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May 2008
Thursday 22 May 2008 9.30am 11.30pm
University of Glasgow
M.Eng., B.Eng. and B.Sc. degrees in Civil Engineering and Civil Engineering with Architecture
STRUCTURAL MECHANICS 2
Attempt THREE questions
Time allowed TWO hours
If an electronic calculator is used, appropriate intermediate steps in calculations should be indicated.
The marks allocated to each question or part question are to be used as a guide only.
Data Sheet, Formulae.
10 pages in total.
2
May-2008
Q1. A simply supported beam is to carry a uniformly distributed load of q = 2 kN/m and is made from an I-section as shown in Fig. Q1.
a) Draw the shear force and bending moment diagrams, indicating key values.
(8 marks)
b) Calculate the 2nd moment of area and elastic section modulus. (8 marks)
c) The bending stress must not exceed 110 N/mm2. Calculate the maximum span allowed.
(7 marks)
d) Given this span, calculate the maximum vertical shear stress in the web and compare it with the average shear stress in the web.
(10 marks)
L
q = 2 kN/m
150
5
5
140
80 5
Fig. Q1
Continued Overleaf /
3
May-2008
Q2. A 5 m long propped overhanging cantilever (statically indeterminate) is loaded by a point load of 10 kN as shown in Fig. Q2.
a) Derive an expression for the bending moment in the beam using Macaulays brackets.
(6 marks)
b) Derive expressions for the curvature, slope and deflection of the beam. (7 marks)
c) Calculate the reactions. (6 marks)
d) Given that E = 200 kN/mm2 and I = 15.0 106 mm4, calculate the tip deflection and the position and magnitude of the maximum deflection between A and B.
(8 marks)
e) Sketch the bending moment diagram, indicating key values. (6 marks)
C
3 m 2 m
A B
EI constant
10 kN
Fig. Q2
Continued Overleaf /
4
May-2008
Q3. a) What is the major principle used when solving a statically indeterminate problem by splitting the structure into substructures? What assumptions does this principle rest on?
(4 marks)
b) Explain why Eulers load is insufficient, in practice, to design against buckling? Show this in the form of a load deflection graph.
(4 marks)
c) Consider a given section along a double-pinned column under uniform compression.
i) Using the bending moment/deflection relationship, derive the ordinary differential equation relating the deflection along the column, the bending moment, the Young's modulus (E) and the second moment of area, I.
(4 marks)
ii) Solve this ordinary differential equation, and deduce the value of Eulers buckling load for a double-pinned column.
(7 marks)
d) Use the Perry-Robertson formula to determine the axial load capacity of an L=2.5m long pin-ended column of hollow circular section of 114mm outside diameter and 5mm wall thickness. Assume that E = 205 kN/mm2 and Y = 275 N/mm2
i) Show that the properties of this section are: second moment of area, I = 2.548 106mm4, cross sectional area A = 1712mm2, radius of gyration, k = 38.6mm. and slenderness ratio 80.64/ == kL
(4 marks)
ii) Using Perry-Robertsons formula (assume a = 2), calculate the critical load. What is the physical interpretation of this critical load?
(10 marks)
Continued Overleaf /
5
May-2008 Q4. a) Consider real struts:
i) Draw the load/deflection curve for a real material, as seen in class. Clearly label the different parts of the curve
(4 marks) ii) Explain the material behaviour for each part.
(6 marks)
b) For the continuous beam in Fig. Q4(b) i) Use the Hardy Cross Method (Moment Distribution) to calculate the
bending moments at all the joints. (15 marks)
ii) Draw the bending moment diagram and shear force diagram (8 marks)
2m
(EI)
50 kN90 kN
2m
C
2m(EI)(2EI) D
3m
BA
3m
20 kN/m
Fig. Q4(b)
END
(+ Data Sheet)
6
Structural Mechanics 2
DATA SHEET
Perry-Robertson formula
( )2120 2.0 YE pi = and ( )0001.0 = a ( ){ }2
1 EY ++=
YE = 2
7
SOME STANDARD BEAM DEFLECTION FORMULA
CANTILEVERS
Loading Beam Displacement
vB at tip B +ive
Slope B at tip B +ive
End moment
M
L A B
EIML2
2
EIML
General moment
M
L a b A B
( )EI
bLMa2
+
EIMa
End point load
W
L A B
EIWL3
3
EIWL2
2
General point load
W
L a b
A B
( )EI
bLWa622 +
EIWa2
2
Uniformly distributed load
q per unit length
L A B
EIqL8
4
EIqL6
3
Partial uniformly distributed load
q per unit length
L a b A B
( )EI
baLLqb24
62 323 ++
( )EI
baLqb6
3 2+
Linearly distributed load
q per unit length
L A B
EIqL
30
4
EIqL
24
3
8
SIMPLY SUPPORTED BEAMS
Loading Beam
Maximum deflection vmax at D
+ive
Central deflection vC
+ive
Slope A at A
+ive
Slope B at B
+ive
Equal and opposite end moments
M M
L
A B
EIML8
2
EIML8
2
EIML2
EIML2
One end moment
M
L
D
A B
x
LLx 577.03
==
EIML
39
2
EIML16
2
EIML3
EIML6
General moment
M
L
A B
a b
-
,for ba > ( )EI
bLM16
4 22
( )EIL
bLM6
3 22
( )EIL
aLM6
3 22
Central point load
W
L/2 L
A B
EIWL48
3
EIWL48
3
EIWL16
2
EIWL16
2
Off-centre point load
W a b
D
L
E x
A B
EILbWa
v3
,Eatloadunder22
=
( )3
,for
bLax
ba+
=
>
( )EIL
xbLWab9
+
,for ba > ( )EI
aabLWb48
48 22
( )EIL
bLWab6
+
( )EIL
aLWab6
+
Uniformly distributed load
L
q per unit length A B
EIqL
3845 4
EIqL
3845 4
EIqL
24
3
EIqL
24
3
Partial uniformly distributed load
L
q per unit length
a b
A B
,for ba > ( )EIL
Lbqb96
32 222
( )EIL
Lbqb24
2 222
( )EIL
baLqb244 22 +
Linearly distributed load
E
q per unit length
0.519L D
L L/3 or 0.577L
A B
EIqL
76801.5 4
EIqL
7685 4
EIqL
3608 3
EIqL
3607 3
9
PROPPED CANTILEVERS
Loading Beam Central
deflection vC +ive
Slope B at B +ive
End moment
M
L
D 2L/3 A B
EILM
2
20
EILM 0
General moment
M
L
D 2L/3
Central point load L/5 or 0.447L
L/2W
L
D
EIWL3
3
EIWL2
2
General point load
L/5 or 0.447L
a W
L
D
Uniformly distributed load
E
q per unit length
L
0.578L 5L/8 or 0.625L
D
EIqL8
4
EIqL6
3
Partial distributed load
E
q per unit length
L
0.578L 5L/8 or 0.625L
D
Linearly distributed load
L/5 or 0.447L
q per unit length
L
D
EIqL
30
4
EIqL
24
3
10