opposite direction anticlastic curvature or saddleshaped These curvature relationships are as follows bend an eraser ARCH 3126 Statics Strength Review 0 O H33 Flexural Stress in Linearly Elastic Beams Now that we have seen how a beam deforms let us look at what stress occur Since the normal strains are a linear function of the distance y to the NS then the normal stresses must also vary linearly See Figure ExO H32 By Hooke s law the following applies Figure ExO H32 lt4 R X J lx a E5 X X p px Remembering that from statics the forcestressarea and M ystressarea positive moment produces compression on y side the following can be derived Fx 2 0di 5L ydA A p Mx yadi j Aysz ARCH 3126 Statics Strength Review Since only moment exist at the section shear too and not normal force the first integral equation is zero What is zero E NO p NO the integral Yes If the integral of ydA is zero then the NS must pass through the centroid got that The second integral equation is y2dA or the moment of inertia about the NS lz This equation can be rewritten as follows MZEZEIK 0 in x p I o O H34 Design of Beams for Strength Let us finally consider that for a given crosssection that the normal stress due to moment only varies as a function ofy the distance to the NS The maximum normal stress will then occur at the maximum y or at the extreme fiber point yc1ctop and yc2cbot For a given section these are constant as is the moment of inertia The value lc is known as the section modulus S the stress equations are then as follows O O O O39 2 top 1 bot 2 1 S1 1 52 These values are given for Timber shapes in Tables 18 through 1D of the NDS Supplement The smallest section modulus is normally given since this produces the higher stress In design we need to select a section and the equation can be written as follows max design req39d 039 allow ARCH 3126 Statics Strength Review Problem O H31 A simply supported timber beam supports a triangularly distributed load as shown in Figure O H31 The magnitude of the allowable stress in tension or compression is omax 800 psi From Table 18 select the lightest timber beam that may be used for this application if the nominal depth of the beam may not exceed twice its nominal width The total beam freebody is drawn first and the reactions found Total FBD lb 1 in 2 M 202 40 12 121 121B121 Z A 2 in ft f 3 f y f By 2 1920lb 192k1 p lb in 12 12 ZB in ft f y 1 ZFy 0 Ay E40 Ay 960lb 096kip Since the loading is linearly varying the shear will be a second degree curve and the moment may be difficult to find area of shear We will use the fact that dMdxV and find the point of zero shear or maximum moment FBD section between AB 1 lb in x F OA 40 12 V 2 y y 2x x Vx 960 20x2lb ARCH 3126 Statics Strength Review EMA 0 A xlx4o12 ij y 2 in ft L Mx Mx 960x 2 30x3 lb ft Zero Shear Maximum Moment Vx 0 960 20x2lb x 6928 ft 693 ft M6928 Mmax 9606928 69283jlb ft Mm 4434lb ft 443k1 p ft Required Section Modulus 4434lb A121 fZ O Mmax max allow re 39d Sreq39d q O allow 1712 S 66511713 2 6651713 req39d For bh2 approximate depth ARCH 3126 Statics Strength Review h ihZ Sbh2 2 h3 6 2 66511713 Selection of choices b x h A in2 8 in3 comments 4 x 12 3938 7383 b lt h2 I6 x 10 5225 827 MES 8x8 5625 703 AgtAfor6x10 ARCH 3126 Statics Strength Review Figure O H31 401bin A B L M L 4 4 y 8 ft b q 28801b J L A 240x 2 L X 12 480 lbft Ir 3 M00 A 39 Vx Y L X L 4 4 U ARCH 3126 Statics Strength Review 0 O H35 Shear Stress and Shear Flow in Beams Now that we have considered how moment affects a beam let us consider how shear force affects a beam First we must separate the two by saying that the distribution of normal stress ids not affected by the deformation due to shear If a beam was made of a series of beams stacked one on top of another than we see if bending occurs that they will slip show this In fact they do not slip therefore a longitudinal force horizontal shear must exist to hold this slip The plane section Whoops must therefore distort with the top fiber stretching and the bottom shortening This is not due to compression and tension since they are in the opposite direction show the distortion If we cut a differential element of a beam of length Ax at a point x along the length then the shear Vx and moment Mx at x can be found