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2022 Exam - 312

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ENGCV312-22A (HAM)
Name:
ID number:
ENGCV312-22A Exam
DEPARTMENT
Engineering
PAPER TITLE
Structural Engineering 2
TIME ALLOWED
Three hours
NUMBER OF QUESTIONS
IN PAPER
Five
NUMBER OF QUESTIONS
TO BE ANSWERED
Five
VALUE OF EACH QUESTION
20 pts
GENERAL INSTRUCTIONS
Show all of your work. Work should be clear,
orderly, and legible. Each question specifies a
method of analysis. You must use this method to get
credit for the problem. You are not permitted to use
computer software programs to complete the
problems. You must complete by hand and show
your work.
SPECIAL INSTRUCTIONS
Allowed a single hand written A4 sheet of notes,
both sides of the paper may be used. Allowed
NZS1170.5.
CALCULATORS PERMITTED
Yes
The calculator memory must be cleared prior to the
test.
THIS TEST PAPER MUST BE HANDED
TURN OVER
2
ENGCV312-21A (HAM)
CONTINUED
ENGCV312-22A (HAM)
1
Lateral Loads from Earthquakes
A developer is considering two location for a new hotel, Hanmer Springs or Paihia/Russell.
The hotel will be designed to match the hotel brand, following a similar floor plan, elevation
design, and structural plan to other hotels owned by the developer. As part of their decision
process for selecting the location for the new hotel, the developer needs know how much the
seismic loads will affect the design.
A Considering the Ultimate Limit State, determine the lateral load distribution for the
hotel at both sites and comment on difference between the two designs.
B Provide advice on how the lateral loads will influence the final design for both
locations.
Building information common to both locations:









A basic layout of the building is shown in Figure 1.
Building is constructed out of structural steel. The lateral force resisting system is a
moment frame.
The dead loads from the structural weight and superimposed loads (HVAC, flooring,
concrete slab, etc) are provided.
The live loads should be determined based on the use of the building and the floor
area. Use a uniformly distributed floor load of 2 kPa.
No part of the building will be used for storage applications and there is no access to
the roof.
The hotel will have 12 apartments (6 on each of the upper floors) that will each be
able to house 10 people at a time.
The ground floor will be retail space.
Importance level should be determined from the use of the structure and is given in
Table 1.
Estimate an initial period based on the structural material and lateral force resisting
system using an appropriate equation based on the structural material and the lateral
force resisting system.
For the ultimate limit state, calculate the lateral loads for a seismic event. Additional
information on the design criteria is provided in Table 2. Fill in Tables 1-5.
TURN OVER
4
ENGCV312-21A (HAM)
Figure 1 Hotel structural layout
Table 1 Total dead and live loads on each floor
Level
Dead loads (kN)
Roof
2000
Level 2
3000
Level 1
3000
Live loads (kN)
Combined loads (kN)
Total
CONTINUED
ENGCV312-22A (HAM)
Table 2 Design Criteria
Design Criteria
Hanmer Springs
Paihia/Russell
Importance Level
2
2
Ductility (assume elastic)
 = 1.0
 = 1.0
Design Life
100 years
100 years
Site Class
C
D
Estimated Period (USL)
Figure 2 NZS1170.5 Supplement estimation of structural period
TURN OVER
6
ENGCV312-21A (HAM)
Table 3 Use this table to calculate the base shear
Variable
Hanmer Springs
Paihia/Russell
Calculations
Calculations
Spectral Shape Factor, 𝑪𝒉 (𝑻)
Z-factor
Return period factor, 𝑹𝒖
Shortest distance to fault, D
Near Fault Factor, 𝑵(𝑻, 𝑫)
Elastic Site Spectrum, 𝑪(𝑻)
Structural performance factor,
𝑺𝒑
0.7
0.7
𝒌_𝝁
Horizontal design action
coefficient, 𝑪𝒅 (𝑻)
Seismic Weight (kN)
Base Shear (kN)
CONTINUED
ENGCV312-22A (HAM)
Table 4 Use this table to distribute the base shear and calculate the lateral loads - Hanmer Springs
Floor
ℎ𝑖
𝑊𝑖
ℎ𝑖 𝑊𝑖
𝐹𝑡
m
kN
kNm
kN
0.92𝑉(𝑊𝑖 ℎ𝑖 )
∑(𝑊𝑖 ℎ𝑖 )
𝐹𝑖
kN
kN
First Floor
Second Floor
Roof
Sum
Table 5 Use this table to distribute the base shear and calculate the lateral loads - Paihia/Russell
Floor
ℎ𝑖
𝑊𝑖
ℎ𝑖 𝑊𝑖
𝐹𝑡
m
kN
kNm
kN
0.92𝑉(𝑊𝑖 ℎ𝑖 )
∑(𝑊𝑖 ℎ𝑖 )
𝐹𝑖
kN
kN
First Floor
Second Floor
Roof
Sum
TURN OVER
8
2
ENGCV312-21A (HAM)
Force Method.
