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ARE Structural Systems Exam Study Seminar

September 17, 2011

Presented by:

Shell + Meyer Associates Inc - Structural Engineers

Ben Van De Weghe, P.E. Chris Meyer, P.E. Joe DeLong, P.E.

SESSION SCHEDULE

I. GENERAL STRUCTURAL OVERVIEW – Joe DeLong II. Wood Construction – Chris Meyer

III. Steel Construction – Ben Van De Weghe IV. Concrete Construction – Ben Van De Weghe V. LATERAL FORCES - Wind Design – Chris Meyer

VI. LATERAL FORCES – Seismic Design – Ben Van De Weghe VII. Structural System Vignette - Gregory M. Spon

Reference:

KAPLAN STUDY GUIDE

20% discount to AIA members. Here is a discount code you can use on the following website:

Go to www.kaplanaecarchitecture.com

Make your product selections and go to checkout

Look for the promotion code field on the right side of the page

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Register as a member (you'll create a user ID and password--hold onto that! It gives you access to the Online Courses and Build Your Own Exam Test Banks-IMPORTANT!) and submit payment with a credit card.

Shell + Meyer Associates, Inc. is not affiliated with KAPLAN Construction Education

Presented by Shell + Meyer Associates, Inc. Page 1 of 43


Structural Systems ARE Seminar 9/17/2011

1

ARE Structural Systems Exam Study Seminar

Presented by:Shell + Meyer Associates Inc - Structural Engineers

Ben Van De Weghe, P.E.Chris Meyer, P.E.Joe DeLong, P.E.

NCARB ARE 4.0 Structural Systems Exam

Ref.: http://www.ncarb.org/~/media/Files/PDF/ARE-Exam-Guides/SS_Exam_Guide.pdf

NCARB ARE 4.0 Structural Systems Exam

Ref.: http://www.ncarb.org/~/media/Files/PDF/ARE-Exam-Guides/SS_Exam_Guide.pdf

ARE Structural Study Resources•Kaplan Construction Education -http://www.kaplanaecarchitecture.com/

•2011 ARE Structural Systems Study Guide•2011 ARE Sample Questions & Answers

•Professional Publications, Inc. –http://ppi2pass.com

2011 ARE R i M l•2011 ARE Review Manual•2011 ARE Structural Sample Problems & Practice Exam

•NALSA Publishing – https://www.nalsa.com/•Archiflash study software, vignette tutorial videos

Statics

Lesson 1Lesson 1

Statics – Forces•Push or pull on an object, measured in units of weight (lbs, kips, N)•External to a body: load; internal resistance to load: stress•Multiple forces acting on the same point may be added together to calculate the resultant force•Analytical method of force addition: Use trigonometry to calculate resultant force & direction

Ref. Kaplan 2011 ARE Structural Systems Guide



2

Statics – Moments•Tendency of force to rotate about a point (not actual rotation)•Moment couple: two equal forces in opposite directions a distance apart


Statics – Equilibrium•Objects at rest are in equilibrium•Sum of all external forces and moments acting on a building is zero


Statics – Free Body Diagrams•Imaginary cut through a structure•Use equilibrium to calculate internal forces


Statics – Properties of Areas•Calculation of internal properties, used later to calculate stress & strain•Centroid, statical moment•Moment of inertia = I = bd3/12 + Ado

2 Parallel axis theorem


Statics – Stresses•Total stress on a section: internal force (measured in lbs, kips, N, kN, etc.)•Unit stress: Force per area (measured in psi, ksi, N/m2, kN/m2, kPa, etc.)•Tensile, compressive, shear


Statics – Strain•Deformation caused by external loads•Total strain: lengthening or shortening (units of length)•Hooke’s Law: Stress is directly proportional to unit strain, up to an elastic limit•Young’s modulus constant: E = stress / strain; “ tiff ”“stiffness”•Stiffness of steel > concrete > wood•Yield point just past elastic limit, ultimate strength at fracture: steel > concrete > wood•Allowable stress accounts for uncertainties in loads, materials, and actual stresses Vs. calculated stresses•Factor of safety = (yield or ultimate strength)/(allowable strength); higher for brittle materials than for ductile materials




3

Statics – Thermal StressesCoefficient of thermal expansion (constant) = unit strain / temperature change; for example: steel ε = 0.0000065/°F


Beams & Columns

Lesson 2Lesson 2

Beams & Columns – Types of Beams and Loads

•Beam: member that supports loads perpendicular to its longitudinal axis i.e. in bending•Types: Simple, cantilever, overhanging, continuous, fixed end•Concentrated & uniform loads•Statically determinate vs. statically indeterminate•Beams impart forces to supports called reactions•Beams impart forces to supports, called reactions


Beams & Columns – Shears and Moments

•Internal forces within a beam•Shear: sum of normal forces on one side of an imaginary cut at any point along the beam•Moment: sum of moments on one side of an imaginary cut, taken about the point of the cut•Shear & moment diagrams•Shear & moment diagrams•Positive moment: top of beam in compression, bottom of beam in tension; negative moment: top in tension, bottom in compression


Beams & Columns – Beam Stresses

•Flexural (bending) stress: f = My/I; maximum stress occurs at outermost fibers•Shear stress: in the plane of the material: f = (VQ)/(Ib); f = 1.5V/(bd) for a rectangular section


Beams & Columns – Columns

•Member subjected primarily to axial compressive load•Sometimes also resists bending moment•Combined axial & bending stress calculation: f = P/A ± Mc/I•Compression members in trusses & diagonal braces can behave similar to columns•The longer a column is the more likely it is to buckle and buckling•The longer a column is, the more likely it is to buckle, and buckling strength can be much less than yield or ultimate strength•Radius of gyration (resistance to buckling): r = √(I/A)•Slenderness ratio = L/r




