Are pocket structures
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This document provides information on structural concepts for the Architect Registration Exam, including: 1) Formulas and diagrams for trigonometry, forces, moments, stress/strain, shear and bending moments, and trusses. 2) Descriptions of key concepts like the four properties of a force, methods of adding forces, and support conditions. 3) Explanations of how to analyze beams, trusses, and other structural elements using methods of sections and joints.
Are pocket structures
- 1. The Non-User's Pocket Guide to the Transient Knowledge Necessary for the Structural Divisions of the Architect Registration Exam- ARE CONCEPT COMMENTS Memory Trick: SOHCAHTOA (Indian Tribe) used when triangle has a 90º angle. Rise (Rise) SIN RISE Slope COS RUN TAN SLOPE Run (Run) a SIN and COS of any angle are between (+/-) 1 Slope 0º < angle < 45º COS > SIN 45º < angle < 90º SIN > COS Run Rise (Slope) b Law of Sines Law of Sines and Cosines are used when triangle has no a = = B right angles. Sin A Law of Sines is used when you are given more angles a than sides. Law of Cosines is used when you are given more sides C than angles Variations in L.O.A. Properties of a Force: A Force is defined by four properties: Transmissibility: Force Addition: Forces FORMULAE AND DIAGRAMS Law of Cosines A B Components of a Force: b2 = a2 + c2 - 2ac (Cos B) c2 = a2 + b2 - 2ab (Cos C) Trigonometry/Math or a2 = b2 + c2 - 2bc (Cos A) or or Variations in Sense: Sin C = or Cos C = or Tan C = or c b 90º triangle Shallower angles (<45º) have larger horizontal components Ø P Px Py OPP HYP ADJ HYP OPP ADJ a b c c a b b Sin B c Sin C c Non- 90º Triangle 1000# = 1 k Ø 2 3 4 1 P Py PX P Py PX Algebraic Method: For finding the resultant of several forces 3 12 For finding the Resultant of several forces. Steeper angles (>45º) have larger vertical components A Ø P Py PX C P Py PX Py PX P PX Py P PX Py P Px Py Graphic Method for Force Addition: 1 2 3 Tail of 2 on Head of 1 Tail of 3 on Head of 2 1 23 R Resultant begins at 1s Tail and ends at last Head 3 12 Tails at same P.O.A. Force Horizontal Vertical 1 +/- +/- 2 +/- +/- 3 +/- +/- R +/- R X = ΣX +/- R y = ΣY 1. 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) 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) The Resultant is also a force and is thus defined by the four properties listed above. The Resultant of several forces is a single force that has the same effect on a body as all the other forces combined. P = P P Algebraic Method for finding the Resultant of several forces is used when force magnitudes and lines of action for each force are known 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 Graphic Method is used when a system is in equilibrium and we need to calculate one or more unknown forces that contribute to equlibrium 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 P PX Py September, 2004 © 2004 David J. Thaddeus, AIA PAGE : 1 OF 4
- 2. The Non-User's Pocket Guide to the Transient Knowledge Necessary for the Structural Divisions of the Architect Registration Exam- ARE CONCEPT COMMENTSFORMULAE AND DIAGRAMS Moment Moment = Distance Summing Moments (∑M = 0) to establish equilibrium To find Beam / Truss reactions To maintain equilbrium of members Overturning Moments due to Wind Loads or Hydrostatic Pressure Couple Unlike a Moment, a Couple is NOT about a certain point, but rather it is about ANY and ALL points. Moment of a A Couple depends on Force (P), and perpendicular distance (d) Force X MomentsandCouples Couple= P x d d d between two Forces that make up the couple. MomentsandCouples (clockwise, CW) P Couple between top Chord (C) and bottom chord (T) in a simply supported truss Couple between compression in concrete ( top ) and tension in rebar ( bottom ) of reinforced beam 1. ELASTIC RANGE: straight line relationship, slope = E P 2. PLASTIC RANGE: increase in strain, no increase in Load / Stress A 3. STRAIN HARDENING: material deforming in section (necking), and in length ΔL 4. FAILURE: Material is gone! Lo 5. YIELD POINT/ YIELD STRENGTH: material is no longer elastic, deformation is permanent F 6. ULTIMATE STRENGTH: material is about to fail ε Unit Strain ( ΔL/ L0 ) 