This Is That 18 Page Illegible Beam Handout You Got

Only the text follows. All omissions (of images, long strings of formulas, etc.) are denotedby a horizontal line.

Architecture, A Five Week Study 3/8/74

Form Function

P.Harding (with some editing by R.Knapp)

"StructureMechanics" – How to calculate the size of a beam.

The purpose of these notes is to give you a very basic introduction and exposure to "structures". Calculation of beam sizes is but one part of a larger design problem within the larger problem of designing the structural system which is, in turn, only one of the considerations in the design of buildings.

These notes will cover the following "steps":

• 1. Calculate the magnitude of the reactions, in pounds, by "taking "moments" about a point along the length of the beam.
• 2. Calculate the vertical shear at some regular internals along the length of the beam
• 3. Draw (accurate freehand) the shear diagram to:
  • a) find the maximum vertical shear
  • b) to locate the point along the beam where the maximum bending moment occurs.
• 4. Calculate the maximum bending moment.
• 5. Draw (accurate freehand) the moment diagram by calculating moments at some regular interval along the beams length.
• 6. [If it is a cantilever, calculate the location for the inflection point(s), of contra flexure. If not, go ahead one step and pay $200.00]
• 7. Select the kind and strength (quality) of wood to be used for the beam. This will give you the values of 'f', the resistance capability of the material to ending. It will be in pounds per sq. inch.
• 8. Calculate the required "section/modulus" (S_x-x) using the flexure formula:
• S=m/f
• 9. Using table #7 (Parker), select a beam size (e.g., a 2x10, etc.)
• 10. Calculate maximum horizontal shear (H) stress in your beam situation and check it against the allowable stress (H) in the beam you have selected.
• 11. [If this were 'real', deflection would be checked and it would likely determine the beam size finally. Either 1/240^th or 1/360^th deflection is usually permitted.

p3

Before we can begin step #1, you ought to know something about where the beam situation is in the building and what produced the loads.

Here is a "floorplan" (actually a "structural plan"):

Here is an "elevation" of the beam:

p4

Each joint exerts a concentrated load (force) on this beam. Thee loads have to be computed, but we will assume them for this exercise – also, each joint exerts the concentrated load of the post out on the end of the beam.

Here is a diagram showing location and magnitude of the loads and the location of the reactions. We will have to calculate the magnitude of the reactions.

*The Joist loads (dotted) at the columns are assumed to be carried by the columns and are not, therefore, producing or resisting bending int he beam. They do have to be considered in the design of the buildings's minimum columns and footings!

Step #1 Calculation of Reactions

We must calculate how much of the total weight on the beam is carried by each column or "reaction." This is a cantilevered beam –R_Right by inspection, will be holding up more than R_Left

What we must do is to "take moments about" a point along the beam, write an equation, and solve for one of the reactions ... what is "taking moments," you ask?? "What is a moment?" [If you ask what is an equation go to jail and pay $200.00]

p5

A "moment" is the tendency of a force (in pounds) to cause rotation about a point!

A moment is described in foot pounds, or :

The distance away from the point about which the tendency to rotate is to be measured times the force.

or: (force)(moment arm) = moment

or: (500)(10) = 5000 ft_lbs

Problem One: Calculate the moment at points 'a' and 'b'

p6

Another illustration:

Suppose we had to remove a bolt which was rusted tight – and we only had a 10" crescent wrench.

We try to loosen it – we are ale to "push" on the wrench with 150# pressure (we are hanging on it!) we are exerting a moment (a tendency to rotate the bolt!) of (150)(10"), or 1,500"#...but it doesn't budge...(isn't this exciting and dramatic!)

We have essentially three options:

1) Forget it*

*We don't hammer on crescent wrenches

2) Cut it off with a torch, or

3) Increas the moment arm and thereby the moment!!

Naturally we opt for #3:

Now! (150#)(30") = 4500"# and yes! it comes off!

p7

Now the same kind of thing is working in a beam, except, it isn't (we hope) rotating!

