CE2164 Quiz Dated 06 Mar 07-Solutions for Marking
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Transcript of CE2164 Quiz Dated 06 Mar 07-Solutions for Marking
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8/6/2019 CE2164 Quiz Dated 06 Mar 07-Solutions for Marking
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CE 2164 - STRUCTURAL DESIGN AND MATERIALS
Open-Note Quiz (6 Mar 2007) Time allowed: 60 minutes
Answer ALL questions in Parts 1 & 2 Max. 20 marks
Matriculation No: _________________________ Group:______________
Part 1 [10 Marks]
From questions 1.1 to 1.6, select ONE or MORE correct statement(s) by writing on tothe space provided below each question the appropriate number(s) as illustrated in the
following example 1.0.
Please note that each correct answer or statement will be given 1 mark and half amark will be deducted for the selection of each incorrect answer (or statement).
1.0 Steel is an alloy of iron with the following material:
(1)water(2)cement(3)carbon(4)zinc
Answer(s): ____________ (if (3) is selected as the correct answer to 1.0)
1.1 In concrete, hydration reaction takes place between
(1)Water, cement and sand(2)Cement, water and granite(3)Water and cement(4)Water and aggregate
Answer(s): _____(3)______
1.2 Chemical compositions of Portland cement causing Alkali-aggregate reaction are:
(1)Sodium Oxide(2)Magnesium Oxide(3)Titanium Oxide(4)Potassium Oxide
Answer(s): _____(1) & (4)___
1.3 In the design calculations of reinforced concrete structures, concrete strength is taken
as its
(1)Actual tested strength(2)Specific Mean strength(3)Characteristic strength(4)The value of concrete strength, which is exceeded by 95% of the standard cube test
results.
Answer(s): __(3) & (4)_______
(3)
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1.4 To achieve the objectives of structural design, various design methods may be
considered. The statement(s), which is/are NOT TRUE about various, design methods:
(1)In ultimate strength design method, the ultimate loads are enhanced by anappropriate load factor
(2)In working stress design method, allowable material stresses are their ultimatestrengths divided by a safety factor.
(3)Limit state design in effect compasses both working stress and ultimate strengthdesign methods.
(4)BS 8110 is based on ultimate strength design methodAnswer(s): ____(4)______
1.5 Serviceability limit state concerns the behaviour of structures under service loads. In a
common building structure, this is associated with
(1)Strength of the structure(2)Durability of the structure
(3)Water tightness , fire resistance etc(4)Deflection, crack width etc
Answer(s): ____(2) , (3) & (4)___
1.6 Which of the following will NOT be the consequence of allowing redistribution of
moment in the design of a structural element?
(1)Tension steel can be reduced(2)Ductility can be improved(3)Additional moment will get redistributed to other parts of the beam(4)Load capacity of the structural element will be reduced
Answer(s): ____(4) __________
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Matriculation No: _________________________ Group:______________
Part 2 [10 Marks]
2. Being a structural engineer in an engineering design office, you are asked to design a
typical internal rectangular beam for a parking lot construction project in Sentosa. The
beams are expected to carry, a service load of 22.5 kN/m and a dead load of 10 kN/m,including its self-weight and the bituminous materials on the slabs. The typical span length
is 8 m and it is simply supported. Material properties are specified below:
Steel : 460=yf MPa,
and
Concrete : 35=cuf MPa
The difference between effective depth and overall depth of the beam can be taken as 50
mm. Take any beam dimensions in nearest 25 mm.
(Design load, W= 1.4 Gk+ 1.6 Qk,)
Due to the headroom constraints, the beam depth shall be limited to 650 mm. Ductile
behavior is required. Perform the beam design and sketch your design with proper rebar
arrangements. The design shall conform to BS 8110.
Solution:(a)
Design load, W= 1.4 Gk+ 1.6 Qk) = 1.4 10 + 1.6 22.5 = 50 kN/m
Design ultimate moment of the beam =2
8
wl=
250 8
8
= 400kNm [1]
Try as asingly reinforced beam
According to BS 8110, to ensure adequate ductility applies when redistribution does not
exceed 10 % 0.775d z 0.95d
In order to find the minimum depth for a singly reinforced sectionz = 0.775dwhich
corresponds tox/d= 0.5 and then
.0 156K K= =
Then 2cu
MK bd f=
{ }.
=
6
2
400 100 156 600 35b [1]
.
=
6
2
400 10
0 156 35 600b .= 203 5 b mm
In nearest 25 mm
= 225 b mm [1]
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Now2
cu
MK
bd f=
{ }. .
= =