Structural analysis and mechanics 3 Exam January 2015, questions PDF

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Structural Analysis and Mechanics 3 (14CVB008) January 2015

Two hours

Candidates should answer FOUR questions, TWO from Section A and TWO from Section B All questions carry equal marks Candidates may use any Loughborough University approved calculator Three data sheets are provided

SECTION A Q1. A steel truss acting as a bracing system is shown in Figure Q1. Assuming that the same steel section is used for all members, and that the Elastic Modulus E of steel is as given in the figure: Calculate the area of the smallest steel section such that the maximum horizontal deflection at joint F does not exceed 3.0mm. [25 marks]

Q2. a)

In a flexibility analysis what is meant by the term fij.

[2 marks]

b)

The frame shown in Figure Q2b is to be analysed using the Compatibility (Flexibility) method assuming that the redundant forces are R1 the horizontal reaction at joint A, and R2 the moment reaction at joint D:

(i)

Carry out a qualitative analysis to establish the flexibility equations that correspond to the released displacements. Use diagrams to indicate clearly all displacements involved. DO NOT PERFORM ANY CALCULATIONS. [14 marks]

(ii)

Draw ALL bending moment diagrams required for the calculation of the displacements.

[7 marks]

Calculate, in terms of EI, the flexibility coefficient f12.

[2 marks]

(iii)

Q3. Use the Moment Distribution method to determine the moments in the structure shown in Figure Q3. Draw the final bending moment diagram. [20 marks] Calculate, in terms of EI, the vertical deflection of point B at the middle of span AC. [5 marks]

Continues/… 1

…/continued SECTION B Q4. The height of the cantilevered column of Figure Q4 is L = 4.00 m. The material is reinforced concrete, with Young’s modulus E = 18.5 GPa and compressive strength σc = 30 MPa. The cross section is a T section with breadth b = 45 cm, depth d = 30 cm and thickness t = 15 cm, as shown within Figure Q4. (i)

Use the Rankine’s formula to get an estimate of the failure load PR of the column. [20 marks]

(ii)

What would be the Rankine’s failure load if the horizontal movements at the top of the column were restrained? [5 marks]

Q5. Figure Q5 shows a Z thin-walled cross section with unsymmetrical flanges and constant thickness t = 4 mm throughout. All dimensions are shown in the same figure, where point G is the centroid of the cross section. (i)

Determine the principal values Imin and Imax of the second moment of area. [16 marks]

(ii)

Determine the inclination of the principal axes of inertia with respect to the reference system of axes Gyz shown in the figure. Use a drawing (not necessarily to scale) to clearly indicate the strong axis and the weak axis. [9 marks]

Q6. A steel plate (Young’s modulus E = 210 GPa and Poisson’s ratio  = 0.30) has a welded connection forming an angle  = 30 with the horizontal axis, as shown within Figure Q6. The plate is subjected to plane state of stress, with  x = 10 MPa, z = 4 MPa and

 xz = 3 MPa. (i)

Construct the Mohr’s circle of the stress in the steel plate.

[6 marks]

(ii)

Use the Mohr’s circle to evaluate the magnitude and direction of the principal stresses min and  max in the element. Draw the state of stress in a properly oriented stress element. [7 marks]

(iii)

Use the same Mohr’s circle to determine the normal stress n perpendicular to the weld and the shear stress nm parallel to the weld. Draw the state of stress in a properly oriented stress element. [7 marks]

(iv) Calculate the maximum normal strain max and the maximum shear strain max in the element. [5 marks] Dr J El-Rimawi Dr Alessandro Palmeri

2

Structural Analysis & Mechanics 3 14CVB008 January 2015 24kN E

F 3.0m

Elastic Modulus E= 200kN/m2

12kN D

A

3.0m

C

B 4.0m

Figure Q1

24kN

3.0m

1.0m

Same EI for all

A

4.0m

C 4kN/m

3.0m

B

D 6.0m

Figure Q2b

15kN

8kN/m A

B 2.0m

15kN

C

4kN/m D

2.0m

2.0m 6.0m

4.0m

Figure Q3

3

E

2.0m 2.0m

Structural Analysis & Mechanics 3 14CVB008 January 2015 PR  ?

