Structural Analysis Cheat Sheet

The Non-User's Pocket Guide to the Transient Knowledge Necessary for the Structural Divisions of the Architect Registration Exam- ARE FORMULAE AND DIAGRAMS

CONCEPT

Memory Trick: SOHCAHTOA (Indian Tribe) c (Rise) Sin C = OPP or Rise or B Slope a HYP

COMMENTS

n n

90º triangle

n

Trigonometry/Math

n

b Run or or Slope a

Cos C = ADJ HYP

(Run)

a

c

n n n

C

Run

OPP Tan C = ADJ

c or or Rise b

(Slope)

A

a = b = c Sin A Sin B Sin C

B

b

Law of Sines

n

Non- 90º Triangle n

a a2 = b2 + c2 - 2bc (Cos A)

Law of Cosines 2

2

2

2

2

2

Variations in L.O.A.

n

A

n

Properties of a Force: Py 3

PX

Shallower angles (45º) have larger vertical components

n

1000# = 1k

1

n

2

The Resultant of several forces is a single force that has the same effect on a body as all the other forces n

Ø

Components of a Force: P

PX

P Py

Py

Py

P

PX

Forces

PX

Py

P

Variations in Sense:

Px P

Transmissibility: Py

P

P

n

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

P Graphic Method for Force Addition: n For finding the Resultant of several forces.

Algebraic Method: n For finding the resultant of several forces Force 1 2

3 R

June 2004

The Equilibrant is also defined as a force that has the same P.O.A., Magnitude and L.O.A. as the

=

PX

Force Addition:

3

n

P

PX

1

The Resultant is also a force and is thus defined by

PX

P Py

2

n

Py

Py

PX

A Force is defined by four properties: 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)

4

Ø

Py

Ø Px

Law of Cosines is used when you are given more sides than angles

C

b

P

P

Law of Sines and Cosines are used when triangle has no right angles. Law of Sines is used when you are given more angles than sides.

c

b = a + c - 2ac (Cos B) c = a + b - 2ab (Cos C)

used when triangle has a 90º angle. SIN RISE COS RUN TAN SLOPE SIN and COS of any angle are between (+/-) 1 0º < angle < 45º COS > SIN 45º < angle < 90º SIN > COS

Graphic Method for Force Addition 1. Arrange all forces Head to Tail then add (independent 2 1 +/+/of order) R 2. Resultant begins with its Tail at the Tail of the 1 st 1 1 +/+/Force +/+/and Head at the Head of the last 3 3. Resultant can be determined through calculation n Resultant begins at 1 s Tail n Tail of 2 on Head of 1 +/- R X = SX +/- R y = SY (All angles are typically known) and ends at last Head n Tails at same P.O.A. nTail of 3 on Head of 2 Horizontal

Vertical

3

2

3

© 2004 David J. Thaddeus, AIA

2

n

PAGE : 1 OF 4

The Non-User's Pocket Guide to the Transient Knowledge Necessary for the Structural Divisions of the Architect Registration Exam- ARE FORMULAE nAND DIAGRAMS Tails at same P.O.A.

CONCEPT

COMMENTS

Tail of 3 on Head of 2

n

A & B are called Centers of Moment, or Centers of Rotation n The perpendicular distance (d) is called the Moment Arm, or Lever n Summing Moments (∑M = 0) to establish equilibrium n To find Beam / Truss reactions n To maintain equilbrium of members n Overturning Moments due to Wind Loads or Hydrostatic Pressure n Unlike a Moment, a Couple is NOT about a certain point, but rather it is about ANY and ALL points. n A Couple depends on Force (P), and perpendicular distance (d) between two Forces that make up the couple. n Couple between top Chord (C) and bottom chord (T) in a simply supported truss n Couple between compression in concrete ( top ) and tension in rebar ( bottom ) of reinforced beam

Moment

n

Moment = Force X

Distance

P

_

CCW +

CW

B Force P creates a Negative Moment about point B

_

+

d

d

Couple

d

P

Moment of a Couple= P x d (clockwise, CW)

d

Moments and Couples

A Force P creates a Positive Moment about point A

P

( CCW ) P

Formulas

Units 2 Fu

5

PSI

A = bd

Area (In2)

3 Ixx = bd

Moment of Inertia (In4)

Ixx

Section Modulus (In3)

bd2 Sxx = C = 6

Roller: 1 Reaction ( V )

n

n n

n n

2 V

5. YIELD POINT/ YIELD STRENGTH: material is no longer elastic,

b = width d = depth c = location of Deflection Neutral Axis Bending

Y

x b

2

CG ; Center of Gravity

1 2

3

1

Indeterminate Loading: 2

2 3

V M

Radius of = r = Gyration

Simply Supported:(Determinate)

