Timber Frame Tension Joinery - Timber Frame Engineering Council
Timber Frame Tension Joinery - Timber Frame Engineering Council
Timber Frame Tension Joinery - Timber Frame Engineering Council
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Plastic hinges occur in the peg due to bending and are located at points of maximum<br />
moment (see Figure 3-5). This derivation is as follows:<br />
From equilibrium:<br />
x<br />
A<br />
P<br />
= (3-23)<br />
DF<br />
em<br />
For failure to occur, two plastic hinges must occur in the peg. Areas under the shear<br />
diagram give<br />
DF ( a − b) x DF ( d − c)<br />
x<br />
+<br />
2 2<br />
em A es D<br />
= 2 M<br />
(3-24)<br />
p<br />
DFem( a − b)<br />
= P<br />
(3-25)<br />
DF ( d − es<br />
c ) = P<br />
(3-26)<br />
Simplifying Equation 3-24, we obtain<br />
Px<br />
A<br />
+ PxD = 4 M<br />
p<br />
(3-27)<br />
x<br />
D<br />
4M<br />
p<br />
= −x<br />
P<br />
A<br />
(3-28)<br />
Total distance between hinges is<br />
x = x + x<br />
(3-29)<br />
A<br />
D<br />
4M<br />
p<br />
4FybD<br />
x = =<br />
P 6P<br />
3<br />
(3-30)<br />
Substituting Eq. 3-22 into Eq. 3-30 we obtain<br />
x = D<br />
⎛ F<br />
2Fyb⎜1+<br />
⎝ F<br />
3F<br />
em<br />
em<br />
es<br />
⎞<br />
⎟<br />
⎠<br />
(3-31)<br />
As a numerical example, typical values for 1” Red Oak Pegs in Douglas Fir give a<br />
total distance between plastic hinges of 3.0 inches (using F yb = 12,601 psi, F em = 2070 psi,<br />
17