# Structures Formula “Cheat Sheet”

GENERAL STRUCTURES

Use the time during the exam tutorial to go over the References they give you. You will not need to remember any of the formulae for Max. Moment, Shear and Deflection that re in the steel manual, they will be given to you on the test. During the tutorial you can write down the basic ones so you have them next to you as a reference, and you don’t have to go back and forth to the reference screen. Also during that time I wrote down the formulae that I did memorize just for reference later. That way you are not double guessing yourself on the units when you are under time pressure. I also marked all the math questions and did them at the end.

Here is some advice from David Thadeus that I found very helpful:

“As far as memorizing the formulas, it is by far more important to know what is involved in a formula, rather than memorize it without fully comprehending how and when it applies. I am trying to emphasize that there is a fair amount of “qualitative” questions about the formulas themselves.

As for the radius of gyration, it is important to recognize that it is strictly a property of the geometry of the section (r = (SQRT (I÷A))) and is critical in the assessment of the slenderness of steel columns. The lesser the radius of gyration, the more likely a steel column is to buckle (the other factors that affect the slenderness of a steel column are its un-braced length and how strongly its ends are connected.”

Make sure you know the concepts behind the formula, which will in turn help you to remember the formula itself better. For example in the deflection formula the span of a beam (L) is to the 3rd power (cubed) where the load (W) is to the 1st power; therefore increasing the span of a beam will impact its deflection much more than if you increase the load. Similarly values in the denominator they will impact the result differently than values is in the numerator.

As far as which formulae to remember, I found that I only needed to remember 3 or 4 and knowing a bit of algebra you can solve for any variable:

# Modulus of Elasticity formula:

E = Stress (F) ÷ Strain or E = (P ÷ A) ÷ (Deformation ÷ L)

and all the variations of that formula

i.e.: Deformation = (P ×L) ÷ (A × E)

# Section Modulus formula:

S = M ÷ Fb and know how to solve for each variable

ie: M = Fb × S & Fb = M ÷ S

Also know that S = I ÷ c and therefore Fb = Mc ÷ I

They will give you (S = b × d3 ÷ 6)

# Formula for Shear

Fv = const. × V ÷ A and same as above know how to solve for each variable

# Coefficient of thermal expansion (n) is the ratio of unit strain to temperature change and is constant for a given material.

n = D / L \ D = n L t

t

D = PL P = DE

AE A L

# They will give you most of the geometry formulae for the area, moment of Inertia (I) and Section Modulus (S) for rectangular shapes and other basic shapes, but not the area of a circle – so remember the formula for the area of a circle and it’s circumference.

# Although they will give the Deflection formulae, know that Delta L = (Const. × Load × Span3) ÷ (E × I)

# Know how to solve for angles and legs of a triangle, all I needed on my test was the basic:

a2 + b2 = c2

Sin = opposite ÷ hypotenuse, Cosine = adjacent ÷ hypotenuse and Tangent = rise ÷ run

# Know how to figure the loads on tributary areas.

# Know how to figure out reactions.

# For the 2 moving load problem, the formula is in the steel manual, but all you need to know is how to figure the reactions at the worst case scenario, for shear it would be with the loads closest to the supports and for bending it would be with the loads at the center, then treat them as regular point loads.

Math problems are really only about 15% to 20% of the test, the rest of it will be questions on general knowledge of how structures work, what is the best system for a given situation, and some on cost in general. Look at the reference list NCARB suggests for this division and go to the library they have a few of them.