H

H

3

8. E: Modulus of Elasticity.Measures material's resistance to deformation

x Y

Moment

Fixed / Moment: 3 Reactions (V , H , M) Continuous: Multiple Reactions

7. RUPTURE: Kiss it Good-Bye

DL = a (DT) L0 DL: Deformation, change in length (in), caused by change in temperature (ºF) DT: Change in temperature

Shear

V

deformation is permanent 6. ULTIMATE STRENGTH: material is about to fail

e

k DL k DL m DL m DL

kP k L0 kA kE

and in length 4. FAILURE: Material is gone!

7

Modulus of Elasticity: E (slope) Unit Strain ( DL/ L0 )

EA36,A-50= 29,000 KSI

Pin / Hinge: 2 Reactions ( V , H )

1

6 8

c

e

1. ELASTIC RANGE: straight line relationship, slope = E 2. PLASTIC RANGE: increase in strain, no increase in Load / Stress 3. STRAIN HARDENING: material deforming in section (necking),

4

d

Modulus of Elasticity= Stress / Strain

F

12

June 2004

3

d/2

in / in

Unit Strain

DL: deformation, changes in Length (in) caused by Axial Load (P) DL= PL0 P : Axial Load (#,k) L0 : Original, undeformed Length (in. not ft.) AE A : Cross Sectional Area (in 2) E : Modulus of Elasticity (PSI, KSI)

Geometry

Axial Loads

PSI

Fy

DL Lo

E:

Support Conditions

Direct Stress

Stress (F=P/A)

Stress / Strain

e:

1

P A

F:

3

3

n n

n

I A

Shortening or Elongation of members along their axis Change (Expansion & Contraction) of shape due to Temperature Examples include Columns, Trusses, Cables, Cross Bracing

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 Statically Determinate (Simply Supported) loading = three unknown reactions, and can be solved using the equation of Static equilibrium. n Statically Indeterminate loading > 3 unknown Reactions Call your engineer. 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. n

1

© 2004 David J. Thaddeus, AIA

PAGE : 2 OF 4

The Non-User's Pocket Guide to the Transient Knowledge Necessary for the Structural Divisions of the Architect Registration Exam- ARE FORMULAE AND DIAGRAMS

CONCEPT

Example 1: n

L< R L = 5' x 12k = 4k

n

15' R = 10' x12k = 8k

Shear and Bending Moment Diagrams

n

COMMENTS

12k

L

10'

5'

P w,W

R

Load/ FBD

15'

15' +

Example 2: V=0 12k

w = 1k/ft. W = 18k

18k

12k

12k

= 6'

6'

6'

+ 6'

18' L = 23k

R = 25k

21k

L= + L = 23 k

2k

21k +

R= R = 25 k

L = 21k

6' 18'

6'

-

6k

12'

6' 18'

R = 21k

4k

L = 2k

R = 4k

M=0

6k

L = 6'/18' x = 2k R = 12'/18' x 6k = 4k

Trusses

C

C

T

C

C

T

C

T

T

T

T

T

T

C

C

C

C

C

C

C T

C T

T

Method of Sections:

C

T

T

C

Method of Joints: C

T

C C T

Top and Bottom Chord Stress

June 2004

n

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 n All joints in an honest Truss are Pinned Joints n 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 Compresion n 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 truss n 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 PAGE : 3 OF 4 n n

Possible Zero Members C

M = Moment V =Shear n Equilibruim = ∑ Fx = 0; ∑ Fy = 0; ∑ MAny = 0 n Sum of Areas in Shear Diagram = Moment n Magnitude of drop = Concentrated Load n Between concentrated loads, Moment Diagram Slopes n Uniform loads create gradual drop in Shear ( straight line ) n Uniform loads create curve (downward cup) in Moment Diagram n Overhangs and cantilevers will always have a negative Moment in Moment Diagram. Simply supported beams always have positive Moments n VMAX always occurs at support Moment is minimum n MMAX occurs where V = 0 n Uniform load coefficient, w, = slope in Shear Diagram n Point of Inflection (P.O.I.) is a point on the Moment Diagram where M = 0 n Point of Inflection only happens when a beam has an overhang n If Loading Diagram (FBD) is symmetrical, then the Shear Diagram and the Moment Diagram are also symmetrical. n Maximum Shear dictates how much Beam area is needed n Maximum Moment dictates how much Bema Depth is needed n 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 n