7. RUPTURE: Kiss it Good-Bye 8. E: Modulus of Elasticity.Measures material's resistance to deformation ΔL = α (ΔT) L0 Shortening or Elongation of members along their axis Change (Expansion & Contraction) of shape due to Temperature Examples include Columns, Trusses, Cables, Cross Bracing b = width d = depth c = location of x b Roller: 1 Reaction ( V ) Pin / Hinge: 2 Reactions ( V , H ) Simply Supported: Statically Determinate (Simply Supported) loading = three unknown reactions, and can be solved using the equation of Static equilibrium. Statically Indeterminate loading > 3 unknown Reactions Call your engineer. Fixed / Moment: 3 Reactions (V , H , M) Continuous: Multiple Reactions Indeterminate Loading: AE Neutral Axis ΔL= PL0 AxialLoadsGeometry Deflection Shear Bending Moment Area (In2 ) PSI Modulus of Elasticity= Stress / Strain ε: MomentsandCouples Units in / in PSI Stress(F=P/A) Formulas F: Direct Stress Unit Strain E: SupportConditionsStress/Strain bd3 12 Radius of = r = I Gyration A Modulus of Elasticity: E (slope) Fy 8 1 2 3 4 5 6 7 Fu ΔL: 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) ΔL: Deformation, change in length (in), caused by change in temperature (ºF) ΔT: Change in temperature α : Coefficient of thermal expansion/contraction A = bd EA36,A-50= 29,000 KSI P ΔL L0 ΔL A ΔL E ΔL Moment of Inertia (In4 ) Section Modulus (In3 ) V V H V H M (Determinate) 1 2 3 2 1 3 3 3 Force P creates a Negative Moment about point B d A _ + d P B Force P creates a Positive Moment about point A P P ( CCW ) ε Gravity CG ; Center of x Y Y c d/2 d 2 2 3 1 If a Member is inadequate in Shear, increasing the Area (either Width (b) or Depth (d)) is effective. 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. 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. Pin/Hinged connections iclude most wood to wood, bolted steel, and precast concrete connections. fixed connections include most welded steel / steel connections and cast-in-place concrete. _+CCW CW A & B are called Centers of Moment, or Centers of Rotation The perpendicular distance (d) is called the Moment Arm, or Lever Ixx = =Sxx = Ixx bd2 C 6 12 September, 2004 © 2004 David J. Thaddeus, AIA PAGE : 2 OF 4
- 3. The Non-User's Pocket Guide to the Transient Knowledge Necessary for the Structural Divisions of the Architect Registration Exam- ARE CONCEPT COMMENTSFORMULAE AND DIAGRAMS Example 1: M = Moment V =Shear Equilibruim = ∑ Fx = 0; ∑ Fy = 0; ∑ MAny = 0 Sum of Areas in Shear Diagram = Moment Magnitude of drop = Concentrated Load Between concentrated loads, Moment Diagram Slopes Uniform loads create gradual drop in Shear ( straight line ) Uniform loads create curve (downward cup) in Moment Diagram Overhangs and cantilevers will always have a negative Moment in Moment Diagram. Simply supported beams always have positive Moments VMAX always occurs at support Moment is minimum MMAX occurs where V = 0 Uniform load coefficient, w, = slope in Shear Diagram Point of Inflection (P.O.I.) is a point on the Moment Diagram where M = 0 Point of Inflection only happens when a beam has an overhang If Loading Diagram (FBD) is symmetrical, then the Shear Diagram and the Moment Diagram are also symmetrical. Maximum Shear dictates how much Beam area is needed Maximum Moment dictates how much Bema Depth is needed If a hole must be punched out of a Beam to allow for passage of pipe or similar reduction, this must happen at a location of low Shear and low Bending Moment Method of Method of Sections: Joints: Web Stresses Stress increases towards middle Stress increases towards end panels ShearandBendingMomentDiagramsTrusses Top and Bottom Chord Stress L < R L = 5' x 12k = 4k Possible Zero Members C Method of Joints is used to analyze Force / Stress in every member of a Truss Method of Joints is also used to analyze Force / Stress in a member that is close to a support (not in middle of truss) Method of Sections is used to analyze only a few (3 max) members of a truss After cutting a truss in 2 segments, each segment is in Equilibrium ΣF X = 0 ; ΣF Y = 0 ; ΣM ANY = 0 Concentrated Loads in a Truss must be applied at panel points; otherwise we have combined stresses ( T or C + V and M ) Joints that have three or less members framing into them, may