b

t

dG  ? G

L 4m

y d E  18.5 GPa   c  30 MPa   

 b  45 cm   d  30 cm  t  15 cm 

z t

Figure Q4

yG

b

t

zG

d G y

     

b  18 mm d  66 mm B  48 mm t  4 mm

yG  7.5 mm   zG  40.5 mm    

z B

Figure Q5

4

Structural Analysis & Mechanics 3 14CVB008 January 2015

x

z  xz z

x       

 x  10 MPa E  210 GPa  z  4 MPa   0.30  xz  3MPa   30 o

x

 nm  ? 

      

n  ?  xz z

Figure Q6

5

Structural Analysis & Mechanics 3 14CVB008 January 2015 DATA SHEET 1 (For use with Section A) L

 M k dl

Product Integrals

v

0

Virtual Mv ( x ) 

I. d

c

Real k (x ) 

II.

III.

c

c

L

L

IV.

c L

L/2

L/2

1. b

a L

L [ a( 2 c  d )  b( 2 d  c)] 6

L c( a  b ) 2

L c( 2 a  b ) 6

L c( a  b ) 4

L a( c  d ) 2

Lac

L ac 2

L ac 2

L a( 2 c  d ) 6

L ac 2

L ac 3

L ac 4

L b( 2 d  c ) 6

L bc 2

L bc 6

L bc 4

L a( c  d ) 3

2L ac 3

L ac 3

5L ac 12

2. a L

3. a L

4. b L

5.

a L

Mv k

= Moment due to virtual load = Curvature due to real load

6

Structural Analysis & Mechanics 3 14CVB008 January 2015 DATA SHEET 2 (For use with Section A)

Member End Moments (Clockwise +ve) I. Both Ends Fixed

End Moment   Load Case 

At End A (MA)

At End B (MB)

PL 8

 PL 8

Pab 2 L2

Pba 2  2 L

wL2 12

wL2  12

1. P A

B L/2 L

2. P

a 3.

b

w /m

4.

6 EI 

 2 L

6 EI

4EI  L

2 EI  L

2

L



5.



II. Far End (end B) Pinned M

M  MA 

7

MB 2

where MA and MB as above.

Structural Analysis & Mechanics 3 14CVB008 January 2015 DATA SHEET 3 (For use with Section B) Elastic Constitutive Law

Jourawski’s Formula



Q V Q  V  ; bI b 2 A

x 





Buckling Load

 2 E I min  2 E PE   2 ; L2e  PR 



1    y    z ; E x E  1          x  x y z 1   1  2 

I A





 xy 

Le



G

PE Pc PE  Pc

Unsymmetrical Bending

 xy



G E

2 1 







;

 xy  G  xy



Mohr’s Circle for Stress State !x

z

My

! x

G y

! xz

x

! xz

z

z



Mz



X   x , xz

 min

!

x   y   z ;

 M I  My I yz    z yy 2 Iyy Izz  Iyz   My Izz  M z I yz     Iyy Izz  I2yz 



Z   z , xz !

 ave



 max 



Mohr’s Circle for Strain State



I mn



Y  Iyy ,I yz !

Imin I!ave



Z  I zz ,Iyz

!2



   X   x , xz  2  ! 

 min



! av e

I!max I!mm



! max



   Z   z , xz  2  



DATA SHEET 3 Continues/… 8

Structural Analysis & Mechanics 3 14CVB008 January 2015 …/ DATA SHEET 3 Continued

Strain-Gauge Rosette 60°Rosette

45° Rosette !x

!x !90

!9 0 z

z

45

! 15 0 !45 !45

! 3 0 60 60 30

0

 xz  0   90  2 45

   2  x    30  90   150  3 2    2   150  30  xz  3 



9

...