Web Stresses

© 2004 David J. Thaddeus, AIA

The Non-User's Pocket Guide to the Transient Knowledge Necessary for the Structural Divisions of the Architect Registration Exam- ARE CONCEPT

FORMULAE AND DIAGRAMS

COMMENTS ( T or C + V and M ) Joints that have three or less members framing into them, may potentially have Zero Members n

June 2004

© 2004 David J. Thaddeus, AIA

PAGE : 4 OF 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 FORMULAE AND DIAGRAMS

CONCEPT

MATERIAL:

Fv , F b , E

GEOMETRY:

L, w, W, P, FBD

LOAD:

FC , FT , F P

DESIGN FOR SHEAR:

COMMENTS

A = bd

f v < F v ; Fv a V MAX A MIN

MMAX f b < F b ; F b =SMIN

DESIGN FOR BENDING:

DEFLECTION:

P

w

w

Shear

I = bd3/ 12 S = (bd2) /6

VMAX, M MAX

Dactual = CONST.x (W or P) (Lx12"/ft.) Dactual < Dallow EI

w

3

w

W = wL L

Beam design must satisfy Shear, Bending Moment and Deflection requirements n 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 n The Actual Stress ( f ) is an outcome of the application of a load ( W , P ) on a member n When a Load is applied perpendicular to the axis of a member ( Normal Loading), Shear and Bending stresses develop n The Strain associated with Bending is called Deflection and the deflected shape of a Beam is the inverse (upside/down) of the Moment Diagram n When a load is applied along the axis of a member, Axial Compression and Tension Stresses develop n The strain associated with Tension is Elongation and the strain associated with Compression is Shortening n 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 beam n For the same Load, Material and Geometry a slight increase in Span will create a huge increase in Deflection n 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 deflection n 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. n

Deflection Bending

W = wL

W/2

L/2

P/2

L/2

W/2

W/2

W/2

W/2

W/2 W/2

P/2

VMAX = P/2

= wL2/ 8

3 4 DMAX = 5 WL = 5 wL 384 EI 384 EI

P

WL/8

PL/4

MMAX = WL / 8

MMAX = PL/4

P

P

P

P

w

L/3

L/4

L/4

L

L/4

0.4W

3P/2

VMAX = 0.6 W 0.4W

0.6W

MMAX = PL/3

0.08WL

MMAX = PL/2

+

PL3

19 DMAX = 348 EI

DMAX = 648 EI

WOOD BEAMS:

Shear:

b

+

MMAX (+) = 0.08 W L

+

MMAX (-) = - 0.1 W L

d

Fb= 24 KSI (full lateral support) Sxx tables

- 0.1WL

f y, f, A v, spacing Bending Concrete: f'c b, d, f 'c Rebars: f y

Fb< 24 KSI (partial lateral support) LUNB , M-Charts

For all beams; Dactual = CONST.(W or P)(Lx12"/ft.)3 EI n Allowable Deflecion is specified by model codes as a fraction of the span Dallow = L / 240, L / 360,... n

Shear: Concrete: f 'c b, d, f 'c Stirrups: f y

bf

AW

A=bxd

.- . . - .

CONCRETE BEAMS:

F V = VMAX

Bending: F b = MMAX SMIN

SMIN

0.08WL

0.025WL P.O.I.

AWEB

Bending: Fb = MMAX

0.5W

- 0.1WL

STEEL BEAMS:

FV = 3 VMAX 2 A MIN

0.4W

0.6W

0.5W

PL/2

PL3

L 1.1W

VMAX = 3P/2

PL/3

Shear:

1.1W

0.4W

VMAX = P

23

L

b

d

L/3

L/4

h

L/3

P

P

Beams

MMAX = WL/8 = wL2/8

PL3 EI

DMAX = 1 48

W=wL P

Columns

VMAX = W/2

WL/8

d

General Beam Design

VMAX = W/2

AV AS

f y, (f, # rebars), A s FC = P/A Long and thin ( slender ) columns tend to be governed by buckling n Short and fat ( chunky ) columns tend to be governed by crushing

WOOD COLUMNS:

STEEL COLUMNS:

n

Slenderness:

Slenderness: kLUNB.

n

slenderness ratio

LUNB./ dLeast

r

kwood= 0.671 E Fc

June 2004

wood 11 steel

k

k=0.5

k=1

k=2

50 L/d 200 KL/r

© 2004 David J. Thaddeus, AIA

PAGE : 4 OF 4