i. Calculate the self weight of the shelter
ii. Describe with diagrams what the lateral loads on the structure would be. Note, you
are not asked to calculate any lateral loads but simply describe what they might
be.
iii. Calculate and resolve the gravity forces in the members based on self weight and
your point load of 10kg
iv. Determine which member in the structure has the greatest load and choose this
member for analysis
v. Calculate the second moment of area of your chosen member
vi. Calculate the capacity of the member to carry the calculated force
vii. Show iterations if necessary
TIMBER POST AND BEAM
3m
2.5m
2m
1m
Front Elevation
1m
1m
Side Elevation
Post Cross section:
30mm
Area = 900m2
30mm
Beam Cross section:
80mm
50mm
i. Calculate the self weight of the shelter
No of posts:
4no. posts @ 2.5m
4no. posts @ 3m
4no. beams @ 2m
2no. beams @ 3m
Timber prop:
structural timber with yield stress of 25MPa
density, ρ = 370kg/m3
Area = 4000m2
Cover material: This structure is covered with a material with a weight of 0.4kg/m2
Weight of posts = (370)(900)/(1000000) = 0.333kg/m
Weight of beams = (370)(4000)/(1000000) = 1.48kg/m
Therefore, self weight of the timber structure only
= (4) (2.5)(0.333)+(4)(3)(0.333)+(4)(2)(1.48)+(2)(3)(1.48)
= 28.046kg
Surface area of shelter = (3)(3)+(3)(2.5)+(2)(2)(2.5)+(0.5)(0.5)(2)+(2)(3)
=33m2
Therefore, self weight of cover = (33)(0.4) = 13.2kg
Therefore, total weight of shelter = 61.046kg.
Question: Is this light enough to be deemed portable?
ii. Calculate and resolve the gravity forces in the members based on self weight and your point
load of 10kg
iii. Determine which member in the structure has the greatest load and choose this member for
analysis
10kg
10kg
3m
2.5m
2m
NB:
1m
1m
Always design for the worst case first
Simplify the design (if the result is close, then go into more detail)
1m
PL = 10kg
SW = 1.88kg/m
2.5m
2m
Self weight of the beam = 1.48kg/m (as calculated in (i))
Self weight of cover material = (0.4)((1) = 0.4kg/m (as the material is transferred half a meter to each
beam)
Beam carries this
weight of material
3m
1m
Therefore, in kN:
1m
1m
PL = 1.02kN
SW = 0.19kN/m
Total force = 1.02 + (0.19)(2) = 1.4kN
Bending moment:
Point Load
PL/4
Bending moment:
Uniformly Distributed Load (UDL)
wL2/8
= PL/4 + wL2/8
= (1.02)(2)/4 + (0.19)(2)(2)/8
=0.605kNm
Total moment on the beam
iv. Calculate the second moment of area of your chosen member
y
80mm
50mm
b = 50mm
d = 80mm
I=bd3/12
=(50)(80)3/12
= 2133333.33mm4
v. Calculate the capacity of the member to carry the calculated force
Moment Capacity:
M = σI/y
σ = 25MPa = 25N/mm2
Therefore,
M = (25)(2133333.33)/(80/2)
= 1333333.333 Nmm
= 1.333kNm > 0.605kNm
therefore, ok
vi. Show iterations if necessary
Q1: Is the self weight of the structure adequate for a person to carry?
Q2: Is the moment capacity greater than the calculated moment?
If yes to both, then no iterations necessary
If no to Q1:
Can you change your cover to a lighter material and what difference will this make?
Do all beams need to carry this moment or just the load associated with the point
load? i.e. can the other beams of the shelter be smaller sections?
If no to Q2:
The section size is not big enough, therefore choose a greater depth to give you
larger moment capacity but don’t forget to rework your self weight
PIN JOINTED FRAME USING BAMBOO
0.866
0.866
1.0
1.0
Cross Section:
Diameter = 42mm
Wall Thickness = 9mm
Area = 1838mm2
i. Calculate the self weight of the shelter
Density of bamboo = 350kg/m3
Assume weight of cover to be = 0.5kg/m2
Total no. of components = 9 (front)+9(back)+5(side1)+5(side2) = 28
Self weight of bamboo = (28)(350)(1.0)(1838/1000000)
= 18kg
Surface Area
= (4)(0.5)(2)(1.732)
= 6.928m2
Weight of cover
= (6.928)(0.5)
= 3.464kg
Total self weight
= 10.392kg
ii. Calculate and resolve the gravity forces in the members based on self weight and your point
load of 10kg
Note: As the structure is symmetric, only need to analyse one side
Weight of one side = (0.25)(10.392) = 2.598kg = 0.26kN
Point Load at centre = 10kg = 1.02kN
Total Load = 1.28kN (For simplicity, it is assumed that this load acts as a point load at the centre of
the structure)
3 support points, therefore at each support point, Reaction = 0.43kN
The structure is symmetric, therefore examine the following four points:
D
1.28kN
C
A
0.43kN
B
0.43kN
0.43kN
Point A
F1
F2
0.43kN
;
0.43 = F1Cos600;
F1=0.86kN
;
F2=F1Sin600;
F2=0.75kN
Point B
F4
F3
0.75kN
0.75kN
0.43kN
Section is symmetrical, therefore, F3 = F4
;
0.43 = 2F3Cos600;
F3=0.43kN
Point C
F5
F6
0.86kN
0.43kN
;
0.86Cos600+0.43Cos600=F5;
F5=0.645
;
0.86Sin600+F6=0.43Sin600+F5Sin600
F6=0.181
Point D
0.645kN
0.645kN
1.29kN
Check:
;
0.645Cos600+0.645Cos600=1.29kN
QED
iii. Determine which member in the structure has the greatest load and choose this member for
analysis
Greatest tension member = 0.86kN
Greatest compression member = 0.75kN
But, bamboo would be much harder to break in compression than in tension, therefore the
compression member is the worst case
iv. Calculate the second moment of area of your chosen member
For circular sections, I = πr4/4;
Therefore for the circular hollow section
I = (π)(214)/4 – (π)(124)/4
= ((π)/4)(194481‐20736)
=136459mm4
v. Calculate the capacity of the member to carry the calculated force
For compression,
Fcap = π2EI/(kL)2
k = effective length, for simplicity, assume that k=1.0 (i.e. pinned either end)
therefore,
Fcap = π2(1784)(136459)/(1000)2
(N/mm2)(mm4)(mm2)
= 2402N = 2.4kN > 0.75kN, therefore ok.