The tendency to rotate about any point along the length of the beam is called thebending moment. This tendency is resisted by the fibers in the wood. If there are great loads involved there will be great tendencies to rotate (bend) and we will have to have lots of fibers (i.e., a bigger beam!)

We are striving for equilibrium – that is why this business is called "statics"...we want the beam not to bend, rather to be static.

If the beam is static – in equilibrium – the summation (algebraic)* of momenta at any point along the beam = 0. *We say 'algebraic' because of the plus and minus signs involved – some forces tend to cause clockwise rotation (+), some counter-clockwise(-)

SO, to "take moments" about a point is to algebraically sum the + and - forces (loads and reactions) times their moment arms (distances from the point.)

We will take moments about the point thru which R_L passes, thus giving it a moment arm of zero and reducing the equation to only one unknown (R_R)

p8

Summation of moments about R_L =

(1) (200)(4) + (200)(8) - (R_R)(12) + (200)(16) + (444)(18) = 0

(2) 800 + 1600 - 12R_R + 3200 + 7992 =0

(3) 2400 + 11,192 =12R_R

(4) 13,592 = 12R_R

(5) R_R = 1,133#

To find R_L we simply add the total load and subtract. R_R ---->because the sum of all the vertical (+&-) forces must =0 for equilibrium. (Sum v=0)

200+200+200+444=1,044#

1,044-1,133=[-089#] =R_L

What have we here? What does it mean if R_R is greater than the Sum of the loads? What does it mean if R_L comes up with a negative value? I'll tell you...it means _RL is tending to produce counterclockwise rotation...that the beam is tipping over andR_L is not holding it up. It is holding it down!!

p9

Now we are "in balance", in equilibrium,...static! Can you see how it goes?

As careful designers, we always check our work, class* *In writing this illustration, I checked and found an error! and had to re-write page 8 and 9.

Lets take moments about a point through which R_R passes and solve for R_L

(1) (R_L)(12) + (200)(8) + (200)(4) - (200)(4) -(444)(6) =0

(2) 12 R_L + 1600 - 2664 =0

(3) 12 R_L = 1064

(4) R_L = 088# = close enough

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STEP #2 Vertical Shear

Vertical shear is the tendency for the beam to be "sliced-off" vertically by the forces operative.

Imagine a beam, say a 2"x4", made of balsa wood. It spans 3'0" and holds up a concentrated load right next to one of its supports of 500#. What is going to happen? How will it fail?

With such a loading situation, the beam will "slice-off" before it can fail in bending; it will fail in vertical shear.

As you move the 500# load outward toward the center of the span the vertical shearing tendency decreases and the bending stresses increase!!

Now, as the vertical shear approaches zero, the bending moment increases. Where the vertical shear = zero (in a simple beam, such as this) the bending moment = maximum!!* *Conversely, where the vertical shear = maximum, the bending moment = zero!!

But we have a cantilevered beam. Where the vertical shear changes from + to = is where the maximum bending moments occur!

p11

What we have to do is draw the shear diagram in order to locate the maximum bending moment. First, we must calculate the vertical shear values (V) at intervals along the beam.

Rule: " V at any point along the length of a beam = the algebraic sum of the reactions minus the loads to the left of the section"

Believe it...

p12

STEP #3 Shear Diagram

Now we have the figures to draw the shear diagram

The question you now have is how to connect up these points – are they just connected to produce a sloped line or are they a series of horizontal lines??

Let's findout by computing V_@6:

V₆ = -89# -200# = [-289#].. the same as V₄

Therefore, they are constant values between the loads and reactions, or horizontal lines* *Distributed loads produce sloped shear diagrams. Concentrated produce sloped horizontal blocks in shear diagrams.

Now, remember the remarks on page 10? Where the vertical shear changes from + to - = the location of the maximum ending moment!

Mmax =@12

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STEP #4 Bending Moment

We all now know what a bending moment is and now we know where bending is maximum.

