Your UC is made up from the product of the alphabetical

Uniqueness Code (UC) – Your UC is made up from the product of the alphabetical
position of the first letter of your forename and surname respectively. Fred Smith has
the UC = 6 x 19
= 114
because F is the 6th letter of the alphabet and S is the 19th letter of the alphabet.
Uniqueness Code calculation
First letter of your forename is Position in Alphabet
First letter of your surname Position in Alphabet
Your Uniqueness Code =
S 19
S 19
361
Q1 Determine, using a table to lay out your calculations, the position of
the centroid for the cross sectional area shown in Figure Q1. Quote x
bar and y bar in mm from the bottom left hand corner of the area.

All dimensions in m
2xUC

10xUC

19xUC

2xUC
1xUC

7xUC

Figure Q1
Q2 The beam shown in Fig. Q2 below has a pinned support at A and roller
supports at B and C. Note that an internal pin is located at point P. Determine
the reactions at A, B and C when the beam is loaded as shown.
8xUC (kN/m)
Pin “P”
15xUC (kN)
A
B
C

45o

4 m 2 m 3 m 2 m
Figure Q2
Q3

(a)	Calculate the reactions HA, VA and VB for the pin jointed plane frame shown in Figure Q3(a).

(b) Calculate the magnitude and sense of the axial force in members AC,
AF, CD, CF, FD and FB using the method of joint resolution
3xUC (kN) 2xUC (kN)
C D E
3 m
F
A B
3 m 3 m
Figure Q3(a)

(c)	Use the method of sections to obtain the forces in the members DE, DB and FB.

ANSWER ALL FOUR QUESTIONS
Q4. A compound beam consists of a standard I section 356 mm deep and 171 mm
wide with a steel plate 200 x 15 mm deep welded to each flange. Given that
the beam is simply supported over a length of 7 m, Ixx for the standard I
section alone is 200 x 106 mm4 and the maximum allowable bending stress
σmax is 165 N/mm2 then:

(i)	Sketch the cross section and calculate the second moment of area (Ixx) for the compound section.

(ii)	Calculate the maximum uniformly distributed load (UDL) the beam can carry if the flange plates extend over the whole 6 m span.

(iii)

With reference to the bending moment diagram explain why the 200 x
15 mm plates are not required over the whole span.

Q5
1xUC (kN)
(a) The beam shown in Fig. Q5(a) below has a pinned support at A and roller
supports at B and C. Note that an internal pin is located at point P. Determine
the reactions at A, B and C when the beam is loaded as shown.
(b) The beam shown in Fig. Q5(b) is simply supported and subjected to the loads
shown. If the reactions VA and VB are 51 kN and 78 kN respectively,
find the distance from the left hand support “A” to the point of
contraflexure.
Figure Q5(b)

3.5 m	75 kN
B

P
60 kN
A

45o

6 m 2 m 4 m 2 m
Figure Q5(a)
A

2 m

4 m
2	m

25 kN

15 kN/m

7 kN/m

Q6 (a) Calculate the reactions HA, VA and VB for the pin jointed plane frame
shown in Figure Q6(a).
30 kN 40 kN
20 kN
3 m
A B
3 m 3 m
Figure Q6(a)
(b) A new load condition is placed on the pin jointed plane frame as
shown in Figure Q6(b). Hence, calculate the magnitude and sense of
the axial force in members AC, AF, CD, CF, FD and FB using the
method of joint resolution given that the vertical reactions VA and VB
are both 75 kN and the horizontal reaction HA is zero.
75 kN 75 kN
C D E
3 m
F
A B
3 m 3 m
Figure Q6(b)

(c)	Use the method of sections to check the values obtained for the forces in the members CD, FD and FB.

Q7. Fig. Q7 shows a beam resting on simple supports which is loaded by a
combination of uniformly distributed and point loads.
Fully analyse the beam and draw the shear force and bending moment
diagrams clearly indicating the position and magnitude of the maximum
bending moment.
4 3 2 3.5
All dimensions are in m
Figure Q7

10 kN/m

25 kN

15 kN/m
A

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