CIVE3013 Steel and Timber Design

PART A – SHORT ANSWER QUESTIONS (25 marks, each question has 5 marks)

A1.

You need to use your student ID to get some parameters. No mark for wrong value Compare the compression capacity of the steel column in Case 1 and Case 2, and explain why they are similar or different in your own words. • Knowing that the column and beam sizes in Case 1 and Case 2 are the same and there is no intermediate restraint along the column length. Assume there is no bending • Only difference between two cases is the bolt connection at each end. • No calculation required (only explain based on AS 4100)

A2.

Answer the following question • If the last digit of your student ID is even (0, 2, 4, 6 or 8), discuss how can you define k4 factor for unseasoned timber in dry environment. • If the last digit of your student ID is odd (1, 3, 5, 7 or 9), discuss how can you define k4 factor for seasoned timber in wet environment.

A3.

A 12m long steel roof beam (410UB53.7) is restrained by the purlins and fly-brace. The purlin spacing is 3m cts and the fly-brace is at the mid-point of the roof beam. Assume that the load combination, 0.9G + Wu (wind is more critical), generated a bending moment (diagram in Fig. 2) on the roof beam. Check whether this beam is fully lateral restrained or not.

A4.

For bending timber member, how many k factors can be more than 1? Provide examples + explain how those factors are more than 1.

A5.

Answer the following question

• If the last digit of your student ID is even (0, 2, 4, 6 or 8), describe briefly how the capacity factor, for timber members is determined.

• If the last digit of your student ID is odd (1, 3, 5, 7 or 9), describe how the grade of MGP timber is determined.

PART B – STEEL MATERIAL (35 marks)

B.

1. You need to use your student ID to get some parameters. No mark for wrong value A simply supported steel beam is subjected by the point loads, P (Check Fig. 3). Note, some data is given based on your student ID. Check the last two digits of your student ID. For example, if ID = 12345678, use 78 to get the below information

• The whole span of the beam, L (m) = 5 + 0.05 x (the last two digits of your student ID). For example, L = 5 + 0.05 x 78 = 8.9 m Each point load, P is the combination of unfactored loads (Dead and live loads)

• Dead load, G (kN) = 80 – 0.5 x (the last two digits of your student ID).

• Live load, Q (kN) = 30 – 0.2 x (the last two digits of your student ID).

Answer the following questions:

a) Draw the shear force and bending moment diagram (with all correct values) for the combined ultimate load case (G and Q). (7 marks)

b) Check whether the given section is adequate for bending. Assume that the two points provide the lateral restraint for the steel beam. (15 marks)

c) Check whether the given section is adequate for shear and combined bending with shear, even if the bending check is not satisfactory in Question b. (13 marks)

PART C – TIMBER MATERIAL (40 marks)

C.

1. You need to use your student ID to get some parameters. No mark for wrong value

A seasoned timber column in a residential house is made from 2 seasoned MGP timber members (see Fig. 4). Knowing that there is no intermediate lateral restraint on the column, and the house is located in the northernmost part of Queensland, whose has very high average humidity.

Note, some data is given based on your student ID. Check the last two digits of your student ID. For example, if ID = 12345678, use 78 to get the below information

Column height (m) = 2 + 0.02 x (the last two digits of your student ID).

For example, column height = 2 + 0.02 x 78 = 3.56 m

The loads acting on the column are:

• Dead load, G (kN) = 125 – 0.2 x (the last two digits of your student ID).

• Concentrated live load, Q (kN) = 25 – 0.2 x (the last two digits of your student ID).

• There is only upward wind load in this case, Wu (kN) = 250 – 0.2 x (the last two digits of your student ID).

Answer the following questions:

a) Define the strength combinations for the column according to AS1170.0:2002. Then find the design compression and tension loads (10 marks)

b) Check whether the given section is adequate for tension. (8 marks)

c) Check whether the given section is adequate for compression. (10 marks)

d) Check whether the given section is adequate for bending. Assume that the compression load is more critical, and its eccentricity is 20mm i.e. M* = N* x eccentricity. (12 marks)