PAGE 1,2,3CONCEPTFORMULAE AND DIAGRAMSCOMMENTSTrigonometry/MathMemory Trick: SOHCAHTOA (Indian Tribe)n used when triangle has a 90 angle.Sin C = orRiseor(Rise)Bn SIN RISESlopen COS RUNn TAN SLOPECos C = orRunor(Run)can SIN and COS of any angle are between (+/-) 1Slopen 0 < angle < 45 COS > SINn 45 < angle < 90 SIN > COSTan C = orRunorARise(Slope)bLaw of Sinesn Law of Sines and Cosines are used when triangle has noa==Bright angles.Sin An Law of Sines is used when you are given more anglesathan sides.Law of Cosinesa2 = b2 + c2 - 2bc (Cos A)b2 = a2 + c2 - 2ac (Cos B)n Law of Cosines is used when you are given more sidesc2 = a2 + b2 - 2ab (Cos C)Cthan anglesForcesVariations in L.O.A.Properties of a Force:n A Force is defined by four properties:Components of a Force:Variations in Sense:Transmissibility:Force Addition:Moments and CouplesMomentMoment =Force XDistancen Summing Moments (M = 0) to establish equilibriumn To find Beam / Truss reactionsn To maintain equilbrium of membersn Overturning Moments due to Wind Loads or HydrostaticPressureCouplen Unlike a Moment, a Couple is NOT about a certain point,but rather it is about ANY and ALL points.Moment of aPn A Couple depends on Force (P), and perpendicular distance (d)Couple= P x dddbetween two Forces that make up the couple.(clockwise, CW)Pn Couple between top Chord (C) and bottomchord (T) in a simply supported trussn Couple between compression in concrete ( top ) andtension in rebar ( bottom ) of reinforced beamStress / StrainFormulasUnits1. ELASTIC RANGE: straight line relationship, slope = EF:=PDirect StressPSI2. PLASTIC RANGE: increase in strain, no increase in Load / StressAStress (F=P/A)3. STRAIN HARDENING: material deforming in section (necking),and in lengthe:=DLUnit Strainin / in4. FAILURE: Material is gone!Lo5. YIELD POINT/ YIELD STRENGTH: material is no longer elastic,deformation is permanentE:=FModulus of Elasticity= Stress / StrainPSI6. ULTIMATE STRENGTH: material is about to faileUnit Strain ( DL/ L0 )7. RUPTURE: Kiss it Good-Bye8. E: Modulus of Elasticity.Measures material's resistance to deformationAxial LoadsDL = a (DT) L0n Shortening or Elongation of members along their axisDL= PL0n Change (Expansion & Contraction) of shapedue to TemperatureAEn Examples include Columns, Trusses, Cables, CrossBracingGeometryb = widthArea (In2)Sheard = depth=Deflectionc = location ofNeutral Axisx=BendingMomentbSupport ConditionsRoller: 1 Reaction ( V )Pin / Hinge: 2 Reactions ( V , H )Simply Supported:n Statically Determinate (Simply Supported) loading = threeunknown reactions, and can be solved using theequation of Static equilibrium.n Statically Indeterminate loading > 3 unknown ReactionsCall your engineer.Fixed / Moment: 3 Reactions (V , H , M)Continuous: Multiple ReactionsIndeterminate Loading:Shear and Bending Moment DiagramsExample 1:n M = Momentn V =Shearn Equilibruim = Fx = 0; Fy = 0; MAny = 0n Sum of Areas in Shear Diagram = Momentn Magnitude of drop = Concentrated Loadn Between concentrated loads, Moment Diagram Slopesn Uniform loads create gradual drop in Shear ( straight line )n Uniform loads create curve (downward cup) in MomentDiagramn Overhangs and cantilevers will always have a negativeMoment in Moment Diagram. Simply supported beamsalways have positive Momentsn VMAX always occurs at support Moment is minimumn MMAX occurs where V = 0n Uniform load coefficient, w, = slope in Shear Diagramn Point of Inflection (P.O.I.) is a point on theMoment Diagram where M = 0n Point of Inflection only happens when a beam has anoverhangn If Loading Diagram (FBD) is symmetrical, then the ShearDiagram and the Moment Diagram are also symmetrical.n Maximum Shear dictates how much Beam area is neededn Maximum Moment dictates how much Bema Depth is neededn If a hole must be punched out of a Beam to allow for passageof pipe or similar reduction, this must happen at a location oflow Shear and low Bending MomentTrussesMethod ofMethod ofSections:Joints:Top and Bottom Chord StressWeb StressesStatically determinate loading = 3 unknown reactionsStatically indeterminate loading = >3 unknown reactions

