1. A timber beam is used for the structure. Knowing []=10MPa for timber, design the size for the beam for (1) a rectangular cross-section with breadth b and height h in the ratio of h=2b =2b; (2) a square cross-section with side length a; (3) a circular cross-section with diameter d .
A
0.5m
420N
300N
C
D
0.5m
0.5m
B
Solution:
190 10 3 95 3 10 3 (1) I z bh b ， b 2 4 I z 2 12 3 b3 b 3 b 30.5mm 1 4 M a 190 10 3 a 190 6 10 3 （2） I z a ， 1 4 12 I z 2 2 a3 a 12 b 48.5mm 1 M d 190 10 3 d 190 32 10 3 4 （3） I z d ， 1 64 I z 2 2 d 3 4 d 64 b 57.8mm 1
3
2
4
M h
2. A W15018-shape beam is to be used for the structure shown in the following figure. For the rolled-steel we have [ ]=170MPa and []=100MPa. From the handbook we can find d =153mm, =153mm, t w=5.8mm, -6 4 -4 3 J z=9.210 m , S z=1.210 m . Determine the allowable intensity of load in the middle of the beam.
f b f
C A
t f
D
2m
3.5m
2m
B
t w
z
d
Solution: M diagram：
V diagram：
M d
J z 2 V S z
J z t w
5.03125 f 10 3 153 9.2 10 6
2
1.75 f 1.2 10 4 9.2 10
6
5.8 10
3
170 f 4.06 kN m
100 10 6 N m 2 f 25.4 kN m
f 4.06 kN m
3. A steel beam (305x165x40) (305x165x40) is to be used for the structure shown in the following Figure. From the handbook we can find: d 305mm , b f 165mm , t w t f 40 mm , J z 8.551 10 5 m 4 , S z 5.63 10 4 m 3 . Calculate the maximum flexure stress and shear stress of this beam.
12kN
b f
36kN/m
t f B
C
A
D
0.8m
E
1.8m
1.2m
t w z d
1.6m
Solution: M diagram：
V diagram：
M d J z
2
V S z J z t w
54.48
0.305
8.551 10
5
2
37.3 5.63 10 4 5
8.55110 0.04
97.2
kN mm
2
6140 kN m 2 6.14 kN mm 2
4. A 10 kN.m couple is applied to a wooden beam, of rectangular cross section 15 by 30 cm, in a plane forming an angle of 45°with the vertical, as shown in Fig. 4. Determine (a) the maximum stress in the beam, (b) the angle that the neutral surface forms with the horizontal plane.
Solution: (a) Dividing the couple into the following two components 1 I bh 3 33750 cm 4 z M z 10 cos 45 7.07 kN.m 12 M y 10 sin 45 7.07 kN.m
I y
1 12
hb 3 8437.5 cm 4
The largest tensile stress due to Mz occurs along AB and is M z y 7.07 10 6 150 3.1422 MPa 1 I z 33750 10 4 The largest tensile stress due to My occurs along AD and is M y z 7.07 10 6 75 2 6.2844 MPa I y 8437 .5 10 4 The largest tensile stress due to the combined load action at A max 1 2 9.4266 MPa (b) Angle of Neutral surface with horizontal plane I 33750 tan z tan tan 45 4 I y 8437.5 75.96
5. A cantilever beam with a uniform load as shown in the following figure has a height h equal to 1/10 of the length L. L. the beam is a steel wide-flange section with E with E =208GPa =208GPa and an allowable bending stress of 130 MPa in both tension and compression. Calculate the ratio δ /L of /L of the deflection at the free end to the length, assuming that the beam carries the maximum allowable load. A
B
L
q Solution The maximum moment at the fixed end is
1
max
qL2 2
The allowable bending stress in tension or compression is max Since y h / 2 L / 20 , 1 max
2
qL2
qL3
y
max
I
L
3 20 qL 130 Pa I 40 I
5200 I Since the deflection at the free end when the beam carries the maximum allowable load is qL4 8 EI qL3 5200 1 Therefore 3 L 8EI 8 208 10 320 Thus
6. For the beam and loading shown in the following figure, determine (a) the deflection at the midspan of portion BC portion BC of of the beam, (b) the slope at the supporting point B.
