Exam January Autumn 2018, questions PDF

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CIV2100

Data Provided: Tables of fixed-end moments (pages A1-2)

DEPARTMENT OF CIVIL AND STRUCTURAL ENGINEERING

Structural Analysis 1

Autumn Semester 2017-18

2 Hours

Answer ALL THREE questions. All questions carry equal marks.

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1

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CIV2100

Q1

The continuous beam shown in Fig. Q1, whose cross-section is uniform throughout, rests on knife-edge supports which restrain uplift at A, B, C and D.

Slope-Deflection Analysis is to be used to find the bending moment diagram.

(A) Briefly state the information which must be noted about the structure before its slope-deflection equations can be used. Give numerical values where appropriate (5 marks) (B) (i) Find the end moments for each span under the load case shown. (8 marks) (ii) Sketch the bending moment diagram, giving values under the point loads and at supports. (5 marks) (iii) Sketch the deflected shape, indicating curvature directions and points of contraflexure. (2 marks)

250kN A

B

D

C

4m 12m

50kN/m

250kN

4m 12m

10m

Fig. Q1

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2

CONTINUED

CIV2100

Q2.

The elastic portal frame shown in Fig. Q2 is of uniform section throughout. Use the Moment-Distribution Method to find the end-moments of each of the members of the frame to within about 1-2% accuracy when it is loaded as shown by the midspan point load on the beam BC. (A) Very briefly describe the sequence of steps in the process used on a problem of this type which includes the possibility of sway deflection. (5 marks) (B) (i) Use the process described in (A) to find the relevant member end moments. (10 marks) (ii) Sketch the bending moment diagram approximately to scale, with values at member ends and the beam mid-span shown. (5 marks)

C

B

50kN

6m

A

4m D 7m

7m

Fig. Q2

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Q3. Fig. Q3 shows a beam of 4 m length subjected to a vertical downward force and an anti-clockwise moment at node 3. At its left hand end (at node 1), the beam is fully clamped. At the middle (at node 2), the beam is pin-supported. At its right hand end (at node 3), the beam is free to move. The 4x4 element stiffness matrix for a beam element, assuming that axial displacements and forces in the beam can be neglected, is:  12  L3  6  2 K  EI  L  12  L3  6  2  L

6 L2 4 L 6  2 L 2 L

 

12 L3 6

L2 12 L3 6  2 L

6  L2  2   L  6  2 L  4   L 

Take the length of the first element 𝐿1 = 2 m, the length of the second element 𝐿2 = 2 m, 𝐸𝐼 = 8x106 Nm2 , the applied moment 𝑀 = 10x103 Nm and the applied force 𝐹 = 15x103 N. Be sure to specify units for numerical answers. (A) Draw a sketch of the deformed structure along with a sketch of the undeformed structure. (1 marks) (B) Write down the assembled global stiffness matrix for this structure. (4 marks) (C) Write down the boundary conditions (for example θ2=0, δ 4=0, F3=10) and then write the reduced system. (4 marks) (D) Resolve the unknown degrees of freedom and compute the reaction forces and moments. (8 marks) (E) Check the vertical and rotational equilibrium. (2 marks) (F) Now, imagine the beam is pin-supported at its left end in Fig. Q3 (i.e. when the beam is pin-supported at nodes 1 and 2 and is free at node 3). If the applied force and moment at node 3 does not change, comment on the size of the reduced system stiffness matrix. (1 mark) NB. Figure Q3 can be found on the next page.

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CONTINUED

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F M (2)

(1)

①

L1 =2 m

②

L2 =2 m

③

Fig. Q3

END OF QUESTION PAPER

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DATA SHEET FIXED-END MOMENTS: General Cases All F.E.M. values = (Formula x Total Load W x Span l) Load Case

MFAB

l  ( 1  )

W

F MBA

2

F MAB

l

l

W

F MBA

2



l

F MAB

l 

W total

 ( 3  2 ) 12

l

30

F MBA

W total

 ( 10  10  3 )

l 

15

F MBA

W total

 ( 10  15  6  2 )

l

F MBA

( 1   )( 1  3  )



 2 ( 5  4 )

 ( 2  3 )

Wl

F MAB

l

 ( 1  )

 2 ( 5  3 )

10

l

F MAB

 ( 3  2 )

30

l

F MAB

( 1  2  2 2 ) 12

12

l 2



F MBA

l

F MAB

 2 ( 4  3 ) 12

l

F MAB

( 1  2  2 ) 12

F MBA

W

 ( 6  8  3 2 ) 12

 2(1 )

l

l 

M FBA

W/2

2

W/2

l

F MBA

 ( 1  ) 2

F MAB

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A1

CONTINUED

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FIXED-END MOMENTS: Special Cases All F.E.M. values = (Formula x Total Load W x Span l) Load Case

M FAB l/2 1

W

F MBA

1

8 F MAB

l

1

12

F MAB

l l/2

 11

W total

 23

W total 240

F MAB

W total

F MAB

12

3

F MAB

(1 

1 ) n2

40

F MBA

l

15

F MBA

1

(parabolic)

10

W/2n

1

F MBA

1

10

W total 1

240

l

W total 1

7

15 F MAB

96

F MBA

l

l/2 2

5

l l/2

12

F MBA

96 F MAB

8

F MBA

W total 1

M FBA

l

W/n equal spacing W/2n F MBA l/n

1

12

10

( 1

1 ) n2

F MAB

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A2

END...