&L&C&"Arial,Bold"&12The Non-User's Pocket Guide to the Transient Knowledge Necessary for the Structural Divisions of the Architect Registration Exam- ARE&R&LJune 2004&C 2004 David J. Thaddeus, AIA&RPAGE : &P OF 4bd312Radius of = r = I Gyration An L < Rn L = 5' x 12k = 4kPossibleZero MembersbModulus of Elasticity: E(slope)Fy81234567FuDL: deformation, changes in Length (in) caused by Axial Load (P)P : Axial Load (#,k)L0 : Original, undeformed Length (in. not ft.) A : Cross Sectional Area (in2)E : Modulus of Elasticity (PSI, KSI)DL: Deformation, change in length (in), caused by change in temperature (F) DT: Change in temperature a : Coefficient of thermal expansion/contractionA = bd90 trianglen Shallower angles (45) havelarger vertical componentsAPPyPXn EA36,A-50= 29,000 KSI

n k P k DL n k L0 k DL n k A m DL n k E m DLCn Method of Joints is used to analyze Force / Stress in every member of a Truss n Method of Joints is also used to analyze Force / Stress in a member that is close to a support (not in middle of truss)n Method of Sections is used to analyze only a few (3 max) members of a trussn After cutting a truss in 2 segments, each segment is in Equilibrium SF X = 0 ; SF Y = 0 ; SM ANY = 0 n Concentrated Loads in a Truss must be applied at panel points; otherwise we have combined stresses ( T or C + V and M )n Joints that have three or less members framing into them, may potentially have Zero MembersMomentof Inertia (In4)Section Modulus (In3)15'15'CPPyPXPyPXPPXPyPPXPyPPxPyGraphic Method for Force Addition:123n Tail of 2 on Head of 1nTail of 3 on Head of 2123Rn Resultant begins at 1s Tail and ends at last Head312n Tails at same P.O.A.Force Horizontal Vertical1 +/- +/-2 +/- +/-3 +/- +/-R +/- R X = SX +/- R y = SYVVHVHM(Determinate)123213331. Point of Application (P.O.A.) 2. Magnitude ( #,kips ) 3. Sense (Arrowhead, Push or Pull, C or T) 4. Line of Action (L.O.A.) , (Angle with horizontal)n The Equilibrant is also defined as a force that has the same P.O.A., Magnitude and L.O.A. as the Resultant but has an opposite sense (Arrow)n The Resultant is also a force and is thus defined by the four properties listed above.n The Resultant of several forces is a single force that has the same effect on a body as all the other forces combined.P=PPForce P creates aNegative Momentabout point BdA_+dPBForce P creates aPositive Momentabout point APP( CCW )eGravityCG ; Center ofxYYcd/2d2231n If a Member is inadequate in Shear, increasing the Area (either Width (b) or Depth (d)) is effective.n If a Member is inadequate in Deflection, increasing the Moment of Inertia (Width (b) is OK; but Depth (d) is cubed and) is much more effective in reducing Deflection.n If a Member is inadequate in Bending, increasing the section modulus (width (b) is OK; but Depth (d) is squared and) is much more effective in reducing Bending.n Pin/Hinged connections iclude most wood to wood, bolted steel, and precast concrete connections.n fixed connections include most welded steel / steel connections and cast-in-place concrete.+-w , WPLoad/FBDV=0M=0n R = 10' x12k = 8k12k10'5'15'LR+6'6'6'18'12k12kL = 21kR = 21k6'12'18'6kR = 4kL = 2kL = 6'/18' x 6k = 2kR = 12'/18' x 6k = 4kCCTTTTTTTTTCCCCCCCCCCCCCCTTTTTCCTn A Truss is inherently stable due to triangulationn Truss is stable in its own plane but needs bridging or cross-bracing perpendicular to its own planen All joints in an honest Truss are Pinned Jointsn Rigid Joints in a Truss will result in less Deflection than Pinned Joints (Advantage)n Rigid Joints in a Truss will result in larger size members than Pinned Joint Trusses since members will have to resist V and M in addition to C or T (Disadvantage)n Members carrying Tension can be much smaller than members carrying Compresionn m + 3 = 2 j ; where m = Number of Members j = Number of Jointsn Algebraic Method for finding the Resultant of several forces is used when force magnitudes and lines of action for each force are known

n Algebraic Method of Force Addition 1. Resolve each force into vertical and horizontal components 2. The algebraic (+/-) sum of all horizontal components gives the horizontal component of the Resultant. 3. The algebraic (+/-) sum of all vertical components gives the vertical component of the Resultant

n Graphic Method is used when a system is in equilibrium and we need to calculate one or more unknown forces that contribute to equlibrium

n Graphic Method for Force Addition 1. Arrange all forces Head to Tail then add (independent of order) 2. Resultant begins with its Tail at the Tail of the 1st Force and Head at the Head of the last 3. Resultant can be determined through calculation (All angles are typically known)PPXPy_+CCWCWn A & B are called Centers of Moment, or Centers of Rotationn The perpendicular distance (d) is called the Moment Arm, or LeverIxx ==Sxx =Ixx bd2 C 612Example 2:6'6'6'18'12k18k=w = 1k/ft.W = 18kL = 21k + 2kL = 23 kR = 21k + 4kR = 25 kL = 23kR = 25k