from statics left side of element On the right side of the element the shear VxAx and moment VxAx is different See Figure ExO H33 This is how we previously derived the following V dx ARCH 3126 Statics Strength Review Figure EXO H33 Y MltXAgtltgt MOO gtltAgtlt LA F1 F2 AH Now let us draw the normal stress on each face of the segment more stress on right side and consider a horizontal plane at a distance y from the NS The forces F1 and the left and F2 on the right are not in equilibrium and thus a shear H must exist on the cutplane as follows AHzFZ Fl The forces can be found from the equilibrium stress equations as follows over the area from y to the top A and the centroid of the area is denoted y E 0di IMSdoA ngny39dA A39 A39 A39 ARCH 3126 Statics Strength Review F 0di jMltx1Axy39dA Mltx1 Ax yd1 MAx 1 AHzFl FZ yd1 A The force H can be converted to a force per length by dividing by the length Ax This is called the shear flow as follows AHMAx 39 AM1 V E ijdA EYJydA dm The integral term is given a term Q and is the first moment of the area from the shear plane to the extreme fiber taken about the neutral axis neutral surface This equation is written below and the stress can be simple found by dividing by the width of the beam t I It Another way to write this is as follows I like this qZVAy TZVAy I It Problem O H32 A beam with a rectangular cross section supports two equalmagnitude loads P 16 kN as shown in Figure O H32 The beam is 200 mm wide by 300 mm high a Determine the maximum flexural stress omax in the beam and b determine the maximum shear stress Tmax in the beam Neglect the weight of the beam The total beam freebody is drawn first and the reactions found ARCH 3126 Statics Strength Review Total FBD 2 MA 0 16kN2m 16kN5m Dy 7m Dy 16kN ZFy 0 Ay 16kN 16kNDy Ay 16kN No Kidding The shear and moment diagrams are drawn to determine the maximum values of shear force and bending moment This should not be a problem and is omitted here Vmax 16kN Mmax 32kNm The section properties be determined as follows draw the areas I bh3 0200m0300m3 Z 12 12 12 000045m4 Sz 0003m3 ARCH 3126 Statics Strength Review The stresses can now be determined Mmax 32kNm kN 0x 3 210670 2 Sz 0003m m 0x 107MPa 16kN0200m0150m 015039quot QVmaxA39y39 2 400kN W I Zr 000045m40200m m2 7W 400kPa ARCH 3126 Statics Strength Review Figure O H32 P 16 kN P16kN l l L 2m 3m 16 V kN 16 32 M kN m ARCH 3126 Statics Strength Review 0 O H36 Shear in BuiltUp Beams If the horizontal shear is resisted by continuous area material or horizontal glue joint the shear stress equation applies If the horizontal shear is resisted by a linearly continuous element line like a weld the shear flows equation applies If the horizontal shear is resisted by discrete element as some spacing IIs points like nails or screws the shear force equation applies added below All three forms are shown below IZQ TZVA It It VQ VA39739 61 1 q 1 FZVQIAS FZVAJIAS Problem O H33 A gluedlaminated glulam timber beam is made from three 2 X 4 112 X 312 in finish dimensions boards as shown in Figure O H33 The strength of the wood in horizontal shear is mallow 80 psi which takes into account a factor of safety What is the required shear strength of the glue joints psi if there is to be a factor of safety against failure of the glue joints of FS 30 That is how strong must the glue joint be so that the joints do not fail before the wood itself Neglect the thickness of the glue joint The maximum allowable shear force can be found from the allowable shear stress in the wood draw the areas ARCH 3126 Statics Strength Review lb 35m45m3 72 T QHV 71 80m2 35m I If y 35m225m2225m V 8401b The shear stress in the glue can now be found 1539 8401b35m15m075m PEI1L It It 35m45m335 m Tglue 7111psi Applying a factor of safety of 3 we get Tglue 213psi ARCH 3126 Statics Strength Review Figure O H33 iiiE iiiE 2 Lzlld L3ilE lgj Problem O H34 A steel plate girder is fabricated by welding of two 16 in X1 in flange plates to a 70 in X 38 in web plate using fillet welds as shown in Figure O H34 The welds have an allowable shear stress is qauow 2 kipsin Determine the allowable vertical shear force V for the plate girder if the four fillet welds run continuously along the length of the girders Let us determine