The rigid frame shown in Figure 3 Building layout (left) and frame for analysis (right)supports
one end of simple building. The building has a depth of 10 meters. The roof beam along BC
supports half of the roof load. The roof consists of dead loads from a concrete slab (200 mm
thick), gypsum plaster suspended ceiling (13 mm thick), ducting and piping. The ducts and
piping can be assumed to have a distribute weight of 0.15 kN/m2. The concrete slab has a unit
weight of 23.6 kN/m3. The weight for the suspended ceilings should be taken from
AS/NZS1170.1 table A2. Assume there is no live load for the structure. Do not factor the loads.
A Determine the distributed load on frame ABCD.
B Determine the vertical and horizontal reactions at A and D using the Force Method.
If you cannot figure out the distributed load in part A, use a load of 10kN/m.
C Draw the shear and bending moment diagrams.
Given: 𝐼𝐴𝐵 = 𝐼𝐶𝐷 = 250𝑥106 𝑚𝑚4 , 𝐼𝐵𝐶 = 500𝑥106 𝑚𝑚4 , 𝐸 = 200 𝐺𝑃𝑎 for all members
Figure 3 Building layout (left) and frame for analysis (right)
CONTINUED
ENGCV312-22A (HAM)
3
Slope Deflection.
The simple frame shown in Figure 4 represents a simple one storey building. The frame
supports a uniform lateral load W. The beam has a second area of moment, I, value twice that
of the column. The frame is pin supported at A and D. The beams and columns have a
constant E value.
A Analysis: Using the slope deflection method, determine the internal moments at each
of the joints.
B Discussion: If the building were to have an additional lateral load applied at B,
Figure 5, discuss how your analysis would change to compensate for the sway in the
structure?
Figure 4 Frame for analysis (left)
Figure 5 Frame with lateral load
TURN OVER
10
4
ENGCV312-21A (HAM)
Stiffness Method
A simple beam is shown in Figure 6. The beam supports a uniformly distributed load of W.
The beam has rigid supports at the two ends and roller supports at B and C. The length of
each beam segment is 3 meters. EI is constant for all members.
A Determine the equivalent beam by transferring the distributed load to point loads at
the joints, develop the Force and Displacement vectors for the structure.
B Construct the 𝐾11 and the 𝐾21 sections of the stiffness matrix for the beam structure
using the beam member stiffness matrix given.
C Solve for the rotational displacements at B and C
D Determine the vertical and moment reactions at supports A, B, C, and D.
Figure 6 Structural Stiffness Matrix for Beam
𝑘𝑏𝑒𝑎𝑚 = 𝐸𝐼
4/9
2/3
−4/9
2/3
2/3
4/3
−2/3
2/3
−4/9
−2/3
4/9
−2/3
[ 2/3
2/3
−2/3
4/3 ]
CONTINUED
ENGCV312-22A (HAM)
TURN OVER
12
5
ENGCV312-21A (HAM)
Plastic Theory:
The frame shown below supports a combination of lateral and horizontal loads, Figure 7.
A The frame has 5 possible failure mechanisms. For each mechanism, determine the
critical load W that can be applied using the plastic moment 𝑀𝑝 . Note that the
columns have an 𝑀𝑝 equal to twice the beam.
 Beam: Hinges in beam at B, C, and E
 Beam: Hinge in beam at B, D, and E
 Sway: Hinge in column at B and E
 Combined: Hinge in beam at C, and hinge in column at E
 Combined: Hinge in beam at D, and hinge in column at E
B State the critical mechanism for the frame.
Figure 7 Plastic theory Frame
CONTINUED
ENGCV312-22A (HAM)
Reference Material
cos 2 𝜃
𝐴𝐸 sin 𝜃 cos 𝜃
𝑘𝑡𝑟𝑢𝑠𝑠 =
[
𝐿
− cos2 𝜃
−sin 𝜃 cos 𝜃
sin 𝜃 cos 𝜃
𝑠𝑖𝑛2 𝜃
−sin 𝜃 cos 𝜃
− 𝑠𝑖𝑛2 𝜃
12
𝐿3
6
2
𝑘𝑏𝑒𝑎𝑚 = 𝐸𝐼 𝐿
12
− 3
𝐿
6
[ 𝐿2
6
𝐿2
4
𝐿
6
− 2
𝐿
2
𝐿
−cos 2 𝜃
− sin 𝜃 cos 𝜃
cos2 𝜃
sin 𝜃 cos 𝜃
12
𝐿3
6
− 2
𝐿
12
𝐿3
6
− 2
𝐿
−
− sin 𝜃 cos 𝜃
−𝑠𝑖𝑛2 𝜃 ]
sin 𝜃 cos 𝜃
𝑠𝑖𝑛2 𝜃
6
𝐿2
2
𝐿
6
− 2
𝐿
4
𝐿 ]
TURN OVER
14
ENGCV312-21A (HAM)
CONTINUED
ENGCV312-22A (HAM)
TURN OVER
16
ENGCV312-21A (HAM)
--END--
CONTINUED
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