1. CITY UNIVERSITY OF HONG KONG
Course code & title : BST10346
Structural Design
Session : Semester B, 2007-2008
Time allowed : Two hours
This paper has SEVENTEEN pages (including this cover page).
1. This paper consists of four questions.
2. Answer Question 1 and ANY OTHER TWO questions.
3. Start each question on a new page.
4. State all the assumptions clearly if necessary
5. ALL STEPS OF CALCULATION AND SIGN CONVENTION MUST
BE CLEARLY SHOWN.
Materials, aids and instruments permitted to be used during examinations:
Approved Calculator
Special materials (other than standard materials e.g. answer book or
supplementary sheet) to be supplied to students:
Nil
Not to be taken away
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2. Question 1 (40 marks)
The bending moment diagram of the continuous beams below from different load cases are
shown below:
190 200
Dead Load kNm
180 220 230
150 90
Live Load kNm
130 180 200
a) Calculate the ultimate design moment at mid-span and support. (5 marks)
b) Calculate the As required and As’ required (if any) of the mid-span (10 marks)
moment only for the two concrete section sizes with the following data
provided.
1.) 400 x 600 mm depth
2.) 400 x 500 mm depth
fcu = 45 N/mm²
fy and fyv = 460 N/mm²
Nominal cover = 30 mm
Link diameter = 12 mm
Main bar diameter = 40 mm
Shear link diameter = 12 mm
c) Suggest three methods in order to avoid provision of compression steel. (3 marks)
d) From the result of b), design the shear reinforcement for the two (5 marks)
sections size if the shear force from DL and LL is 300 kN and 200 kN
respectively.
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3. e) Calculate the vc of the below section. (5 marks)
Design data:
fcu = 40 N/mm²
fy and fyv = 460 N/mm²
d = 900 mm
Link diameter and spacing
= T12@250c/c 2 legs
+
T16@200c/c 1 leg
f) Calculate the maximum shear capacity in kN of the beam. (10 marks)
g) If only two legs are allowed in this beam, design the other shear (2 marks)
reinforcement for this beam.
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4. Question 2 (30 marks)
a) Prove that 254 x 254 x 73 kg/m Grade 55C UC and 305 x 165 x 54 (18 marks)
kg/m Grade 55C UB are adequate against the following ultimate load
and ultimate moments about X-X axis in low shear load case.
Mxx = 100 kNm
Myy = 30 kNm
F = 700 kN
b) From the question a), discuss why UB has sufficient structural capacity (2 marks)
against the same loadings as UC although UB has lighter self weight.
c) Design the bolt connection detail according to the diagram below to (10 marks)
fulfill the following ultimate load. The shear plane will be happened in
the threads.
(Remarks: The no. of bolts should not be less than 6)
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5. Question 3 (30 marks)
a) Use force method to determine all reactions for the beam shown below. (15 marks)
b) Sketch the bending moment and shear force diagrams, indicating (12 marks)
maximum and minimum values, and exact location of contra-flexure
point(s), where appropriate.
M (kNm)
L (m) L (m)
c) If the roller support is removed from the beam, sketch the (3 marks)
corresponding bending moment and shear force diagrams.
(Hint: No further calculation shall be required)
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6. Question 4 (30 marks)
A single-bay frame is subject to a vertical UDL of 5 kN/m and a horizontal point load of
20 kN as shown in the figure below. Using an approximate approach, determine, under the
combined vertical and horizontal loads:
a) the reaction forces at the hinge supports D and E (indicate clearly the (15 marks)
direction of reaction force);
b) the bending moments and shear forces at point B (middle of beam) and (15 marks)
point C (right end of beam). Indicate clearly whether the moment is
sagging or hogging, and whether the shear is clockwise or anti-
clockwise.
L = 4m
5 kN/m
20 kN
A B C
H = 4m
D E
Hint:
♦ For lateral load analysis, you can assume a point of contra-flexure at mid-span of the
beam.
♦ For vertical load analysis, you can assume a point of contra-flexure located at 0.1L from
each end of the beam of span L.
