联系我们: 手动添加方式: 微信>添加朋友>企业微信联系人>13262280223 或者 QQ: 1483266981
Copyright 2022 v02 University of Southampton Page 1 of 7
UNIVERSITY OF SOUTHAMPTON CENV2031W1
______________________________________________________
SEMESTER 1 FINAL ASSESSMENT 2022/23
STRUCTURAL ANALYSIS
DURATION: 2 Hours (On-Campus Traditional)
______________________________________________________
This paper contains 4 questions
Answer ALL questions.
Each question carries 25 marks out of a total of 100 marks for the exam paper.
Note that marks will only be awarded when appropriate working is given.
An outline marking scheme is shown in brackets to the right of each question.
Only University approved calculators may be used, but all stages of working must be
shown.
A foreign language direct ‘Word to Word’ translation dictionary (paper version) ONLY
is permitted provided it contains no notes, additions or annotations
Some useful equations and the Structures Data Sheet are provided at the end
of this paper.
CENV2031W1
Copyright 2022 v02 University of Southampton Page 2 of 7
Q1. The steel cross-section in Figure Q1 is subjected to a
compressive force = 400 kN applied at C and a shear force =
50 kN along the y-axis. The yield stress is 355 MPa.
i. Determine the second moment of area of the cross-section
about the z- and y-axis.
[6 marks]
ii. Calculate the shear stress xy at A.
[7 marks]
iii. Calculate the longitudinal stress σx at A due to combined
bending and axial load.
[8 marks]
iv. Find the safety factor at A using the Von Mises criterion.
[4 marks]
[Total 25 marks]
Figure Q1. Steel cross-section.
G
20 mm 10 mm 20 mm
50 mm
20 mm
10 mm
[Point of action of
compressive force P]
A
5 mm
C
10 mm
CENV2031W1
Copyright 2022 v02 University of Southampton Page 3 of 7
Q2. Consider the truss shown in Figure Q2.
i. Examine whether the truss is statically determinate or
indeterminate. If the latter condition applies, determine the
degree of indeterminacy.
[5 marks]
ii. Set up the equilibrium equations at joint D. Note that there is
a roller support at D.
[4 marks]
iii. Set up the compatibility equations at joint D.
[6 marks]
iv. Determine the axial force in each truss member and the
reaction force at D.
[10 marks]
[Total 25 marks]
Figure Q2. Truss.
TURN OVER
CENV2031W1
Copyright 2022 v02 University of Southampton Page 4 of 7
Q3. Use the flexibility method to analyse the beam shown in Figure Q3.
i. Choose the releases.
[2 marks]
ii. Calculate the release displacements due to the external
load.
[4 marks]
iii. Calculate the flexibility coefficients.
[8 marks]
iv. Solve the flexibility equation.
[8 marks]
v. Draw a qualitative sketch of the bending moment diagram.
[3 marks]
[Total 25 marks]
Figure Q3. Beam.
CENV2031W1
Copyright 2022 v02 University of Southampton Page 5 of 7
Q4. The portal frame in Figure Q4 is loaded by a 5 kNm clockwise
bending moment at B. Axial rigidity may be assumed. The flexural
rigidity EI is 11,550 kNm2
. Use the direct stiffness method to
analyse the structure.
i. Identify the freedoms.
[5 marks]
ii. Determine the stiffness coefficients by releasing each of the
freedoms in turn.
[16 marks]
iii. Set up the structure stiffness equation.
[4 marks]
[Total 25 marks]
Figure Q4. Portal frame
TURN OVER
CENV2031W1
Copyright 2022 v02 University of Southampton Page 6 of 7
Useful equations
Second moment of area about the z and y axes:
= ∑[ , +
( G i
)
2
]
= ∑[ , +
( G zi
)
2
]
Shear stress :
=
′
The equivalent stress (Von Mises criterion):
= √
2 + 3
2
Truss member elongation:
=
Degree of indeterminacy (only for beams and frames):
= 3 + 3
Flexibility equation:
=
Stiffness equation:
=
CENV2031W1
Copyright 2022 v02 University of Southampton Page 7 of 7
Structures Data Sheet
Flexibility coefficients due to a unit moment
1,1 =
3
2,1 =
6
Rotations due to external loads
1= 2=
16
2
1= 2=
24
2
Stiffness coefficients due to unit displacements and rotations
1,1 =
3
1,1 = 2,1 =
3
2
1,1 =
4
2,1 =
2
1,1 = 2,1 =
6
2
1,1 = 2,1 =
6
2
1,1 = 2,1 =
12
3
1,1 =
3
2
1,1 = 2,1 =
3
3
Volume integrals
2
( + )
2
2
3
(2 + )
6
2
6
( + 2 )
6
( + )
2
(2 + )
6
(
6
2 + )
+
(
6
+ 2 )
( +4 + )
6
( + 2 )
6
(
6
+ 2 )
+
(
6
2 + )
END OF PAPER
1.
2.
4.
5.
6.
7.
3.


发表评论