This section is from the book "Safe Building", by Louis De Coppet Berg. Also available from Amazon: Code Check: An Illustrated Guide to Building a Safe House.
It being understood that at the point of greatest bending moment - or where the greatest tension strain exists - that the part is designed to resistor exert a stress = (t/f) per square inch. If this greatest tensional stress is less than (t/f ) insert its value in its place in Formula (85). Of course, it must never be greater than (t/f).
For a beam, girder or truss with a load concentrated at the centre, but with flanges or chords of uniform cross-section throughout, the average strain would be just one-half that at the centre; for, the bending-moment graphical-figure will be a triangle, and inserting the values in Formula (83) would give for the compression member: v = 1/2. (c/f) (86) and for the tension member: v= 1/2. (t/f) (87)
The meaning of letters being the same as in Formulae (84) and (85), but the total safe load being concentrated at the centre instead of uniformly distributed.
Where w = the average strain, in pounds per square inch, in either chord or flange.
Where l = the length of span, in inches.
Where e = the modulus of elasticity of the material, in pounds-inch.
Where x = the total amount of extension or contraction, in inches, of the chord or flange.
Figure 145 shows the same, after the deflection has taken place. We can now assume approximately, that C A is equal to one-half the difference between the contraction of G C and the elongation of HB, or, what amounts to the same thing, that is equal to one-half the sum of the contraction of the one and the elongation of the other.
Further, we can assume that approximately, A B = d or the depth of beam, and C D = l/2 or one half the span.
The curve C E C will approximate a parabola, so that if we draw a tangent C F to the same at C, we know that D E = E F=DF/2 or DF= 2. D E. But as DE represents the deflection ( δ ) of the beam, we have
Now as C F is normal to C B, and C D normal to A B, we know that angles D C F = A B C; further, as both triangles are right angle triangles, we know that they are similar, therefore:
D F: CA...D C: A B, or
2. δ: C A ...1/2: d or
δ = CA. 1/2/2.d = CA.l/4.d
If now we assume the sum of the extension and contraction of the two flanges or chords to be = x.
We have C A= x/2 or
δ = x.l/8.d (89)
Where δ = the deflection, in inches, of a beam, plate girder or truss, with parallel flanges or chords.
Where x = the sum of the amount of extension in tension chord, plus the amount of contraction in compression chord.
Where / = the length of span, in inches.
Where d = the total depth of beam, girder or truss in inches.
Take the case of a wrought-iron plate girder or beam of uniform cross-section throughout carrying its full uniform load, we should have the strain at the centre on the extreme fibres =12000 pounds per square inch. Now the average strain on both upper and lower flanges would be, Formulae (84) and (85). v = 2/3. 12000 = 8000 pounds per square inch. Therefore amount of contraction in upper flange
Formula (88), (and remembering that, from Table IV.e = 27000000) x = 8000.l/27000000 = l/3375
The elongation of the bottom flange would be an equal amount, therefore the sum of the two x1 = 2. x= 2.l/3375 .
= l/ 1687,5
Inserting those values in Formula (89) we have the deflection
δ = l2/8.1687,5. d = l2/ 13500. d and inserting for l2=144. L2, we have δ = 144.L2 13500. d. = L2/93 3/4. d. Had we assumed that the area of flanges or chords diminished to-wards the supports in proportion to the bending moment or actual stresses required, the average strain would, of course, be 12000 pounds per square inch throughout the entire length, no matter how the load might be applied.
Inserting this value in Formula (88) we should have had, for the amount of contraction of top flange x = 12000.l/27000000 = l/2250
The same for the extension of bottom chord, or x = 2. l/2250 = l/1125
Inserting this in Formula (89) we have for the deflection:
δ = l2/8.1125. d = l2/ 9000. d Inserting 144 L2 = l'2 we have δ = 144. L2/9000.d or
δ = L2/62 1/2.d (90)
Where δ = the greatest deflection, in inches, of a wrought-iron plate girder, or wrought-iron truss, with parallel flanges or chords, and where the areas of flanges or chords are gradually diminished towards supports, and no matter how the load is applied; in no part however must the stresses, per square inch exceed respectively either
(c/f) or (t/f)
Where L = the length of span, in feet.
Where d = the total depth (height) in inches, from top of top flange or chord to bottom of bottom flange or chord.
If girder or truss is of steel, use 53 2/3 instead of 62 1/2
From Formula. (90) and Formula (28) we get the rule that (no matter how the load is applied) if we want to carry the full safe load and not have deflection enough to crack plastering the length in feet must not exceed 1 7/8 times the total depth in inches. For:
L. 0,03= L2/62 1/2.d or
L = 62 1/2.0,03.d
= 1,875.d or say
L = 1 7/8.d
Safe Length, Diminished Cross-section, any Load, Parallel Flanges or Chords.
Where L = the length, in feet, of a wrought-iron plate girder or wrought-iron truss, with parallel flanges or chords and with area of flanges or chords diminishing gradually towards supports and no matter how the load is applied; in no part however must the stresses, per square inch, exceed respectively either (c/f) or (t/f).