PAGE 4CONCEPTFORMULAE AND DIAGRAMSCOMMENTSGeneral Beam DesignMATERIAL:DESIGN FORDESIGN FORDEFLECTION: Dactual = CONST.x (W or P) (Lx12"/ft.)3SHEAR:BENDING:E IBeamsWOOD BEAMS:STEEL BEAMS:CONCRETE BEAMS:Shear:Shear:Shear:Concrete: f 'cb, d, f 'cStirrups: f yf y, f, A v, spacingBending:BendingBendingConcrete: f'cb, d, f 'cFb= 24 KSIRebars: f y(full lateral support)f y, (f, # rebars), A sColumnsWOOD COLUMNS:STEEL COLUMNS:Slenderness:Slenderness:LUNB./ dLeastkLUNB.k=0.5kwood= 0.671 E

&L&C&"Arial,Bold"&12The Non-User's Pocket Guide to the Transient Information Needed to Successfully Pass the General Structures Division of the Architect Registration Exam - ARE&R&L&"Lucida Grande,Regular"&12June 2004&C 2004 David J. Thaddeus, AIA&RPAGE : 4 OF 4I = bd3/ 12 Deflection: F b = MMAX SMIN5 wL4FcW = wLFv , F b , ELOAD:GEOMETRY: A = bd ShearS = (bd2) /6 Bendingf v < F v ;V MAXA MINFv aL, w, W, P, FBDf b < F b ; F b =MMAXSMINDactual < DallowFV = 3 VMAX 2 A MINFb = MMAXSxx tablesLUNB , M-Charts(partial lateral support)woodsteelrMMAX = WL / 8VMAX = W/2MMAX = PL/4VMAX = P/2VMAX = 3P/2MMAX = PL/2MMAX = PL/3VMAX = P23 PL3DMAX =648 EI19 PL3DMAX =348 EI1 PL3DMAX =48 EI384 EI5 WL3DMAX =384 EI=VMAX, M MAXW = wLW = w LMMAX = WL/8VMAX = W/2VMAX = 0.6 WMMAX (+) = 0.08 W LMMAX (-) = - 0.1 W LW/2AWEBF V = VMAXhA VA SdbSMIN= wL2/ 8= wL2/8Fb< 24 KSIFC , FT , F PA = b x dslendernessratio50 L/d200 KL/rk=1k=2bdn Beam design must satisfy Shear, Bending Moment and Deflection requirementsn The Allowable Stress (F) of a species of wood or a Grade of steel depends on the material itself and is tabulated in Manuals and Building Codesn The Actual Stress ( f ) is an outcome of the application of a load ( W , P ) on a membern When a Load is applied perpendicular to the axis of a member ( Normal Loading), Shear and Bending stresses developn The Strain associated with Bending is called Deflection and the deflected shape of a Beam is the inverse (upside/down) of the Moment Diagramn When a load is applied along the axis of a member, Axial Compression and Tension Stresses developn The strain associated with Tension is Elongation and the strain associated with Compression is Shorteningn For the same Magnitude and span, a Uniform Load will cause less Deflection than a Concentrated Load for the same material and geometry n The the same Load and Span, a Cantilever will deflect more than a simply supported beamn For the same Load, Material and Geometry a slight increase in Span will create a huge increase in Deflectionn For the same Load and Span, an increase in the Modulus of Elasticity, E, ( a stronger material), will result in less Deflection n For the same Load and Span, an increase in the Moment of Inertia, I , (a deeper member) will result in less deflectionn The Points of Inflection on the Moment Diagram of the Continuous beam (Left) indicate the locations of curve reversal, and are the locations where reinforcing steel would be flipped from bottom to top of the beam.L/3 L/3 L/3PPPPL/3PPPL/4 L/4 L/4 L/4PP3P/2PL/2PL/2 L/2P/2PL/4P/2LwW/2WL/8W/2W/2wW/2W/2WL/8W/2wL L L0.4W 1.1W 1.1W 0.4W0.6W0.5W0.4W0.4W0.5W0.6Wdb fA Wk11n For all beams; Dactual = CONST.(W or P)(Lx12"/ft.)3 EIn Allowable Deflecion is specified by model codes as a fraction of the span Dallow = L / 240, L / 360,...n FC = P/An Long and thin ( slender ) columns tend to be governed by bucklingn Short and fat ( chunky ) columns tend to be governed by crushing n Short and fat ( chunky ) columns tend to be goww- 0.1WL0.08WL0.025WL0.08WLP.O.I.--+++- 0.1WL....