potentially have Zero Members 15' 15' + - w , W P Load/ FBD V=0 M=0 R = 10' x12k = 8k 12k 10' 5' 15' L R + 6' 6' 6' 18' 12k 12k L = 21k R = 21k 6'12' 18' 6k R = 4k L = 2k L = 6'/18' x 6k = 2k R = 12'/18' x 6k = 4k C C T T T T T T T T T C C C C C C C C C C C C C C T T T T T C C T A Truss is inherently stable due to triangulation Truss is stable in its own plane but needs bridging or cross-bracing perpendicular to its own plane All joints in an honest Truss are Pinned Joints Rigid Joints in a Truss will result in less Deflection than Pinned Joints (Advantage) 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) Members carrying Tension can be much smaller than members carrying Compresion m + 3 = 2 j ; where m = Number of Members j = Number of Joints Example 2: 6' 6' 6' 18' 12k 18k = w = 1k/ft. W = 18k L = 21k + 2k L = 23 k R = 21k + 4k R = 25 k L = 23k R = 25k September, 2004 © 2004 David J. Thaddeus, AIA PAGE : 3 OF 4
- 4. The Non-User's Pocket Guide to the Transient Information Needed to Successfully Pass the General Structures Division of the Architect Registration Exam - ARE CONCEPT COMMENTS MATERIAL: DESIGN FOR DESIGN FOR DEFLECTION: Δactual = CONST.x (W or P) (Lx12"/ft.)3 SHEAR: BENDING: E I WOOD BEAMS: STEEL BEAMS: CONCRETE BEAMS: Shear: Shear: Shear: Concrete: f 'c b, d, f 'c Stirrups: f y f y, φ, A v, spacing Bending: Bending Bending Concrete: f'c b, d, f 'c Fb= 24 KSI Rebars: f y (full lateral support) f y, (φ, # rebars), A s WOOD COLUMNS: STEEL COLUMNS: Slenderness: Slenderness: kLUNB. k=0.5 kwood= 0.671 E Columns FORMULAE AND DIAGRAMS BeamsGeneralBeamDesign LUNB./ dLeast I = bd3 / 12 Deflection : F b = MMAX SMIN 5 wL4 Fc W = wL Fv , F b , E LOAD: GEOMETRY: A = bd Shear S = (bd2 ) /6 Bending f v < F v ; V MAX A MIN Fv α L, w, W, P, FBD f b < F b ; F b = MMAX SMIN Δactual < Δallow FV = 3 VMAX 2 A MIN Fb = MMAX Sxx tables LUNB , M-Charts (partial lateral support) wood steel r MMAX = WL / 8 VMAX = W/2 MMAX = PL/4 VMAX = P/2 VMAX = 3P/2 MMAX = PL/2MMAX = PL/3 VMAX = P 23 PL3 ΔMAX = 648 EI 19 PL3 ΔMAX = 348 EI 1 PL3 ΔMAX = 48 EI384 EI 5 WL3 ΔMAX = 384 EI = VMAX, M MAX W = wL W = w L MMAX = WL/8 VMAX = W/2 VMAX = 0.6 W MMAX (+) = 0.08 W L MMAX (-) = - 0.1 W L W/2 AWEB F V = VMAX h A V A S d b SMIN = wL2 / 8 = wL2 /8 Fb< 24 KSI FC , FT , F P A = b x d slenderness ratio 50 L/d 200 KL/r k=1 k=2 b d Beam design must satisfy Shear, Bending Moment and Deflection requirements 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 Codes The Actual Stress ( f ) is an outcome of the application of a load ( W , P ) on a member When a Load is applied perpendicular to the axis of a member ( Normal Loading), Shear and Bending stresses develop The Strain associated with Bending is called Deflection and the deflected shape of a Beam is the inverse (upside/down) of the Moment Diagram When a load is applied along the axis of a member, Axial Compression and Tension Stresses develop The strain associated with Tension is Elongation and the strain associated with Compression is Shortening For the same Magnitude and span, a Uniform Load will cause less Deflection than a Concentrated Load for the same material and geometry The the same Load and Span, a Cantilever will deflect more than a simply supported beam For the same Load, Material and Geometry a slight increase in Span will create a huge increase in Deflection For the same Load and Span, an increase in the Modulus of Elasticity, E, ( a stronger material), will result in less Deflection For the same Load and Span, an increase in the Moment of Inertia, I , (a deeper member) will result in less deflection 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/3 P P P PL/3 P P P L/4 L/4 L/4 L/4 PP 3P/2 PL/2 P L/2 L/2 P/2 PL/4 P/2 L w W/2 WL/8 W/2 W/2 w W/2 W/2 WL/8 W/2 w L L L 0.4W 1.1W 1.1W 0.4W 0.6W 0.5W 0.4W 0.4W 0.5W 0.6W d b f A W k11 For all beams; Δactual = CONST.(W or P)(Lx12"/ft.)3 EI Allowable Deflecion is specified by model codes as a fraction of the span Δallow = L / 240, L / 360,... FC = P/A Long and thin ( slender ) columns tend to be governed by buckling Short and fat ( chunky ) columns tend to be governed by crushing w w - 0.1WL 0.08WL 0.025WL 0.08WL P.O.I.- - +++ - 0.1WL ... . September 2004 © 2004 David J. Thaddeus, AIA PAGE : 4 OF 4