4

Trusses

Lesson 11Lesson 11

Trusses - Introduction•Truss: assembly of structural members, arranged so that each member is loaded axially for efficient use of material•Commonly steel or wood•Rule of thumb for optimization: Span / depth = 10•Truss forms triangular panels since triangular arrangement is stable•Perimeter members: chords – analogous to beam flanges•Interior members: web members – analogous to web members•Interior members: web members – analogous to web members


Trusses – Method of Joints•Draw a free body diagram for each joint•Use equations of equilibrium to solve for unknown forces


Trusses – Method of Sections•Cut an imaginary section through truss, draw free body diagram of one portion•Use equations of equilibrium to solve for unknown forces


Trusses – Truss Design•Compression members designed similar to columns•Tension member stress calculation must account for net area excluding holes for fasteners•Steel trusses: Common configuration is to have double angle members with long legs back-to-back, gusset plates at joints, forces transferred through bolts or welds•Centroidal axis of members should meet at a point at joints to avoid•Centroidal axis of members should meet at a point at joints to avoid eccentricity and bending


Conventional Structural Systems

Lesson 9Lesson 9



5

Conventional Structural Systems – Building Loads

•Dead load: weight of permanent components•Live load: assumed maximum loads based on occupancy of building, plus 15 psf minimum for partitions where applicable•Live load may be reduced for members supporting large floor areas (main girders, columns, and footings)•Roof live loads may be also be reduced for steep slopesRoof live loads may be also be reduced for steep slopes•Wind & earthquake loads•Retained earth lateral loads: varies by soil type and wall type, but generally equivalent to a fluid weighing 30 pcf.•Hydrostatic pressure from fluid: Unit weight of a liquid x depth; for underground structures with high ground water, hydrostatic pressure acts on walls on floor


Conventional Structural Systems – Building Loads

•Temperature change: restraint on expansion/contraction will induce force in a member•Impact loads: Increase of static loads by as much as a factor of 2 to account for actual dynamic loads greater than static load for elevators, cranes, machinery, etc.y•Handrail live load: 50 lbs/ft or 200 lbs at a point•Vibration: shock absorption, adding mass to reduce effects•Blast loads on critical secure facilities


Conventional Structural Systems – Basic Concepts

•Flexural design criteria: shear stress, flexural stress, deflection•Strength and stiffness: Depth of a member is much more important than width•Narrow, deep flexural members: ideal, but depth may affect other economies of construction and may require more framing for stabilityy q g y•Wide, shallow members may deflect excessively


Structural Systems – Efficient Flexural Members

•Cross-sectional area as far as possible from neutral axis is efficient, hence I-shaped beams & trusses; T-beams & composite beams•Flanges or chords resist flexure, web resists shear•Another example: ribbed elements, e.g. steel deck (stronger & stiffer than flat plate) and concrete joists•Concrete slabs: two way slab more efficient than one way•Concrete slabs: two-way slab more efficient than one-way•Members can be constructed with profile to match moment diagram e.g. bowstring truss or tapered steel plate girder


Structural Systems – Continuity

•Continuous beams: negative moment at supports•Have less deflection and lower maximum moments than simple span beams•Common in cast-in-place concrete construction


Structural Systems – Rigid Frames

Rigid moment-transferring beam-column connectionsFlexural stress in beams is decreased, flexural stress in columns is increasedCan resist horizontal loads in the plane of the frameSteel rigid frames: special detailing, bolted or welded connections




6

Conventional Structural Systems•One-way structures: secondary members span between primary members•Wood systems: joists, plank-and-beam, trussed rafters•Steel systems: Open web joists, beam-and-girder, stub girder•Concrete: one-way systems ideal for rectangular y y gbays, two-way systems for square bays

•One-way: slab-and-beam & joist slab•Two-way: Flat plate, flat slab, waffle slab

•Multistory frames•Cast-in-place concrete: monolithic beam-column joints with reinforcing steel transferring forces•Structural steel: rigid or semi-rigid frames with moment connections at beam column joints, or shear walls or diagonal bracing if beam-column connections are not rigid


Conventional Structural Systems – Selecting a System

Considerations: Safety in supporting loads, allowance for mechanical systems, serviceability: minimize possibility of damage and minimize incon enience to occ pantsinconvenience to occupants, accommodation of architectural design, fire resistance, constructability

Cost considerations: Simplicity of structure, repetition, span length, building height, structural material; structure represents approximately 25% of total construction cost


Long-Span Structural Systems

Lesson 10Lesson 10

Long Span Structural Systems•Once a major part of ARE exam, now just one part of general structures section•Arches and vaults: historical long-span structures•19th & early 20th century: long-span one-way trusses based on railroad bridge designs•Long span: a span in excess of 60 ft•Building types: sports arenas & enclosed stadiums, theaters, convention centers, warehouses, industrial, aircraft hangers, retail•Special requirements: temperature effects, shipping, field-assembled connections


Long Span Structural Systems –One-Way Flexural Systems

•Steel beams: W sections up to 80 ft, plate girders for longer spans•Trusses: more economical; roof framing, floor framing, transfer truss for discontinuous column•Vierendeel truss: no diagonals (thus not really a truss), rigid joints; chosen for architectural reasons•Open web steel joists: SJI standard joists can be chosen from load

ftables; economical for light loads over long spans•Joist girders: shop-fabricated steel truss to support steel joists


Long Span Structural Systems –One-Way Flexural Systems

•Glulams: span up to 40 ft for floors, 100 ft for roofs•1-2”-thick laminations, bonded with adhesives; dried before assembly, thus more dimensionally stable than other structural wood pieces; laminations free of defects, thus stronger than solid lumber/timber•Prestressed concrete beams: Pretensioned & post-tensioned; single T, double T; span-depth ratio double that of conventional reinforced

fconcrete; shipping weight can be a factor




7

Long Span Structural Systems –Two-Way Flexural Systems

•Grid of rigidly connected members•Complex to analyze, complex to build•Main members along column lines•Steel space frames