[SOURCE:http://www.areforum.org/up/GeneralStructures/]

Formulas Downloads

[wpfilebase tag=list id=41]

Change of Length due to Temperature

ΔL = α(ΔT)Lo

α  = Coefficient of thermal expansion

ΔT  = Change in temperature (oF)

Lo  = Original length (inches)

ΔL  = Change in length due to temperature

Deflection

I = bd3/12

Modulus of Elasticity (E)

E = rise/run = stress/strain  =  P/A         or,     ΔL =  P Lo

ΔL/Lo                          A E

where,

ΔL = change in axial length (inches)

P = axial load (pounds)

Lo = original length (inches)

A = cross-sectional area (in2)

E = modulus of elasticity (psi, ksi)

Moment of Inertia

Ixx = Sum (A)(y2)

In which:

Ixx = the moment of inertia around the x axis

A = the area of the plane of the object

y = the distance between the centroid of the object and the x axis

Hint: Moment of Inertia is an important value which is used to determine the state of stress in a section, to calculate the resistance to buckling, and to determine the amount of deflection in a beam. For example, if a designer is given a certain set of constraints on a structural problem (i.e. loads, spans and end conditions) a “required” value of the moment of inertia can be determined.  Then, any structural element which has at least that specific moment of inertia will be able to be utilized in the design.

Slenderness

= k*L/r

Stress (F)

= P/A  where P = pressure, or load (psi, ksi) and

A = area of load (in2)

Trigonometry (90° triangles only)

Sin = Opposite/Hypotenuse

Cosine = Adjacent/Hypotenuse

Tangent = Opposite/Adjacent

Unit Strain (Elongiation Stretch)

= ΔL/Lo

where ΔL is the change in length and Lo is the pre-load length, in inches

Wind

qz = 0.00256*Kz*Kzt*Kd*(V2)*I

p = (q*GCp) – (q*GCpi)

q= qz for load to the windward wall

q= qh for load to the leeward wall

G= gust effect factor

Cp= External pressure coefficient

(GCpi)=Internal pressure coefficient; can be positive or negative. Positive acts towards

the surface and negative acts away from surface.

Variables and Consistences

Columns

k is the effective length factor. Taken as 1.0 for columns that do not depend on their own bending stiffness for stability. k is greater than one when the column is dependent on its own bending stiffness for

stability. k values are dependent on the end restraints of the columns.

L is the actual length of the column

kL is the effective length of the column

r is the radius of gyration of the column. A column has an rx and an ry. For square members the value is the same. For wide flange, channels, built up and similar manners rx is greater than ry.

r = sqrt(I/A) for rx use Ix in the formula and for ry use Iy.

For a rectangular member sqrt (I/A) reduces to sqrt (d/12).

D = dx or dy

Wind

V = Basic wind speed, 3-second gust = 90 mph for almost all of the continental United States (xceptions are the east and west coasts)

h = mean roof height

Kh = velocity pressure coefficient at mean roof height. Adjusts the pressure based on exposure and height

Kz = velocity pressure coefficient at height z. Adjusts the pressure based on exposure and height

Kzt = Topographic factor

p = design pressure – basically q * modification factors for windward. Leeward or sidewalls

qz = Velocity pressure evaluated at height z. This is the basic pressure that is modified in all the other formulas to determine design pressures. Also called the stagnation pressure.

qh = velocity pressure at the mean roof height

MWFRS = Main Wind Force Resisting System

z = height that is being evaluated

FORMULAS TO REMEMBER

Area:[units = inch2]

Area of a Rectangle = b d

Area of a Circle = r2

Area of a Triangle = 1b h

2

Area of a Bolt, = d2 [where d = diameter]

Cable, Tube, Bar 4

Equilibrium:

M = 0; V = 0; H = 0

Force: {units = kips & pounds}

F = M[Force = Moment ]

d distance

RETAINING WALL DESIGN{units = kips & pounds}

F = w h2[Force exerted on the = (fluid pressure provided) X (height)2]

2 Retaining Wall2

Remember: pcf = psf [pounds per cubic feet = pounds per square foot]

ft one foot width of wall

SHEAR DIAGRAM SHEAR FORCE{units = kips & pounds}

R = V = w l[Shear Resisting Force= (uniform load per ft) X (distance)]

