Design of Timber Joists
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description
timber design
Transcript of Design of Timber Joists
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Design of timber joists
Task:
Determine the size of timber joists for the first floor of a typical dwelling house. The
joists are spaced at 400 centres.
Assume the design load to be 2.25 kN/m2.
Modulus of elasticity of timber = 10 700 N/mm2
Strength of grade C24 timber = 7.5 N/mm2(in bending)
Solution:
The sectional and the plan views are shown in Figure App8.1. The spacing between the
joists is 400 mm or 0.4 m. Assume the size of the joists to be 50 200 mm for
calculating the dead load of the joists. Every 1 m length of a typical joist supports the
loading from an area of 0.4 m 1.0 m, shown as the hatched area in Figure App8.1b.
Figure App8.1 Timber floor
50 x 200 mm
Timber joist
plasterboard
12 mm thick
a)
18 mm thick floor boards
0.4 m x 1.0 m
section
b)
Joists
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udl per metre length of a joist = 0.4 1.0 2.25 kN/m2
= 0.9 kN
Figure App8.2 shows how this udl acts on the joist.
Figure App8.2
= 1.8 kN= 1.8 kN R2R1
C BA
4 m
0.9 kN/m (udl)
Maximum Bending moment occurs at the middle of the joist, i.e. point C.
B.M. at C =8
lw 2
=8
0.49.0 2
= 1.8 kN-m or 1.8 1000 1000 N-mm
M = f Z, where M is the maximum B.M., f is the strength of timber in bending
and z is the section modulus.
M = f z =6
dbf 2 where
6
db2
= z (b is the width and d is the depth of joists)
Let us assume the width of the joist to be 50 mm
1.8 1000 1000 =6
d507.5 2 (f = 7.5 N/mm2, b = 50 mm)
d2= 28 800 or d = 169.7 mm
Use 50 200 mm joists at 400 c/c
Check for deflection: Maximum allowable deflection = 4000 360 = 11.1 mm
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Actual deflection =IE384
lw5 4
Second moment of area, I =
12
db 3=
12
20050 3
= 3.33 107mm
4
Actual deflection =7
4
1033.370010384
40009.05
= 8.4 mm < 11.1 mm, therefore Safe.
