1. WORKSHEET 5 PROPERTIES OF SECTIONS AND DEFLECTIONS

2. Q1
Find the Moment of Inertia, I, and the Section Modulus, Z, of the following timber sections: (include units) a)
50 mm width x 200 mm deep X
(i) Ix = bd3/12 = 50 x 2003 / 12
= x 106 mm4
(i) Zx = bd2/6 = 50 x 2002 / 6
= x 103 mm3 b)
100 mm width x 100 mm deep X
(i) Ix = bd3/12 = 100 x 1003 / 12
= x 106 mm4
(i) Zx = bd2/6 = 100 x 1002 / 6
= x 103 mm3

3. Q2 Referring to 1 (a) and 1 (b): a) which is heavier? b)
both have the same cross-sectional area
so both have the same weight per unit length b)
which is stiffer and by what proportion?
I(a)x =33.3 x 106 mm I(b)x = 8.3 x 106 mm4
200 x 50 is 4 times stiffer (about the X-X axis) c)
which is stronger and by what proportion?
Z(a) x =333.3 x 103 mm Z(b)x = x 103 mm3
200 x 50 is 2 times stronger (in bending)

4. Q3 (a) a) You want a beam as stiff as a 200 x 50 mm
beam, but you only have space for 150 mm depth.
How wide does the beam have to be?
I x = 33.3 x 106 mm4
I x = bd3 / 12
33.3 x 106 = b x 1503 / 12
b = 33.3 x 106 / (1503 / 12)
= 118 mm

5. Q3 (b) b) You want a beam as strong as a 200 x 50 mm
beam, but you only have space for 150 mm depth.
How wide does the beam have to be?
Z x = x 103 mm3
Z x = bd2 / 6
333.3 x 103 = b x 1502 / 6
b = x 103 / (1502 / 6)
= 89 mm
(in both cases it requires a substantial increase in width)

6. Q3 (c)
c) why don’t we use an even deeper and narrower beam, e.g. a 300 x 20 mm beam? a)
the beam may buckle b)
the depth may be a problem

7. Q4 You could try using Multiframe for the following questions
Use Douglas Fir (the Modulus of Elasticity in the Wood Sections Library
is 4500KSI = ~7000MPa)
a span-to-depth ratio for timber floor joists - 18:1
Check this out for 200 x 50 mm softwood joists at 600 mm centres
spanning 3.6 m in the upper floor of a house
Assume a live load of 1.5 kPa and a total dead load (incl self-weight
of the joists) of 0.4 kPa

8. Q4 (cont.) tributary area 200 x 50 mm timber joists @ 600mm crs 3600

9. Q4 (a) a) (i) What is the maximum bending stress in the joists?
Tributary area = 3.6 x 0.6
= 2.16 m2
Total loading = ( )
= 1.9 kPa (kN/ m2)
Total load (distributed)
on one joist (W = wL) = 1.9 x 2.16
= kN
Maximum bending moment = WL/8 (where W =wL)
= x 3.6 / 8
= 1.85 kNm
Bending stress, f = M / Z = 1.85 / x 103
bring it all to N and mm - multiply top line x 1000 (kN to N) and 1000 (m to mm)
= 5.6 N/mm2
= 5.6 MPa
(ii) Is F8 strong enough? (Softwood grade F8 is capable of taking 8 MPa)
5.6 < 8.
So this is within the capacity of F8 timber

10. Q4 (b) b) (i) What is the maximum deflection?
the Modulus of Elasticity of F8 timber can be taken as 9000 MPa
For simply supported beam with UDL
Maximum deflection = 5 W L3 / 384 EI
= 5 x 4104 x / 384 x 9000 x 33.3 x 106
(using N and mm everywhere)
= 8.4 mm
(ii) Is the calculated deflection within the allowed limit
Allowed limit = Span / 300
= 3600 / 300
= 12 mm
So this is within the allowed limit

11. Q5
In a two-storey house, you want to span across a double garage 7m wide.
You decide to do this with steel beams at 3.6 m centres. (The timber joist
system in Q1 spans between them)
3.6 m
7 m
steel beams
timber joists
@ 600mm crs
tributary area
for beam = 7 x 3.6
= 25.2 m2

12. Q5 (cont1.) a) Find a Universal Beam section strong enough?
Tributary area for one beam = 7 x 3.6
= 25.2 m2
Total load one beam = 1.9 x 25.2
= kN
Max bending moment of beam = WL/8
= x 7/ 8
= 41.9 kNm a)
Find a Universal Beam section strong enough?
(assume a maximum allowable stress of 200MPa for Grade 300 steel.
Use the Table given or a BHP catalogue)
Bending stress f = M / Z (want f to be 200 or less)
Z must be at least Z > = 41.9 x 106 / 200
>= x 103 mm3
Looking up the Table of Universal Beams we find that a 200UB25.4
has Z = 232 x 103. The depth of the beam d = 203 mm

13. Q5 (cont2.) b) Is this section stiff enough? If not, upsize it.
span-to-depth ratio = 7000 / 203
= 34.5
this looks a bit optimistic. Let us check the deflection
From the table, Ix of the section = 23.6 x 106 mm4
deflection = 5 WL3 / 384 EI
= 5 x x / 384 x x 23.6 x 106
= 45.3 mm
allowable deflection = span / = 7000 / 500
= 14.0 mm
clearly the section is not stiff enough. Working with defl = 14, we get
14 = 5 WL3 / 384 EIxnew
(alternatively) Ixnew / Ix = 45.3 / 14
Ixnew = 3.24 x Ix = 3.24 x 23.6
= x106mm4
From table find that 310UB40.4 has Ix = 85.2 x106 mm4

14. Q5 (cont3.) c) What span-to-depth ratio do you end up with
for the steel beams? (310UB40.4)
depth of beam d = 304 mm
span-to-depth ratio = 7000 / 304
= 23