Long Span Structural Systems –Axial Systems - Arches

•Today, mostly limited to long-span roofs•Pure arch: loaded purely to compression owing to parabolic geometry; most arches must resist some bending due to variable loads•Loading results in horizontal reaction, indicating thrust load that must be resolved with a tie rod or abutments; steep arch - lowermust be resolved with a tie rod or abutments; steep arch lower thrust•Vaults, arched ribs, lamella


Long Span Structural Systems –Axial Systems - Suspension

•“Reverse of an arch” – tension-only cables, pull reactions at ends•Greater sag in a cable lower tension•Circular structure: compression ring/tension ring/tension radial cables


Long Span Structural Systems –Axial Systems – Rigid Frames


Long Span Structural Systems –Axial Systems - Domes

•Dramatic and symbolic, but also structurally efficient•Series of arches (meridians), and series of hoop rings, which prevent thrust loads on supporting structure




1

Wood Construction

• Lesson 3 – Wood Construction• Lesson 7 – Wood Connections• Lesson 6 – Stud Walls

L 11 T• Lesson 11 - Trusses

Positive Construction Qualities• High Strength to Weight Ratio• Readily Available• Economical in Light Construction• Easily installed / modified

Photo – APA The Engineered Wood Association

Disadvantages• Moisture Changes Result in Shrinkage or

Swelling• Combustibility• Insects• Fungus

Properties of WoodMoisture Content (%MC)Fiber Saturation Point (~30%)Allow for shrinkage or swelling in members

• Especially perpendicular to grain• Decrease is MC Increase in E

Photo Credit : Understanding Wood –R. Bruce Hoadley ,The Taunton Press c. 2000

Photo Credit : Understanding Wood – R. Bruce Hoadley ,The Taunton Press c. 2000

Properties of WoodKEEP WOOD DRY• Provide ventilation• Proper drainage• Separation from the ground



2

Allowable Stresses• Fiber Bending Stress, Fb (psi)• Tension Parallel to Grain, Ft (psi)• Compression Parallel to Grain, Fc

• Compression Perp. Grain, Fc

H i t l Sh F• Horizontal Shear, Fv

• Obtain Values from NDS (National Design Specification) • Or Kaplan 2011 study guide page 382

Reference Values

Beam Forces

Understanding Wood – R. Bruce Hoadley ,The Taunton Press c. 2000

Brace Compression Side

Adjustment Factors• CD = Load Duration• CF = Size FactorDecrease Allowable Bending Stress for

members greater than 12”• CL = Beam Stability FactorCL Beam Stability Factor

• 1.0 if laterally braced

• Example• F’b = Fb*CD*CF*CL*…

National Design Specification (2005 NDS, AF&PA)



3

Load Duration• Allowable Stresses (Except for Fc )

can be increased by a load duration factor, CD

» 15% for two month duration (SNOW)» 25% for seven days duration (Construction)» 33 33% for wind or earthquake33.33% for wind or earthquake» 100% Impact

• Creep – Deflection over time

http://timber.ce.wsu.edu/

Design of Wood Beams• Bending Stress, fb = Mc/I = M/S• Horizontal Shear Stress, fv1.5*V / bd

• Design for DeflectionDeflection formulas

**U it i **• **Unit conversion**

Code deflection limits (Floor)• L/360 (Live Load Only)• L/240 (Total Load)

Increase Depth• Bearing – Compression Perp. Grain

Notching• Avoid notching, if possibleDon’t notch in middle third of spanEnd notches should not exceed 25% of the

member’s depth


Glue Laminated Beams• Laminations

24FV4 – SP/SP 24 2400 psi Bending Stress, Fb

V Visually Graded 4 Combination SP/SP So thern Pine (O ter and SP/SP Southern Pine (Outer and

Inner Plies)• Volume Factor Cv

• 5 1/8” x 12” x 21’-0” Long• Applied to Fb


Glulam Reference Values



4

Volume Factor, Cv


Wood Columns• Axial CompressionSlenderness Ratio, L/d

• Short Columns (Small L/d)– Crushing

• Long Columns (Max. L/d = 50)– Buckling

• Intermediate• Intermediate– Combination of Crushing and Buckling

• Reference Kaplan Chart on Page 58

• Combined Bending and Axial Compression


Studs Wall• Bearing, Non-Load Bearing, Shear• Wood columns, with sheathing

bracing the weak axis• Platform Framing/ Balloon Framing

Stud Walls• AdvantagesEconomical, simple to build, light weight,

simple installation of plumbing

• DisadvantagesCombustibilityCombustibilityNot as rigid or durable as masonry or concrete

counterpartSimple installation of plumbing

Too simple to install plumbing? Wood Connections• Single Shear and Double Shear

Connections

Number of fasteners = Load RequiredCapacity of FastenerCapacity of Fastener



5

Connection Design Factors• Lumber Species Increased Density Higher Load Cap.

• Critical SectionDoes the fastener reduce the cross sectional

area of the wood members?• YES – Must use net cross section

A l f L d t G i• Angle of Load to GrainWood strongest in bearing parallel to Grain,

weaker perpendicularHankinson Formula

• Spacing of FastenersGroup Factor – Reduces Allowable Load

Wood Connections• Edge and End DistancesEdge distance, loaded edge, unloaded edge,

end distance

• Condition of Loading Duration Factors also apply


Types of connectors• Timber connectors• Bolts

• Single shear, double shear, load parallel, load perp., wood – wood, wood – steel.