The Mmax occurs at x=12'

(1) Mmax = (89)(12) + (200)(8) = (200)(4)

(2) M12 = 1078 + 1600 + 800

(3) M12 = 3,478 (foot pounds)

This is how much "twisting" is going on. Remember this tendency to rotate is resisted by the internal fibers of the wood beam. This is difficult for some people to visualize.

p14

Lets look, for a moment, at thee "internal stresses" which are produced by the external forces acting on the beam.

First: bending = a combination of tension (stretching) and compression (crushing).

From this definition, it follows that some of the internal fibers of the wood are in tension and some are in internal "resisting moment."

By inspection, we can see that the top part of the beam is "stretching" (in tension) and the bottom is "crushing" (in compression.) ?? We can see that , can't we?

Here are the internal resisting moments!! (Force)(Arm) = Moment!

By observation, we can see that by making the beam deeper (bigger), we can increase the length of the mometn arm and thereby increase the magnitude of the internal moments and so, increase the resistance capability of the beam!!...We make it stronger by making it deeper.

p15

Back to our purposes...

If we are to have a beam of uniform depth, we must design it for the maximum bending moment.

We have calculated this. It = 3,478 foot pounds. And we all understand what this means now...

Step #5 The Moment Diagram

The moment diagram is important, especially in the design of concrete beams. It is worth a few pages here. It is done by calculating the moments at intervals along the length of the beam – much like the shear diagram.

M0 =0

M2 = (89)(2) =178

M4 = (89)(4) = 356

M6 = (89)(6) + (200)(2) = 534 + 400 = 934

M8 = (89)(8) + (200)(4) = 712 + 800 = 1,512

M10 = (89)(10) + (200)(6) + (200)(2) = 890 + 1200 + 400 = 2,490

See how they are getting bigger as we approach M12! Isn't this exciting!!

M12 = (89)(12) + (200)(8) + (200)(4) = 1068 + 1600 + 800 = 3,468

I lost 10# somewhere...?

M14 = (89)(14) + (200)10) + (200)(6) - (1133)(2) = 1246 + 2000 + 1200 -2266 = 2180

M16 = (89)(16) + (200)(12) + (200)(8) - (1133)(4) = 0

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There it is! Remember, distributed loads produce curved diagrams, concentrated loads produce straight line diagrams! If you draw and plot accurately, you won't have to remember this.

STEP #6 Inflection Point

We have a cantilever, but no inflection point. Go ahead one step and recieve $200.00.

STEP #7 Wood Selection

For convenience (i.e., the stuff is everywhere, and so are tables of its properties) I (Knapp) select "Dense No. 2 Douglas Fir / Larch"

This wood has an allowable extreme fiber stress in bending (this mouthful is abbreviated f) of 1700 pounds per squaqre inch (1700 psi)

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STEP #8 Calc. of Req'd Section Modulus (Sx-x)

Now we know what external forces must be resisted (M = 3,478 foot pounds)* and we know our capacity to resist (f=1,700/11squared) *Since f is in pounds per square inch, we must not forget to convert our moment from foot pounds to inch pounds – multiply by 12.

The last remaining question: How big must the beam be?

We use "the flexure formula" S = M/f

S = section modulus = "a property of the section of a beam" Live with that for now... If you wish to calculate it, see me...

S = (3,478)(12) over (1700/11squared) = 41, 736 over (1700/11squared) = 24.5"cubed

STEP #9 Beam Selection

From the attached table, "Properties of structural lumber," llok down the column headed S. Find a value somewhat higher than 24.5"cubed.

Study this table very carefully. Notice how the value increases with depth!!

3x8 ---> 21.901 <--too small

2x12 -->31.641

4x8 --->30.661 <--Phil goes for this one

6x6 --->27.729 <--cutting it a bit close

p18

I would go for the 4"x8" beam. It would probably make it through a check on deflection; the 3x8 would not. Also, torsion, or twisting of the beam is a factor and would normally be calculated. Fatter beams are better in this respect.

So, now we have a beam!