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