22

BEARING TYPE SHEAR CONNECTIONS {units = kips & pounds}

R = Fv Abolts

[Resistance = (allow.shear stress) X (A of bolt cross sections. Remember

###### to Shear Failure to multiply A by total # of bolts)]

Remember: Stress = P Therefore, P = Stress X Area

##### A

Moment: {units = (k ft); (lb ft); (k in); (lb in)}

TAKING MOMENTS ABOUT A POINT TO FIND EQUILIBRIUM

M = Fd[Moment = force X distance]

### UNIFORM LOAD{units = (k ft); (lb ft); (lb in)}

M = w L2[Moment = uniform load X (length)2]

88

### POINT/CONCENTRATED LOAD AT THE CENTER OF A MEMBER

M = P L[Moment = Point Load X length]

44

Remember: w l 2 + P L when Point & Uniform loads combine.

1. 4

Watch out: There are various types of Point loads.

ECCENTRIC LOAD{units = (k ft); (lb ft); (lb in)}

M = Pe [Moment = force X eccentricity] {Same as M=Fd}

Section Modulus: [units = inch3]

S = b d 2in3[Section Modulus]

6

S = M[Section Modulus = Moment in Inches

Fb Bending Stress]

Watch out: Both Moment & Stress should be in # or kips.

Remember: For a Roof Beam,S = M

Fb X 1.25

S = I [Section Modulus = Moment of Inertia

c(Just know this) (dist. from extreme fiber to nuetral axis)]

Understand that S contains Moment of Inertia and c.

Moment of Inertia: [units = inch4]

Remember: Moment of Inertia occurs by default about the Centroidal axis.

I = b d 3in4 [Moment of Inertia]

12

I = b d 3in4 [Moment of Inertia of a rectangle about its base]

3

Ibase= I + A y2 in4 [I @ Base = I + Area X (dist. from centroid to base)2]

Center of Area:

Use the formula M = A d derived from M=Fd to find X and Y

A = Sum of Areas of ALL members

Stress:{units =ksi or psi}

### BENDING / FLEXURAL STRESS {units =ksi or psi}

Remember: Max. Bending stress occurs at the extreme fibers.

# fb = M c[Bending Stress = Moment X (dist. from extreme fiber to N/A)]

I Moment of Inertia

Remember: Greater the c, greater is the Bending Stress.

### AXIAL STRESS {units =ksi or psi}

Remember: Max. Axial stress occurs along the entire cross-section.

# A Compression Stress Area in in2

### SHEAR STRESS {units =ksi or psi}

Remember: Max. Shear stress occurs at the Nuetral Axis

Remember: Shear Stress is the same at both Vertical & Horizontal axis.

fv = 1.5 V[Actual Shear Stress = 1.5 X Shear Force]

A Area

Just understand the fol. 2 formulas. No need to memorize:

fv = V Q

I b (Statical moment about the

[Shear Stress = (Shear force) X nuetral axis of the area above the plane)]

#### (Moment of Inertia) X (width of beam)

Q = (section Area) X (dist. from centroid of rect. to the centroid of section above neutral axis)

fv = 1.5 V = 3 V

A 2 b d

Notching on Tension side of a Wood Beam

fv = 1.5 V X d[d = overall d of beam]

b d’ d’ [d’ = d of the beam that is notched]

Use Actual dimensions of the b and d, NOT Nominal dimensions

Short heavily loaded Beams & Beams with large loads at supports

fv = V[Actual Shear Stress = Shear Force

dt (depth of beam) X (thickness of beam)]

Deflection: {units =inches}

= P L

# A E

[Shortening / Elongation = Force X Length

A of cross-section of member X Modulus of Elasticity]

#### A

for change in length, Multiply Stress by Length

##### Modulus of Elasticity

DEFLECTION OF A BEAM

= 5 w L4

# 384 E I

[Deflection = 5 X (w in pounds) X (Length in feet X12”)4

(384) X 12” X Modulus of Elasticity X Moment of Inertia]

#### ft 12”

Length in inches = Length in feet X 12”

Strain:

= [strain = Deflection

LOriginal Length]

Modulus of Elasticity:

E = f[Modulus of Elasticity = Stress

Strain]

Thermal effects on structures: {units = inches}

SHORTENING OR ELONGATION DUE TO CHANGE IN TEMPERATURE

= e L t

[Thermal Elongation = (Coeff.of thermal linear expansion) X (Orig.Length) X (Temp Change)]

THERMAL STRENGTH IN A RESTRAINED MEMBER

ft = E e t

[Thermal Stress = E X Coeff. of linear expansion X Change in Temp]

in a Restrained member

Slenderness Ratio (Loading Capacity):{units =inches}

STEEL COLUMN

kl is the effective length in feet.

SR = k l[Slenderness Ratio = (end cond.) X Unbraced length in inches]

r Radius of gyration

Remember: Slenderness Ratio should be 200 for a steel column.

WOOD COLUMN

SR = k l[Slenderness Ratio = (k =1) X (Unbraced Length in inches)

b (cross-section width of rectangle)]

Remember: Slenderness Ratio should be 50 for a wood column.

r = I[Radius of Gyration = Moment of Inertia]

A Area

### Retaining Wall

F = w h2[Force exerted on = (fluid pressure at top of soil) X (height)2]

2 the Retaining Wall 2

RM = 1.5 MOT

[DL Resisting Moment = 1.5 (Overturning Moment of the Retaining Wall)]

Factor of Safety (FS) for the Resisting Moment requires it .

FS = RM[Factor of Safety = Resisting Moment

MOT Overturning Moment]

Remember: FS 1.5

SLIDING OF RETAINING WALL

FS against Sliding = Sliding Resistance (#)

Force causing tendency to Sliding (#)

# Sliding Resistance = (Total Vert. Load in # on Ftg) X (Coeff. of Friction)

Force causing Sliding = (Earth pressure in # @ Base of Ftg) X (h in ft)

(2) ft

M= (F) h[Bending Moment = Force X (ht at resultant force)

3

Remember: Bending Moment occurs at 1/3rd the height of the retaining wall, where resultant force occurs.

Weld:{units =inches}

# Throat of Weld = Weld Size X (.707) [.707 = 2 ]

2

Capacity of Weld = (Allow. Stress)(Throat)(Weld Size)(Total Weld Length)

Allow. Stress = 18 ksi for E60 electrode weld for ASTM A-36 base plate.

21 ksi for E70 electrode weld for ASTM A-36 base plate.

ft= P[Stress in the = (Compressive/Tensile Force of the Weld)

Athroat of the weld (.707) X (Weld Size) X (Total Weld Length)]

Ultimate Strength Design for Concrete:

# U = 1.4DL + 1.7LL [Ultimate Load = 1.4(Dead Load) + 1.7(Live Load)]

MU = 1.4 MDL + 1.7 MLL

[Ultimate Moment = 1.4(Dead Load Moment) + 1.7(Live Load Moment)]

MU = As fy(d – a )

2

##### Remember: As is available in a table, “ASTM STD REINFORCING BARS”

[Moment = (strength reduction factor = 0.9) (cross-sectional area of tensile reinforcemnt) (specific yield strength of reinforcemt) {(dist. from extreme compression fiber to centroid of tensile reinforcement) – (depth of rectangular stress block) / 2}]

= AS[Percentage of steel to = (area of tensile reinforcemnt)]

bd achieve a Balanced Design (beam width) X (d)

min = 200 min should be 3 f ’c

fyfy

Live Load Reductions:

# [Live Load Reduction = (rate of reduction) X {(Tributary Area) – 150}

Remember: rate of reduction = 0.08 for Floors

See table 16-C Roofs

# Rmax = 40% for single level floors

Rmax = 60% for multi-level floors

# R = 23.1 (1 + DL)

LL

Remember: Do all 3 checks and then select the lowest value as your final live load reduction.

Thrust in a 3 hinged Arch:

Thrust = w L2[Thrust in a 3 hinged arch = uniform load X (length)2]

8 h8 X height

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