• Lag Screws• Good withdrawal

• Wood Screws• Wood Screws• No end grain installation• Better holding power over nails

• Nails• Common Wire • No Withdrawal Value

• Miscellaneous

TRUSSES• Beam AnalogyMoment and Shear

• Used for long roof spansDepth / Span = 0.10

• Bracing Bracing Bracing!!• Bracing, Bracing, Bracing!!Good for vertical loads, not out of plane

• Triangular configuration most stable

TRUSSES• ComponentsChords (Flanges)Web Members, Diagonals, Verticals

• Panel Points

Redundant members• Not required for stabilityq y

Zero Force Members

Method of Joints• Determine the reactions at the supports –

You usually need to draw the free-body diagram of the entire truss and determine the reactions at its supports

• Analyze the joints – Draw free-body diagrams of the joints and apply the equilibrium equations to determine the axial forces in the members. Recognize “Zero-Force” members Assume all unknown forces are in tension (away from

the joint) Choose joints that require you to solve for no more than

two unknown forces



6

Method of Sections• Determine the reactions at the

supports – You usually need to draw the free-body diagram of the entire truss and determine the reactions at its supports

• Choose a Section –Choose a Section Obtain a free-body diagram of part of the truss by cutting

members, including those whose axial forces you want to determine.

Some trial and error (and cleverness) may be need to find a free-body diagram that leads to the results you need.

Method of Sections



1

Steel Construction

Lesson 4Lesson 4Lesson 7 – Steel Connections

Properties of Steel

Modulus of Elasticity, E = 29,000 ksi Yield StressASTM A36 Fy = 36 000 psi plate L CASTM A36 Fy = 36,000 psi plate, L, CASTM A992 Fy = 50,000 psi W beams

Beam designations

W24x76 M8x6.5

S24x100 C12x20.7

L6x4x3/8”

Design Methods LRFD (Load Resistance Factor Design)

Rn nominal strength multiplied by a resistance factor. Use factored loads for a factor of safety.ASD (All bl St St th D i ) ASD (Allowable Stress Strength Design)Rn nominal strength divided by a factor of safety . Usually 1.67 for bending. AIA likes ASD.

Steel Beam Design for Flexure

Find Ma, the applied moment Compute Zrequired > Ma / Fy

Convert to all inchesConvert to all inches Look-up in Table 3-2 pages 3-11 to 3-19

Pick the bold beam as that’s cheapestAssumes you’re using 50 ksi steel

Example

Lateral Support

Simple Span Beams Top Flange is in Compression – hopefully braced by

metal deck or other beam/joist Bottom Flange is in Tension

Cantilever Beams Top Flange is in Tension Bottom Flange is in Compression

Unsupported LengthWhen the compression flange is free to twist then the

beam fails easily.Allowable Moment in Beams – Table3-10 on Pages 3-96 to 3-139 of AISC

pick the solid line for the cheapest beam



2

Example

Deflection

Roof Beams Span 12( )

240

Floor Beams Span 12( )

Snow load

Floor Beams p ( )

360

Uniform Loads 5 w L

4 1728( )384 29000000 I

Live load

This is the deflection based on a known I (Moment of Inertia), but if you don’t know I yet then solve and figure out what I will be required. See Table 3-3.Rough guess try beam depth to be > L/22

Example

Composite Beams

Concrete slab acts as part of top flange

Steel Columns Column strength is Pn / , where

= 1.67. But this is tough to calculate.

AIA recommends to use Table 4-22 which depends on the most slenderwhich depends on the most slender axis of the column: KxLx / rx or KyLy / ry which is

bigger? K – shape of buckling, usually 1.0 L – height of column, inches r – Radius of gyration, the square

root of I/A

Examples of K

Function of end rotation and ability to translate (move sideways). Most demanding condition is the cantilever/flagpole condition. Column

buckles in a shape more than double its actual height.



3

More typical, use tables W8, W10, W12, W14 Other column issues

Combined Axial Load and Flexure Not likely to be on the exam. Columns cannot be at 100% capacity in both axial and bending

and once, but you can be 50%/50%, 30%/70%, etc. Once columns start moving there are additional load magnifiers.

Column Base Details Need to keep pressure down to avoid crushing CMU or concrete

below.Bearing Pressure = Axial Load

Base Plate Area

Steel Connections

Bolts A325, A490 Snug Tight, Pretensioned, Slip-Critical Tension

Double Shear

Tension Single Shear or Double Shear Threads (N)included or (X)excluded from shear plane

Single Shear Single Shear

Bolt Tension Pick A325 or A490 high strength

bolts Table 7-2 has strength per bolt Try to avoid prying on bolts, that will

i th fincrease the force Example:

Four 1” DIA A325 bolts Pn / = 4 x 35.3 kip/bolt = 141.2 kips

Only have room for two bolts to carry 47 kips tension.

Try 47 kips / 2 =23.5 kip/bolt – 3/4” A490? Pn / = 2 x 25.0 kip/bolt = 50.0 kips

Works!

Bolt Shear Pick A325 or A490 high strength

bolts Figure out if you’re in double or

single shearPi k if th d (N) l d d Pick if threads are (N)encluded or (X)excluded

Example: Four 1” DIA A325 bolts, double shear, N Pn / = 4 x 37.7 kip/bolt = 150.8 kips

Bolt Shear Also check if bolts will tear out or

crush base metal in bearing Need to check both spacing and

edge distance.U ll 3” i f ¾” b lt Usually use 3” spacing for ¾” bolts

Usually use 1-1/2” edge distance for ¾” bolts

Example: 7/8” DIA bolts at 3”o.c. in ½” thick A36 steel plate (Fu = 58ksi)

Rn / = 60.9 kips/in x 0.5 inches . = 30.45 kips/bolt

If there are 6 bolts, then 6 x 30.45 is a total of 182.7 kips



4

Welds Welds

Fillet or Groove

Pick Fexx to exceed Fu, normally 60 ksi for A36 and 70 ksi for A992

Fillet Weld strength Rn / = Length * 0.707 * tw * (0.6 Fexx ksi) / 2

, y

Weld example: 6” long weld x 3 sides, how big to resist 75 kips?

