This examination paper contains FOUR (4) questions and comprises SEVEN (7) printed pages.
2.
Answer ALL FOUR (4) questions.
3.
All questions carry equal marks.
4.
This is a CLOSED-BOOK EXAMINATION.
5.
Answer Questions 1 and 2 in one booklet and Questions 3 and 4 in another booklet.
6.
Programmable calculators are NOT allowed for this examination.
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ME2113
QUESTION 1 (a)
A beam of flexural rigidity EI is loaded by a uniformly distributed load of intensity w as shown in Figure 1a. Using a graphical method, show that the tangential deviation (tAB) of point A from the tangent at point B is given by (usual notations apply):
t AB = ∫A x B
M dx EI
Hints: Consider a segment CD of length dx on the beam deflection curve and construct the tangents at points C and D. (10 marks)
(b)
A beam AB of length 2L and flexural rigidity EI is fixed at A and is loaded by a uniformly distributed load of intensity w over the right-hand half of the beam as shown in Figure 1b. Draw the M/EI diagram along the length of the beam and determine the slope and deflection at point B using the moment-area method. (15 marks)
Y
Load intensity w per unit length
X A
x
D
C
B
dx
Figure 1a
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ME2113
Y
Load intensity w per unit length
X A
L
B
L
Figure 1b
The following information (usual notations apply) may be used: The area A0 under the parabolic curve PQ in Figure 1c is given by:
wL3 A0 = − 6 EI
M/(EI) L/4 P
X Centroid
-wL2/2EI
Q L
Figure 1c
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ME2113
QUESTION 2 A beam AB simply supported at A and B is loaded by a trapezoidally-distributed load and a concentrated moment M0 = 10 kNm at the mid-span as shown in Figure 2a. The intensity of the distributed load varies linearly from 30 kN/m at support A to 60 kN/m at support B. (a)
Draw the shear force distribution along the length of the beam, showing the values of shear force at x = 0, 0.8 m, and 1.6 m only (you need not determine the intermediate shear force values). (7 marks)
(b)
If the beam has a hollow circular cross-section with inner radius R1 = 90 mm and outer radius R2 = 100 mm as shown in Figure 2b, determine the maximum shear stress in the beam. (18 marks)
The following information (usual notations apply) may be used: 1 Fxy Ay b I
τ xy = .
(i)
The shear formula is given by:
(ii)
For a semi-circular area of radius R, the centroidal distance y (as shown in Figure 2c) is given by: y=
4R 3π
60 kN/m
R1 = 90 mm R2 = 100 mm
30 kN/m B
A M0 0.8 m
0.8 m Figure 2b
Figure 2a
Centroid of semi-circular area y=
R Figure 2c
4R 3π
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ME2113
QUESTION 3 (a)
Show that the solid circular rod of radius r under combined bending moment M, and torsion T, in Figure 3a will not yield according to Tresca criterion if,
where σy is the yield stress of the rod material. (12 marks)
(b)
Figure 3b shows a right-angled L-shaped rod ABC clamped perpendicularly to a wall at A. A force P=300N is applied at C at an angle θ as shown. The force P lies on a plane parallel to the wall. Using the expression in (3a) or otherwise, determine the range of θ that will not cause the rod to yield according to Tresca criterion. The rod has a circular cross-section of radius 20mm and a yield stress of 100MPa. (13 marks) M
M
T
T
Figure 3a
2m 1m
A L
B C P
θ
Figure 3b
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ME2113
QUESTION 4 An axial force P is applied to a simply-supported beam-column of length L at a distance of e directly above the centroid of its cross-section as shown in Figure 4. The maximum deflection occurs at B. Given that the beam-column has a square cross-section of dimension , and that where , determine (a)
the reaction forces at A and C, (5 marks)
(b)
the value of α and the deflection at B in terms of e and L, and (10 marks)
(c)
the maximum compressive stress at where the maximum deflection occurs in terms of P, e and a, taking into consideration the beam deflection. (Hint : Sketch the free body diagram of a section of the deformed beam-column from one end to the where the maximum deflection occurs) (10 marks)
P A
B
e C
αL
(1−α)L Figure 4
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ME2113
LIST OF EQUATIONS Unless otherwise stated, the following expressions may be used without derivation. All symbols have their usual meaning.
Deflection of beam-columns Qc Q sin kc x≤L−c Pk sin kL sin kx − PL x , v= Q sin k ( L − c) x≥L−c Q( L − c) sin k ( L − x) − ( L − x) , PL Pk sin kL
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