Solution
Using ABC as a free body M C 0 ;
wL 3L L ( ) R B L ( wL)( )( ) 0 5 2 2 4 R B wL 5 For portion BC only, (0
wL L 4 x ( x) wLx ( wx)( ) M 0 5 2 5 2 3 1 1 M wLx wx 2 wL2 5 2 10
EI EI
d2y dx dy dx
2
3
1
1
5 3
2 1
10 1
wLx wx 2
10
wLx 2
wx 3
wL2x C 1
10 1 1 1 EIy wLx 3 wx 4 wL2x 2 C1x C 2 10 24 20 C2=0 x 0, y 0 ; 0=0-0-0+0+C 2
x L, y 0 ;
6
wL2
0(
1 10
1 24
1 20
) wL4 C1L 0
C1=
1 120
Elastic curve w 1 1 4 1 2 2 1 3 y ( Lx 3 x Lx L x) EI 10 24 20 120 dy w 3 1 1 1 3 ( Lx 2 x 3 L2 x L) dx EI 10 6 10 120 (a) y at x=L/2 w 1 L 3 1 L 4 1 2 L 2 1 3 L wL4 y M L ( ) L( ) ( ) L ( ) EI 10 2 24 2 20 2 120 2 EI
wL3
1 1 1 1 80 384 80 240 13wL4 1920 EI
dy (b) at x=0; dx dy w 1 3 wL3 (0 0 0 L ) dx B EI 120 120 EI
7. The wooden T-beam shown in the below Figure is made from two 200mm×30mm boards. If the allowable bending stress is σ allow allow=12 MPa and the allowable shear stress is τ allow allow=0.8 MPa, determine if the beam can safely support the loading shown. 1.5kN
0.5kN/m B
C
2m
Solution: V max and M max max=1.5kN and M max=2kNm Bending stress
D
2m
The neutral axis will be located from the bottom of the beam. yA 0.1575m y A
Thus I 60.125(10 6 ) m 4 Since c=0.1575m, we require allow
Thus
12(103 )kPa
M max c
2 0.1575 6
60.12 .125(10 )
I
5.24(103 )kPa (ok)
Shear stress Maximum shear stress occur at the neutral axis and for simplicity, we will use the rectangular area below the neutral axis to calculate Q, we have 0.1575 Q [(0. [(0.15 157 75)(0. (0.03)] 03)] 0.3 0.372( 72(103 )m3 2 So that V Q allow max It 1.5 0.372(103 ) 309kPa (ok) 800kPa 60.1 60.125 25((106 )(0. )(0.03 03))
8. A W250×45 wide-flange beam DB has the cross-sectional dimensions and loading shown in the below Figure, E= Figure, E=210 210 GPa, GPa, d =266 =266 mm,, mm,, t w=7.62 mm, mm, 6 4 3 3 J z =71.1×10 =71.1×10 mm , S z =535×10 =535×10 mm , the load q is 36 kN/m. Determine the deflection at point C
Solution: Supporting forces: q 9 9 / 2 27 F A q 6 4 9 F B q 9 F A q 4
The bending moment diagram and shear force diagram are shown in the above figure. We can find that the maximal moment and shear force are 9 / 2 q and 15 / 4 q , respectively. The bending moment of beam BA 9 q m x qx x 2 4 2 The deflection of beam BA q 9 q q 9 1 y ( qx x 2 )dx C 1 x C 2 ( x 3 x 4 ) C 1 x C 2 EJ z 4 2 EJ z 24 24
y (0) 0 , C 2 0 y (6) 0 , C 1
9 q 2 EJ z
So, the deflection of point C is q 9 1 9 yC y (3) ( 33 3 4 3) 0.001989 m EJ z 24 24 2 9. For the beam and loading shown in the figure below, determine (a) the slope at end A, (b) the maximum deflection. (10 marks)
Solution: Differental Equation of the Elastic Curve
Integrating the equation above twice:
Boundary Conditions From both ends of the beam, we know
and
, thus:
Integrating the equation above twice:
And the boundary conditions show that Equation of Elastic Curve (a) the slope at end A For , we have
Thus we have
(b) the maximum deflection For , we have
Thus we have
and