Pick Fexx to be 70 ksi rn / > 75 kips / 18” = 4.17 kips per inch tw > 4.17 kips/in x 2 / (0.707 * 0.6 * 70ksi) = 0.281 inches 5/16”

Fillet Weld strength Rn / = Length * 0.707 * tw * (0.6 Fexx ksi) / 2

p ( ) 5/16” + 1/16” minimum = 3/8”, fits on ½” plate

6”

Plate ½” x 6”tension

Connections Use the simple shear connections to

fasten beams to girders and columns. Connections need to rotate slightly or

else they will pry and pull apart. Minimum number of bolts need to fill

upper half of beam Use tables for double angles, single Use tables for double angles, single

angles, single plates, seat angles, etc.



1

Concrete Construction

Lesson 5Lesson 5Lesson 6 – Concrete Walls

Lesson 8 - Footings

Reinforced Concrete

•Adding rebar will take care of tension, but now we’re mixing materials. Non-homogeneous.•Good news, similar coefficient of thermal expansion. .00055 vs .00065

Concrete Mix – “Hydration”Concrete is combination of:Binding agent Portland cement, fly ashFine aggregate SandCoarse aggregate GravelWater Potable waterWater Potable waterAdmixtures Water reducer, accelerator, air

Keep water/cement ratio low. Improves strength. Use water reducer to increase slump if required.

Air entrained concrete survives freezing better, try for 4% to 6% entrained air.

ASTM Portland Cement Types All cements are fine powders to make the

chemical reaction with the water. HYDRATION Big difference is rate of reaction, slower reaction

means less heat at once. Type I – normal cement Type I normal cement Type II – slower set, less heat, very hot weather

or moderate sulfate attack. Type III – extra fine, high early strength Type IV – very slow strength, mass concrete Type V – Severe sulfate exposure

Slump More slump = lessMore slump less

strength, don’t just add water

More slump = easier to pump

Can also get “sticky” mix, tough to finish

Testing Concrete Need to know strength at 28 days. If existing use destructive test – 4”

diameter cores Non-destructive is not as accurate –

Windsor Probe



2

Flexural Design Not symmetric like steel and wood. Concrete and steel going to resist as a

moment couple. Concrete f’c= 3000 psi to 14000 psi

E= (57,000 f’c)0.5 approx Rebar Fy = 60,000 psi ASTM A615

E = 29,000,000 psi Most basic relationship M = Rbd2

depth is most important variable

Code Limits All members should have shrinkage and

temperature minimum As&t = 0.0018 Agross Going to use ultimate strength design, very

similar to Steel LRFD.ACI 318-05

L d 1 2D 1 6LLoads 1.2D+1.6LFlexure 0.9Shear 0.75Unreinforced 0.55

minimum 200/fy or 3ef’c /fy Maximum 0.851(f’c/fy)[87000/(87000+fy)]

Modes of Failure•ACI 318 limits amount of rebar to provide a smooth, gradual failure of beams to warn theto warn the public.

•Shear failures are never graceful, ACI 318 is extra stringent to avoid this situation. is reinforcement ratio, Arebar/Aconcrete

Beam examplef’c = 4,000psi Fy = 60,000psia = 3*0.79in2*60,000psi = 2.32”

.85 *4,000psi*18inMn = (As Fy)[d – a/2]

(3)-#8

18”

n ( y)[ ]= 3*0.79in2*60,000psi [20” – 2.32”/2]= 2,679,048 in-lb = 223 ft-kips

Mn = 0.9 * 223 ft-kips = 201 ft-kipsCheck reinf. limits: = 3*0.79in2/(18in*20in)

= 0.0066 Okay .0033 & .0214 are limits

Tee Beams & Joists

More efficient, gets rid of dead weight at bottom.

Keep ‘a’ in top flange!!



3

Shear Capacity Total Vn = Vc + Vs concrete + stirrups Vc = 2 (f’c)0.5 b d

if concrete capacity is double what is required, then no stirrups needed.

Vs = Av Fy d / s s v ymaximum spacing is typically d/2

s

Shear ExamplesDeep Beams

run bars both directions on each face

Concrete Cover Must protect rebar from weather, epoxy

coat reinforcement or increase cover if chlorides (salt) are present.

Development Length

Very complex formula involving strength of concrete and rebar, unit weight of concrete, concrete cover, spacing between bars, diameter of rebar, critical locations, depth of concrete below the bar, epoxy coating on bar, presence of hook on end, etc.

Simplest form for 3000 psi concrete and 60 ksi rebar, plenty of cover and spacing, no epoxy, not critical, bottom bars, and #7 or larger simplifies to: ld = 55 db

Also use this length to splice bars. But unless you stagger the laps between adjacent bars and make sure you lap the bars where stress is low you’ll need to use a longer Class B Lap = 1.3 x ld

Deflection Steel beams stay a constant size as loaded.

Concrete beams crack as load is applied, therefore moment of inertia is constantly getting smaller (vicious circle).

Ieff = less than Igross but more than Ifully cracked

Creep After the initial deflection the concrete will keep

on deflecting for years and years to come. This secondary deflection is called “creep”.

Sustained loads may cause 2x to 3x more deflection than your initial deflection.



4

Continuity BenefitsSmaller moments, less deflection

But now require top rebar over support

Tee beams don’t work over supportswl2/10

wl2/11wl2/8

Continuity Drawbacks

Prestressed Concrete

Use 270 ksi wire draped through member. As it tries to straighten itself out a moment is created to counteract the gravity load.

