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

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1 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

2 TIMBER POST AND BEAM 2.5m 3m 2m 1m 1m 1m Front Elevation Side Elevation Post Cross section: 30mm Area = 900m 2 30mm Beam Cross section: 80mm Area = 4000m 2 50mm i. Calculate the self weight of the shelter No of posts: 4no. 2.5m 4no. 3m 4no. 2m 2no. 3m Timber prop: structural timber with yield stress of 25MPa density, ρ = 370kg/m 3

3 Cover material: This structure is covered with a material with a weight of 0.4kg/m 2 Weight of posts = (370)(900)/( ) = 0.333kg/m Weight of beams = (370)(4000)/( ) = 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) = kg Surface area of shelter = (3)(3)+(3)(2.5)+(2)(2)(2.5)+(0.5)(0.5)(2)+(2)(3) =33m 2 Therefore, self weight of cover = (33)(0.4) = 13.2kg Therefore, total weight of shelter = kg. 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 2.5m 3m 2m 1m 1m 1m NB: Always design for the worst case first Simplify the design (if the result is close, then go into more detail)

4 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 1m 1m Therefore, in kn: PL = 1.02kN SW = 0.19kN/m Total force = (0.19)(2) = 1.4kN Bending moment: Point Load PL/4

5 Bending moment: Uniformly Distributed Load (UDL) wl 2 /8 Total moment on the beam = PL/4 + wl 2 /8 = (1.02)(2)/4 + (0.19)(2)(2)/8 =0.605kNm iv. Calculate the second moment of area of your chosen member y 80mm b = 50mm d = 80mm 50mm I=bd 3 /12 =(50)(80) 3 /12 = mm 4 v. Calculate the capacity of the member to carry the calculated force Moment Capacity: M = σi/y σ = 25MPa = 25N/mm 2 Therefore, M = (25)( )/(80/2) = 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?

6 Q2: Is the moment capacity greater than the calculated moment? If yes to both, then no iterations necessary If no to Q1: If no to Q2: 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? 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

7 PIN JOINTED FRAME USING BAMBOO Cross Section: Diameter = 42mm Wall Thickness = 9mm Area = 1838mm 2 i. Calculate the self weight of the shelter Density of bamboo = 350kg/m 3 Assume weight of cover to be = 0.5kg/m 2 Total no. of components = 9 (front)+9(back)+5(side1)+5(side2) = 28 Self weight of bamboo = (28)(350)(1.0)(1838/ ) = 18kg Surface Area = (4)(0.5)(2)(1.732) = 6.928m 2 Weight of cover = (6.928)(0.5) = 3.464kg Total self weight = kg

8 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 C 1.28kN A B 0.43kN 0.43kN 0.43kN Point A F 1 ; 0.43 = F 1 Cos60 0 ; F 1 =0.86kN F 2 ; F 2 =F 1 Sin60 0 ; F 2 =0.75kN 0.43kN

9 Point B F 3 F kN 0.75kN 0.43kN Section is symmetrical, therefore, F 3 = F 4 ; 0.43 = 2F 3 Cos60 0 ; F 3 =0.43kN Point C F 5 F kN 0.43kN ; 0.86Cos Cos60 0 =F 5 ; F 5 =0.645 ; 0.86Sin60 0 +F 6 =0.43Sin60 0 +F 5 Sin60 0 F 6 =0.181 Point D 0.645kN 0.645kN 1.29kN

10 Check: ; 0.645Cos Cos60 0 =1.29kN QED

11 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 = πr 4 /4; Therefore for the circular hollow section I = (π)(21 4 )/4 (π)(12 4 )/4 = ((π)/4)( ) =136459mm 4 v. Calculate the capacity of the member to carry the calculated force For compression, F cap = π 2 EI/(kL) 2 k = effective length, for simplicity, assume that k=1.0 (i.e. pinned either end) therefore, F cap = π 2 (1784)(136459)/(1000) 2 (N/mm 2 )(mm 4 )(mm 2 ) = 2402N = 2.4kN > 0.75kN, therefore ok.

UNIVERSITY OF BOLTON WESTERN INTERNATIONAL COLLEGE FZE. BEng (HONS) CIVIL ENGINEERING SEMESTER ONE EXAMINATION 2018/2019

OCD030 UNIVERSITY OF BOLTON WESTERN INTERNATIONAL COLLEGE FZE BEng (HONS) CIVIL ENGINEERING SEMESTER ONE EXAMINATION 2018/2019 ADVANCED STRUCTURAL ANALYSIS AND DESIGN MODULE NO: CIE6001 Date: Tuesday 8