♦ Moment: Sagging Hogging
♦ Shear: Clockwise Anti-clockwise
- END -
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7. Formulae
K = M / fcu bd²
z = (0.5 + √(0.25 - K/0.9) )d
As req. = M / (0.87fyz)
As′ = (K - 0.156) fcu bd² / ((0.87 fy (d - d′)
As = (0.156 fcu bd²) / (0.87 fy 0.775d) + As′
Mu (conc) = 0.156 fcu bd²
Mu (steel) = 0.87fyAsz
z = d - 0.45x
x = (0.87fyAs) / (0.45fcub0.9)
v = V/bvd
vc = 0.79(100As/bvd)1/3(400/d)1/4/1.25(fcu/25)1/3
Asv/sv = 0.4bv/(0.87fyv)
Asv/sv = bv(v-vc)/(0.87fyv)
Msx = αsx nlx²
Msy = αsy nly²
Fs/Ps + Ft/Pt ≤ 1.4
Ft = Pe y1 / no. of bolt column x Σy²
F/Aepy + Mx/Mcx + My/Mcy ≤ 1
Mc = S py but ≤ 1.2pyZ for plastic or compact sections
Mc = pyZ for semi-compact and slender sections
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8. Appendices
Load Ultimate Limit State
Combination Dead Imposed Wind Earth & Serviceability
water Limit State
pressure
Dead & Imposed 1.4 (or 1.6 - 1.4 1.0
(& earth & water 1.0)
pressure)
Dead & wind (& 1.4 (or - 1.4 1.4 1.0
earth & water 1.0)
pressure)
Dead & imposed 1.2 1.2 1.2 1.2 1.0
& wind (& earth
& water
pressure)
Conditions:
1. The area of each bay exceeds 30m²,
2. Spans are considered to be approximately equal if they do not differ by more than 15% of
the longest span.
3. Qk/Gk < 1.25,
4. Qk < 5 kN/m² excluding partitions
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9. ly / lx 1.0 1.1 1.2 1.3 1.4 1.5 1.75 2.0
αsx 0.062 0.074 0.084 0.093 0.099 0.104 0.113 0.118
αsy 0.062 0.061 0.059 0.055 0.051 0.046 0.037 0.029
Bending moment coefficient for slabs spanning in two directions at right-angles, simply-
supported on four sides
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10. Basic span / effective depth ration for rectangular or flanged beams (BS 8110, Table 3.10)
Support conditions Rectangular sections Flanged beams with bw/b ≤ 0.3
Cantilever 7 5.6
Simply supported 20 16
Continuous 26 20.8
Modification factor (m1)for tension reinforcement (BS 8110, Table 3.11)
Service stress M/bd²
N/mm² 0.5 0.75 1.00 1.5 2.0 3.00 4.00 5.00 6.00
100 2 2 2 1.86 1.63 1.36 1.19 1.08 1.01
150 2 2 1.98 1.69 1.49 1.25 1.11 1.01 0.94
(fy = 250) 156 2 2 1.96 1.66 1.47 1.24 1.1 1 0.94
200 2 1.95 1.76 1.51 1.35 1.14 1.02 0.94 0.88
250 1.9 1.7 1.55 1.34 1.2 1.04 0.94 0.87 0.82
(fy = 460) 288 1.68 1.5 1.38 1.21 1.09 0.95 0.87 0.82 0.78
300 1.6 1.44 1.33 1.16 1.06 0.93 0.85 0.8 0.76
Modification factor (m2) for compression reinforcement (BS 8110, Table 3.12)
100 As′ prov./bd Factor 100 As′ prov./bd Factor
0 1 1 1.25
0.15 1.05 1.5 1.33
0.25 1.08 2 1.4
0.35 1.1 2.5 1.45
0.5 1.14 ≥3 1.5
0.75 1.2 - -
Table 6, Design strength py for steel to BS 4360
BS 4360 Grade Thickness (mm) less than orSections, plates and hollow sections
equal to (flange thinkness forpy (N/mm²)
rolled section)
43 A, B and C 16 275
40 265
63 255
100 245
50 B and C 16 355
40 345
63 340
100 325
55 C 16 450
25 430
40 425
63 400
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