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.
inches. The end of girder should be deep enough to resist vertical shearing. In our case it is trifling, and we need not consider it. In all of these examples we have omitted the weight of the girder, to avoid complication. This should really be taken into account, in such a long girder, and treated as an additional hut uniform load.
"When girders run over three or more supports in one piece, that is, are not cut apart or jointed over the supports, the existing strains and reactions of ordinary girders, are very much altered. These are known as "continuous girders." If we have (Fig. 143) three supports, and run a continuous girder over them in one piece and load the girder on each side it will act as
shown in Fig. 143; if the girder is cut it will act as shown in Fig. 144. Very little thought will show that the fibres at A not being able to separate in the first case, though they want to, must cause considerable tension in the upper fibres at A. This tension, of course, takes up or counterbalances part of the compression existing there, and the result is that the first or continuous girder (Fig. 143) is considerably stronger, that is, it is less strained and considerably stiffer, than the sectional or jointed girder (Fig. 144). Again we can readily see that the great tension and conflict of the opposite stresses at A would tend to cause more pressure on the central post in Fig. 143, than on the central post in Fig. 144, and this, in fact, is the case.
Amount of Reactions.
Amount of Greatest Bending Moments.
Amount of Greatest Deflections.
Two equal spans each carrying a central load but loads not equal.
w1<>w11 l = l1
Left reaction. p =(13.w1-3.w11)/32 Centre reaction.
r = 11/16.(w1+w11)
Right reaction. 13 . to., - 3 . to, q=(13.w11-3.w1)32
Located at r m = 3/32 .l.(w1+w11)
Deflection in left span l δ = (23.w1-9.w11.)/1536.e.i. l3
Deflection in right span l1 δ11 = (23.w11 -9.w1)/1536.e.i. l3
Two equal spans each carrying a central load = w, loads equal.
w =w1 l=l1
p =q = 5/16.w or
r = 11/8.w or
Located at r m = 3/16 .l.w or
= 3/32.l. (w+w1)
Deflection in either span
δ= (w.l3)/110.e.i or
= l3/220.e.i. (w+w1)
Two equal spans each loaded with a
p = q =3/8 .u or
= 3/16 .(u+u1)
Located at r m= (u.l)/8 or
Deflection in either span δ = (u.l3)/185.e.i or
Where δ, δ1δ1= the amount of deflection in inches, if girder of uniform cross-section throughout. " e=:the modulus of elasticity of the material, in pounds-inch, (see Table IV). " i= the moment of inertia of the cross-section, in inches, (see Table I).
uniform load = u u = u1 l = l1
r = 5/4. u or
= 5/8. (u +u1)
= 13/370.e.i. (u+u1)
Three equal spans each carrying a central load = w, all loads equal w = w11 = w11 l=l1 = l11
p =q = 7/20. w or
= 7/60. (w+w1+w11)
r = s = 23/20. w or
= 23/60. (w + w1 + w11)
Located at r or s m=(3.w.l)/20 or
= 1/20. (w + w1 + w11)
Deflection in central span
δ= (w.l3)/480.e.i or
= l3/1440.e.i (w+w1+w11)
Deflection in cither end span
δ = (w.l3)/87.e.i or
= l3/261.e.i. (w +w1 + w11)
Three equal spans each loaded with a uniform load = u u = u1 = u11 l = l1=l11
2 p = q = 2/5 . u . or
= 2/15. (u+u1+u11)
Central reactions. r = s = 11/10. u or
= 11/30. (u+u1+u11)
Located at r or s
. m = (u.l)/10 or
= 1/30. (u+ u+ u11)
Deflection in ventral span
δ = (u.l3)/1920.e.i or
= l3/5760.e.i. (u+u1+u11)
Deflection in either end span
δ = (u.l3)/ 145. or
= l3/435.e.i. ( u + u + u11)
Where to, w,w1 w11 = central concentrated loads in pounds, on either span, being equal, when so stated. " 1, l1,l11, = the length of respective spans, in inches, all being equal. " u, ulu11= uniform loads on each span, in pounds, all being equal. " p, r, s,q, = the amount of respective reactions, in pounds. " m = the bending moment, in pounds-inch.