Precambers the member for dead load. Add regular rebar for live load. Mild Reinforcing Pre-tensioned – pull wire before placing concrete, hollowcore planks Post-tensioned – pull wire after placing concrete, parking garages

Concrete Column1% to 8% reinforcing steel

Concrete Walls

Combination of flexure and axial load. Emperical Design ACI 318-05 Chapter 14Pn = 0 55 f’c Agross [ 1 (kL/32h)2]Pn = 0.55 f c Agross [ 1 – (kL/32h) ]

Can be cast-in-place, tilt-up, or precast. Tilt-up and precast often see worst loads

while being moved and erected. ACI 318 requires minimum vertical and

horizontal reinforcement.

Walls and Dirt Soil will react with walls as either:

Active pressure, wall is free to move such as a retaining wall. P = 45 psf/ft typical

At-rest pressure, wall is restrained top and bottom, such as a typical basement wall. P = 60 psf/ft typical

Passive pressure, wall actually pushes horizontal against the soil, support at base of wall. P = 150 psf/ft typical

Water pressure will increase horizontal load, therefore provide foundation tile to sump or weep holes.



5

Example

Ft-kip/ft

Big reaction on floor

Slab on grade takes bottom reaction

Assumes well-drained foundation, no additional loads on ground outside, f’c=4ksi, fy=60ksi, soils engineered determines 60 psf/ft at-rest soil pressure on wall.

Retaining Walls Check overturning, 1.5 factor of safety. Check sliding, 1.5 factor of safety. Check bearing pressure on soil. Check strength of concrete Check strength of concrete. Provide weep holes to avoid water

pressure overloading wall. Expect top of wall to move under active

pressure, otherwise at-rest pressure will develop.

Masonry Nearly the same as concrete, uses all the same design principles. Traditionally still service loads instead of LRFD factored loads. Lots

of empirical tables and design procedures still in code. Rebar is often centered which really cuts down effective depth. Much lower strengths, f’m =1,350 psi most common. Strength is combination of C90 block and either type M, S, or N

mortar.mortar.

Clay brick Concrete masonry units CMU

Masonry Beams Same procedure as concrete, except replace [d-a/2]

with jd, where j will typically be 0.8 to 0.9. Many of concrete’s reinforcement limits will not apply

to masonry.

Soil Capacity Take soil borings such as with Shelby Tubes. Goal is to identify strata and soil types. Clay,

sand, organics, garbage. Goal is to record “blow count” as tube is driven.

I di ti f b i itIndication of bearing capacity. Density, moisture, plasticity also calculated.

Shear waves for seismic classification. Determine the optimum compaction for

placement under structures to prevent future settlement.

Historically all service load, not LRFD factors.



6

Soil CapacityShortcut bearing capacity tables from Reference Books and from Ohio Building Code.

Foundation Types

Bottoms of all footings must be kept below frost depth.

Square Footing Given: Column reaction = 147 kips

Soil capacity = 4,500 psf Find minimum footing size.

Aminimum = 147 kips / 4.5 ksf = 29.4 sfTake square root – 5.4’x5.4’Use 5’-6”x5’-6”, or if rectangle 4’x7’-6”

Begin concrete design of footing for thickness and reinforcing. As soil capacity increases the footing thickness and reinforcing will also increase.

Square Footing Design

Turned upside-down, its just like designing a floor slab for several thousand psf load!

Flexural Design like a concrete beam

Flexural Shear (One-Way) and Punching Shear (Two-Way). Just make thick enough so stirrups are not required.

Combined Footing Pile Foundation Timber, concrete, steel, precast.

Groups of piles for a single column are covered with pile cap.

Often need grade beams spanning between caps.



1

Lateral Forces

Lesson 13 – Wind Design

CODES• Based on ASCE 7-05 (IBC 2006)

– Use this for ‘ARE’ Exam• DO NOT USE UBC 1997

– (older study guides may reference this though)

Structural Loads• Wind Loads MWFRS

• (Main Wind Force Resisting System)– V, Basic Wind Speed, 3 sec gust– I, Wind importance factor– Wind exposure– GCpi, Internal pressure coefficient– Cp or GCp, External pressure coefficient

G G t Eff t F t– G, Gust Effect Factors– Cladding loads: walls, roof, uplift

• Components and Cladding

Velocity Pressure• qz=0.00256*Kz*Kzt*Kd*V2*I (psf)

qz = velocity pressure evaluated at height ‘z’ above the ground (psf)

Kz = velocity pressure exposure coefficient evaluated at height ‘z’ (Varies with height off ground)

Kzt = topographic factor

K di ti lit f t ASCE T bl 6 6Kd = directionality factor, ASCE Table 6-6

V = Basic wind speed (3s gust) at 33 ft (10m) above ground in Exposure Category C

I = Importance Factor

Calculating Design Pressures• Gust effect factors, G

• Enclosure classification

• Internal pressure coefficient, GCpi

• External pressure coefficient, Cp or CpfCpf

ASCE 7-05



2

ASCE 7-05 ASCE 7-05

ASCE 7-05 ASCE 7-05

ASCE 7-05 ASCE 7-05



3

ASCE 7-05 ASCE 7-05

ASCE 7-05 ASCE 7-05

Wind Exposures

•Exposure B

Suburban Residential

• Exposure B

Urban

Wind Exposures

•Exposure C

Open Areas

Scattered Obstructions < 30’ h



4

Wind Exposures

•Exposure DShorelines and inland

waterways

(excluding hurricane regions)



1

SEISMICUsing IBC 2006Using IBC-2006

Lesson 12

Plate Tectonics

Measures magnitude of seismic waves.

Logarithmic scale, 7 to 8 is 10 times more magnitude.

Energy 33 times gyper step.6 (Virginia, 2011) to 9 (Japan, 2011) is 333=36000 times worse.

Mercalli Intensity Scale

Subjective scale based on visual perception of damage.

Earthquake Force Unlike live load, snow load, wind load; the

seismic load is not an applied force. It is the result of inertia.

• Ground accelerates and top of building attempts to follow per Newton’s Second Law of Motion--- F = ma

The building’s own natural period combines with the acceleration back and forth of the earth. Whiplash and such.