the moment of inertia first Element A an I On d in Ad2 On Top flange 161 161312 3512 20164 16 1333 355 Web 037570 037570312 0 0 2625 1071875 Bot flange 161 161312 3512 20164 16 1333 355 Total 5825 10721416 40328 1 Z Ad2 2 510494161714 The maximum allowable shear force can be found from the allowable shear flow in the welds draw the section ARCH 3126 Statics Strength Review 22Wj51050m4 VQ VA39y39 q in q V 1 1 A39 7 16m10m3 55m V 360k1 ps Figure O H34 J Problem O H35 A wood box beam is fabricated by placing wood screws at regular intervals Ax in the four locations as shown in Figure O H35 Draw this The vertical boards are t1 20 mm X d 250 mm and the horizontal boards are t2 40 mm X b 150 mm and the screws have an allowable shearforce capacity of Vs 12 kN Determine the maximum longitudinal screw spacingljx ifthe beam is designed for a maximum vertical shear force V 10 kN Let us determine the moment of inertia first draw the section ARCH 3126 Statics Strength Review Element A mm2 I mm4 d mm Adz mm4 Top flange 15040 15040312 125 402 666150000 6000 800000 105 Left web 20250 20250312 0 0 5000 26041666 Right web 20250 20250312 0 0 5000 26041666 Bot Flange 15040 15040312 125 402 666150000 6000 800000 105 Total 22000 53683333 132300000 1 Z Ad2 185983333 mm4 The maximum allowable shear force can be found from the allowable shear force in the screws draw the areas l l 4 F QM VA y Ax Ax 2 F1 212kN185983333mm 1 1 VA39 739 10kN150mm40mm105mm Ax 709mm ARCH 3126 Statics Strength Review Figure O H35 40 mm 41 850 mm 40 mm J7 4t 80 mmj L150 mmJ LEO mm El Chapter O H4 Deflection of Beams o O H41 Introduction The equation of the elastic deflected curve can be derived by looking at a point on a deflected shape as shown in Figure ExO H41 Remembering the following derivation for moment vs curvature as follows Ex1x M 00 pltxgt Ex1xK ARCH 3126 Statics Strength Review Figure ExO H41 pgtltgt MOO yvltgtltgt quotlt The slope or inclination angle is 6arctan and for small angles tan99 thus we get 6 From calculus we remember to following radius of curvature relationship right and for small slopes it becomes simpler as follows Combining to two equations we get the following 2 Z i M JV or THEM or V EIM p Exx dx dx E The remaining equations come from previous derivations V39EI 639 V39EIM V 39EIV V EI w ARCH 3126 Statics Strength Review Now we will look at how to use this Remember it is just a continuation of loadshear moment relationships with the 1El added when we go to slope and deflection Problem O H41 Determine the equations of the elastic deflected curve and the maximum deflection in the simply supported beam shown From the Total FBD Pa ZMAZOZPa RBL R32 Pb ZFyZOZRARB P RAZT From the FBD05xlta ZFyZOZPTbVx1 Ic1PTb 2M0 OzMx1 PTbx M ZP beEIVIH L x1 2 E191E1V139IPTI dxPZbZ C1 ARCH 3126 Statics Strength Review 2 3 E1A1E1V1IP2b CII P61 C1xC3 From the FBD a5xltL ZFyZOZT P sz Vx2Z T M OM P bxP x a Z 0 x2 L M PTlm Px aEII2quot x2 2 2 E162 EIVZ39IPTbx Px adv 1321 wc2 2 2 3 3 E1222mwgazgxg Summary of equations 2 Ezalzpbx C1 2L 3 E14212 C1xC3 2 2 EIQZZPbx Px a C 2L 2 2 ARCH 3126 Statics Strength Review Pbx3 Px a3 Em2 6L 6 C2xC4 Boundary Conditions At 1 xa G1 9 2 slopes equal at a 2 xa A1 A2 deflection equal at a 3 x0 A10 deflection is zero at left support 4 xL A L0 deflection is zero at right support 2 2 2 Pba C1 Pba Pa a C2 C 0 2L 2L 2 1 2 Pba3 Pba3 Pa a3 C a C C a C 2 6L 1 3 6L 6 2 4 since C1Cz then CsC4 Pb03 3 0 C1OC3 030 so C40 6L 3 3 2 2 4 owmcw C2 39 6L 6 6L Final of equations 2 2 2 Pb E191 m 91L2b23x2 2L 6L 6LEI 3 2 2 E1A1 mx Al Pbx LEE 8 6L 6L 6LEI ARCH 3126 Statics Strength Review 2 2 2 2 EIQZZPbx Px a PbL b 2 92 PbL2 b2 3x2M 6LEI 2E1 3 3 2 2 EIAZ Pbx Px a PbL b x 6L 6 6L Pbx 2 3 L2 b2x2m 6LEI 6E1 The point of maximum deflection occur where the slope is zero 0ltxlta LZ b2 Oz lz P bL2 b2 3x2 x 6LEI 3 Substituting this x back into the deflection equation yields LZ b2 Pb 2 2 2 3 A1 3 LZ bz ALL 1 2 6LE1 3 1 9J LE1 ARCH 3126 Statics Strength Review Figure O H41 Y P Q hb A a X i i4 gtlt RAL QJltbi RE 0 igtMn O xlto X Q WE Q XltL Many equations are already derived and we can make use of them by superposition We can add effects at any point as long as we can describe that point

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