# ME2114 Solution 1011-1

September 24, 2017 | Author: OliverQueen | Category: Bending, Beam (Structure), Structural Engineering, Deformation (Mechanics), Physics & Mathematics

#### Short Description

ME2114 exam paper Solution...

#### Description

ME2114

NATIONAL UNIVERSITY OF SINGAPORE

ME2114 – MECHANICS OF MATERIALS II (Semester II : AY2010/2011) Time Allowed : 2 Hours

INSTRUCTIONS TO CANDIDATES:

1.

This examination paper contains FOUR (4) questions and comprises SIX (6) printed pages.

2.

Answer ALL FOUR (4) questions.

3.

Answer questions 1 and 2 in one booklet and questions 3 and 4 in another booklet.

4.

All questions carry equal marks.

5.

This is a CLOSED-BOOK EXAMINATION.

6.

Programmable calculators are NOT allowed for this examination.

PAGE 2

ME2114

QUESTION 1 A steel column ( L = 3000 mm) loaded eccentrically by a point load 50 kN as shown in Figure 1 is fixed at its base and free at the top. The column has a T-section and is made of steel which has a Young’s modulus of 200 GPa. Determine the maximum tensile and compressive stresses in the column. (25 marks) The following equation (usual notations apply) may be used without derivation: The secant formula for a pinned-pinned column is given by:

P A

 max   1 

ec  L P   sec r2  2r EA 

50 kN

x

y All dimensions in mm y Point load 50 kN x

x

120

x

3000 20 20

120

Column cross-section

Figure 1

PAGE 3

ME2114

QUESTION 2 A beam made of an elastic-perfectly plastic material has a tensile and compressive yield stress of 250 MPa. It has a cross-section as shown in Figure 2. Determine: (a)

the bending moment about the z-axis required to cause first yield in the beam, and (10 marks)

(b)

the bending moment required to cause the beam to be fully plastic and also the shape factor of the beam. (15 marks)

y

100

130

z

250

All dimensions in mm

35

Figure 2

PAGE 4

ME2114

QUESTION 3 The doubly-bent rod abc of circular cross-section in Figure 3 is fixed perpendicularly to a wall at a and loaded by a force P at c. The wall is normal to the x-axis and the force acts in the yz-plane. Determine the displacement of c in the x, y and z directions (i.e. ux, uy and uz). Consider strain energy due to bending and torsion only. Use E and G for the Young’s and shear moduli, and I and J for the second moment of area and polar moment of area, respectively. (25 marks)

L a b L c P  x y z

Figure 3

PAGE 5

ME2114

QUESTION 4 Figures 4a and 4b show a truss system comprising 4 truss members of cross-sectional area A and Young’s modulus E.

(a)

In Figure 4a, a load P is applied vertically downwards at d. Determine the stiffness matrix [K] in [K][u]=[f], if

uc  v  u   c  , u d    vd 

and [f] is the force vector; uc and vc represent the x and y displacements of joint c, and ud and vd represent the x and y displacements of joint d. (17 marks)

(b)

In Figure 4b, joint d is constrained from displacing horizontally. Determine the displacement of c if d is pulled downwards a small distance . (8 marks)

y

y

c

c

x

x L

L d

d a

a L

L

P L b

Figure 4a

L b

Figure 4b

PAGE 6

ME2114

LIST OF EQUATIONS Unless otherwise stated, the following expressions may be used without derivation. All symbols have their usual meaning.

1.

Strain energy in slender rods due to axial loads 1 N2 1  u  dx   EA  dx  2 0 EA 2 0  x  L

U

2.

2

L

Strain energy in slender rods due to bending 2

L L   2v  1 M2 1 U  dx   EI  2  dx 2 0 EI 2 0  x 

3.

Strain energy in slender rods due to torsion L

1 T2 U  dx 2 0 GJ

4.

Stiffness matrix of a truss member

u  c2  k local   k  cs2 c    cs

v

u

cs  c 2 s2  cs  cs c 2  s 2 cs

where c  cos  , s  sin 

v  cs  u   s 2  v cs  u   s 2  v  and k 

EA L

- END OF PAPER -

-L

pg..

NATIONAL UNIVERSITY OF SINGAPORE Department of Mechanical Engineering

M"Gltl'tE;ME 2, ,t

Module Code: _ _ _ _Module Title:

Solution to Question No. --=-­

Point load 50 kN

120

A,.-­ -­

x ----------­

r---t--....::;.,-----I

--------- x

~~~~-~~~~~~~r~--r20 A~ A, = (d-o -i;ro

A

'K."Po ==-;;L1-o

(d-6

2-:::

A"A ~ :::- tm ~lte WloWtfJllt"

(A,-t-f!t2-) -;:

= 2trov

Wp n'I v

VVl r'l1 ").­

V1-1J'h)..

:J 'j

D.b~

= AI (/ ihho) T A2. (60

4~;:

::

d

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