Assignment 1.pdf

Assignment 1 1–7. The pre-cast T-beam has the cross-section shown. Determine its weight per meter length if it is made from reinforced stone concrete and eight 20-mm cold-formed steel reinforcing rods.

Ans: = 9.16 kN/m

1-8. The wall is 2.5 m high and consists of 51 mm 102 mm studs plastered on one side. On the other side is 13 mm fiberboard, and 102 mm clay brick. Determine the average load in kN m of length of wall that the wall exerts on the floor.

Ans: = 6.20 kN/m

1–12. A three-story hotel has interior columns that are spaced 6 m apart in two perpendicular directions. If the loading on the flat roof is estimated to be 1.5 kN/m2, determine the live load supported by a typical interior column at (a) the ground-floor level, and (b) the second floor level.

Ans: a) 141.2 kN b) 97.6 kN

ERT 256-SEM II 17/18

ASSIGNMENT 1

©MAISARA AZAD BT MAT AKHIR

2–1. The steel framework is used to support the reinforced stone concrete slab that is used for an office. The slab is 200 mm thick. Sketch the loading that acts along members BE and FED. Take a = 2 m, b = 5 m. Hint: See Tables 1.2 and 1.4.

2–24. Determine the reactions on the beam. The support at B can be assumed to be a roller.

Ans: NB = 70 kN Ay = 110 kN Ax = 0

3–8. If the maximum force that any member can support is 8 kN in tension and 6 kN in compression, determine the maximum force P that can be supported at joint D.

Ans: P = 5.20 kN

ERT 256-SEM II 17/18

ASSIGNMENT 1

©MAISARA AZAD BT MAT AKHIR

3–11. Specify the type of compound truss and determine the force in each member. State if the members are in tension or compression. Assume the members are pin connected.

Ans: FAB = 3 kN (T) FFD = 4.24 kN (C) FED = 0 FAF = 4.24 kN (C) FEA = 4 kN (C) FEF = 0 FED = FDC = 0 FEF = FCG = 0 FDF = FDG = 4.24 kN (C) FAF = FGB = 4.24 kN (C) FAB = 3kN (T) FEA = FCB = 4kN (C)

Assignment Format: i. ii.

You must submit a handwritten report. Dateline = Before 4 pm on 16/3/2017 (Friday).

ERT 256-SEM II 17/18

ASSIGNMENT 1

©MAISARA AZAD BT MAT AKHIR