In order to quantify a design force NEHRP has developed the Response Spectrum. Not a total time history, but an approximation of the maximum response.

Actual force will depend on mass of building and the entire system’s ability to dissipate energy.

Building will not survive with only elastic movements Building will not survive with only elastic movements. Plastic movement is expected meaning permanent damage, but hopefully not complete failure.

Lots of special details are required since the typical ones only survive up to elastic movements. In order for beam to column connections to survive plastic deformation special plate thicknesses, extra weld and stiffeners, design connections for full capacity of member. We want beams to be fully loaded and rip off building while the columns are tough enough to remain standing.



2

The period of a building is the time to sway back and forth one cycle. Taller buildings have longer periods.

Longer periods also mean the acceleration at the top is less, therefore F=ma is less.

The taller your building the less the seismic forces will be, but also the higher the wind forces will be.

Mass is also a big factor. The heavier the building is the more force it generates when it sways back and forthmore force it generates when it sways back and forth.

Soft soils cause the acceleration to increase.

Worst case = short, heavy building on poor soils All CMU and brick 2 story school built on drilled piers. High seismic forces.

Best case = very tall steel frame building on good soils 20 story glass office tower bearing directly on bedrock. Low seismic forces.

Code Requirements IBC 2006 Index Force (0.01 W)

(ASCE 7-05,NEHRP 2000) Equivalent Lateral Force

Modal Response Spectrum Analysisp p yCode won’t Linear Response Historyeven explain how Non-Linear Response History

In each case there is no applied force to the building (like wind blowing). But when the building slides left, stops, and then goes right again all this acceleration generates force. Let’s pretend it’s applied from the outside.

Z Seismic Zones

Out of date now, but 30 years ago the USA was grossly divided up into zones.

MCE – Maximum Considered Earthquake

Worst expected acceleration anticpiated once every 2500 years (California uses worst ever expected).p )

United States Geological Survey (USGS) has mapped the whole country.

Graphed both short (Ss = 0.2sec) and long (S1=1.0sec) for pretty good soil (Category B).

Ss and S1

Easier to go to USGS website and enter zip code

Site Class

Site Class ‘B’ is used on all the maps. ‘A’ is better and all others are worse.

Site Class ‘D’ is the default. But ask a Geotech! Site Class ‘F’ is so bad you need special analysis. Such

as building in bogs or on quicksand.



3

Design Accelerations

Take Ss and S1 from the maps. These are the MCE values.

Modify the MCE values by your Site Specific Modify the MCE values by your Site Specific values. SM1 = FvS1 SMS = FaSs

Code permits the design accelerations to be 2/3 of the modified values: SD1 = 2/3SM1 SDS = 2/3SMS

http://earthquake.usgs.gov/hazards/designmaps/

Design Parameters Select Soil FactorJust like UBC-97, use the soil type to increase or decrease the acceleration.

MCE Parameters x Soil Factors IBC Seismic Design Categories

Tri-state area

If you come up with SDC A you may use the simpler Index Procedure, otherwise you’ll most likely use

Tri-state area

most likely use Equivalent Lateral Force procedure.



4

Equivalent Lateral Force• IBC 2006/ASCE 7-05/NEHRP 2000• F = a * m• V = Cs * W Cs min = 0.01

Cs = SD1/(R/I)Cs = SDs/(TR/I)

I is importance factor, varies 1.0, 1.25, or 1.5.SDC A, B, C – check one direction at a time.SDC D, E, F – check 100% one way plus 30%

perpendicular direction

or 0.5S1 / (R/I) in higher rated areas

Natural Frequency - Period

ASCE 7-05 uses T = TaCu

Ta = 0.1N number of storiesTa = Ct (hn)x additional building typesa t ( n) g yp

Response Modification Coefficient Based on the energy dissipating ability of

the structural system. Flexible, ductile systems do best. Also

systems that provide redundancy. R= largesystems that provide redundancy. R large Brittle systems do worst. R= small For steel, most of Ohio tries to use very last

entry “Not specifically detailed for seismic” R = 3. No fabricators are experienced and special inspections are much less.

Resistance comes from bending, shear, or axial loading. The most ductile systems use bending. Eccentric joints inEccentric joints in your frames allow a short bending zone where energy can be transformed into heat and released.

Vertical Distribution Load is spread so building is top heavy. When checking if the building overturns you can only use

(0.6- 0.14SDS) of the weight to resist the overturning.



5

Use the Computer

Just use the dead load weights of your building plus fixed equipment, 25% of your storage allowance, assume cars will roll freely and not add load, minimum 10 psf for architect’s interior partition walls (or actual weights), and in high snow zones assume that 20% of the snow is present.

Horizontal Load Distribution The force at each level ends up in the floor

or roof diaphragm.

Diaphragm Forces This is just a big flat plate that can transfer the

load to the moment frames or X-brace/shear walls.

Slab needs to be able to transfer shear forces to X-braces/shear walls, plus the edges of the flat plate are in tension and compression. These “diaphragm chords” will also need reinforcing.

Rigid Diaphragm

When the center of mass is not the center of rigidity the building starts swinging and twisting.

All codes add extra to this eccentricity, typically 5% of the building width to account for accidental torsion.

Example Building Eccentricity exists in both directions



6

ShearwallRigidity

The smaller the h/d is, the stiffer the wall will be.

Stiffness is the Stiffness is the inverse of rigidity.

A 10’ long wall next to a

Wall 1 is 40’ long, but it carries more load than all four Wall 2 which are 20’ long.

Architectural Components ASCE 7-05 Chapter 13, but you can get out of

most of this if you’re SDC A or B. Partition walls, storage shelves shouldn’t break

loose and crush owner. Access floors! Typically 5psf interior wind load will control. Fp = 0.4apSDSWp[1+2z/h] / (Rp/Ip) Whole section is only 6 pages, simple! Lots of ceiling requirements. They tend to fall on

people when not installed correctly. Same rules apply to the mechanical systems.

Lists the apand Rp for the most common components.

Other concerns Irregular

structures Soft story at base Discontinuing

shearwalls on lowest level

Stopping columns Stopping columns on transfer beams

Three sided boxes, one face all glass or all open

Holes in diaphragm

If in doubt try to add a seismic break/expansion joint in building.

Tubular Concept For very tall buildings it can be efficient to build a

structural “tube” around the perimeter. Make the façade out of closely spaced vertical

columns. Provide a deep edge girder beam. You’ll end up with something like a very dense

moment frame, almost more like a shear wall tube around the outside.

The former World Trade Center towers where tubes and performed quite well under impact loads.


How to Approach the Structural Systems Vignette:

I’ve read much on areforum.org that the ‘column and beam’ approach is easier and less complicated than the ‘bearing wall’ approach. So, this guide outlines how to solve the structural vignette using the column and beam approach.

1. Start with the upper level

Starting with the upper level allows you to see where the columns are transferred down to. This helps to eliminate what could be a fatal error with the vignette.

2. The program states that the common area is the only double story space in this layout 3. Lets start by drawing the columns in each of the corners in the upper level.


This gives us an ideas of the span between columns.

So, we have 52’-8” between our drawn columns. Thinking ahead to the beam requirement, we need to take into consideration the spacing between beams. The consensus on areforum.org is that the maximum span between beams should be 40’-0”… which means to be efficient, we need to add one more column in between these columns…not two, but one.

With the column added to the north wall and the south wall, we have the appropriate spacing of columns.

4. Next, lets add the beam.

A commonly asked question is: Should I draw beams continuously, or stop and start the beam at intermediate columns? Either approach is OK. Drawing continuous beams is quicker.


When we draw the beams, draw it in the centerline of the wall that it is being placed into.

Now lets draw the joists and decking. Decking always run perpendicular to the joists. Both the joists and decking should be drawn to the centerline of the wall. This is important for all decking, joists and beams you draw in this vignette! The program states that you cannot cantilever beams, so sticking to this method avoids doing that.


Keep in mind that we use 48” o.c. joists because the program says the deck can support a span up to 4’-0” you can use 32” o.c. but that wouldn’t be as efficient as 48” o.c. joists.

What is the maximum span of joists? 30'-0”

Ok now we have the upper level complete.

5. Now lets move to the lower level.


Using the principles discussed above, lets add joists, columns, beams and decking to the lower level.

Keeping in mind the spacing of the beams, add columns so they do not exceed 40’-0” in length, but without drawing too many columns.

By drawing the upper level first, we are able to see an overlay of the upper floor plan, which allows us to place the columns directly under (as seen in the red box)

So we add the remaining columns and then draw the beam for both sides:

We have 3 beams within 15’-0” of each other! This is necessary because: The column on the left most within the red box is supporting a column from above (beams cannot transfer column loads) The middle column in the red box is supporting the column from above and the column to the right most in the red box needs to support the south east corner of the roof above the toilet room, seating area, kitchen and prep & stor. rooms.

6. Now let’s draw the decking and joists.

Looking at the toilet room, seating area, kitchen area and prep & stor. rooms, it is highly efficient to draw the joists and decking as one continuous element spanning the rooms together.


Now lets draw the decking, joists, beams and columns for shop 1 and shop 2 using the using the techniques listed above.

Lets look for a second at the area in the red box above. I have seen this mistake on the forum often. There is NO BEAM required in this area. It does not carry any joists above it.

A simple thing to remember for beams is: DOES THE BEAM CARRY ANYTHING? YES – THEN I NEED A BEAM. It’s simple.. beams carry joists. Beams on this vignette DO NOT act as drag struts..just beams. Yes JUST BEAMS

Remember: decking is supported by joists, joists are supported by beams and beams are supported by columns. Simple, yes, but when you’re in a panic in the exam, this can ease the thought process. DO NOT CANTILEVER BEAMS


7. Now we have the covered entry left to draw.

Since we have two areas of beams in-between the covered entry, (In the red box above) the most efficient way to approach this solution is to span the joists between this entry. Keep in mind also that the span between these beams is 26’-0”, which is less than the 30’-0” maximum desired span length. Yes the joists spanning east and west would carry more load, but this approach requires extra beams.

All we need to add are joists and the covered decking. No more beams or columns. Efficient.

Completed lower level:


DOUBLE CHECK YOUR WORK – You should have plenty of time to double and triple check your work! Trace the decking to see that its perpendicular to the joists, trace joists to a beam and trace the beams to both sides to make sure they go to a column on both sides!

Questions / Concepts (from areforum.org)

What is the maximum span of joists? 30'

What is the maximum span of beams? 40'

Where should I draw the columns, beams, bearing walls, joist boundaries, and decking boundaries? Draw all elements to the center lines of the base plan walls.

Should I use bearing walls and/or columns? All-column and beam solutions are recommended, because there are less rules to remember.

Should I include a beam underneath a clerestory window (if the lower roof joists run parallel to the clerestory)? Coach says, "yes." lug-nut says "no." Lengthy discussion here: http://www.areforum.org/forums/showthread.php?t=203291

If a rectangular decking section can span over multiple joist areas, should I draw the decking as one or two elements? Either approach is OK.

Should I draw beams continuously, or stop and start the beam at intermediate columns? Either approach is OK. Drawing continuous beams is quicker.

Can joists and/or beams extend from an interior space to an exterior "covered entry" space? Coach says "yes"

Can I cantilever joists and/or beams? No, per the vignette's program.


Documents

ARE Structural Systems Exam Study Seminartheaudioarchitect.weebly.com/uploads/1/0/4/4/104475147/aiaohio_ss...•Shear stress: in the plane of the ... varies by soil type and wall type,