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UNSYMMETRICAL BENDING
Moment applied along principal axis
dAyMM
dAzMM
dAFF
AZZR
AyyR
AxR
0 ;
0 ;
0 ;
If y and z are the principal axes. ∫ yz dA = 0(The integral is called the product of inertia)
UNSYMMETRICAL BENDING (cont.)
• Moment arbitrarily applied
• Alternatively, identify the orientation of the principal axes (of which one is the neutral axis) • Orientation of neutral axis:
y
y
z
z
I
zM
I
yM= +
= +
tantany
z
I
I
EXAMPLE
The rectangular cross section shown in Fig. 6–33a is subjected to a bending moment of 12 kN.m. Determine the normal stress developed at each corner of the section, and specify the orientation of the neutral axis.
EXAMPLE (cont.)
• The moment is resolved into its y and z components, where
• The moments of inertia about the y and z axes are
Solutions
mkN 20.7125
3
mkN 60.9125
4
z
y
M
M
433
433
m 10067.14.02.012
1
m 102667.02.04.012
1
z
y
I
I
EXAMPLE (cont.)
• For bending stress,
• The resultant normal-stress distribution has been sketched using these values, Fig. 6–33b.
Solutions
(Ans) MPa 95.4102667.0
1.0106.9
10067.1
2.0102.7
(Ans) MPa 25.2102667.0
1.0106.9
10067.1
2.0102.7
(Ans) MPa 95.4102667.0
1.0106.9
10067.1
2.0102.7
(Ans) MPa 25.2102667.0
1.0106.9
10067.1
2.0102.7
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
E
D
C
B
y
z
z
z
I
zM
I
yM
EXAMPLE (cont.)
Solutions
m 0625.02.0
95.425.2
z
zz
(Ans) 4.791.53tan102667.0
10067.1tan
tantan
3
3
y
z
I
I
• The location z of the neutral axis (NA), Fig. 6–33b, can be established by proportion.
• We can establish the orientation of the NA using Eq. 6–19, which is used to specify the angle that the axis makes with the z or maximum principal axis.
GENERAL CASE
GENERAL CASE
GENERAL CASE
z
SHEAR FORMULA
'' where
'
AyydAQ
It
VQ
A
SHEAR IN BEAMS
Rectangular cross section• Shear –stress distribution is parabolic
A
V
yh
bh
V
5.1
4
6
max
22
3
SHEAR IN BEAMS (cont)
Wide-flange beam• Shear-stress distribution is parabolic
but has a jump at the flange-to-web junctions.
EXAMPLE 1
A steel wide-flange beam has the dimensions shown in Fig. 7–11a. If it is subjected to a shear of V = 80kN, plot the shear-stress distribution acting over the beam’s cross-sectional area.
EXAMPLE 1 (cont)
• The moment of inertia of the cross-sectional area about the neutral axis is
• For point B, tB’ = 0.3m, and A’ is the dark shaded area shown in Fig. 7–11c
Solutions
4623
3
m 106.15511.002.03.002.03.012
12
2.0015.012
1
I
MPa 13.13.0106.155
1066.01080
m 1066.002.03.011.0''
6
33
'
''
33'
B
BB
B
It
VQ
AyQ
EXAMPLE 1 (cont)
• For point B, tB = 0.015m, and QB = QB’,
• For point C, tC = 0.015m, and A’ is the dark shaded area in Fig. 7–11d.
• Considering this area to be composed of two rectangles,
• Thus,
Solutions
MPa 6.22
015.0106.155
1066.010806
33
B
BB It
VQ
33 m 10735.01.0015.005.002.03.011.0'' AyQC
MPa 2.25
015.0106.155
10735.010806
33
max
C
cC It
VQ
SHEAR FLOW IN BUILT-UP BEAM
• Shear flow ≡ shear force per unit length along longitudinal axis of a beam.
I
VQq
q = shear flowV = internal resultant shearI = moment of inertia of the entire cross-sectional area
SHEAR FLOW IN BUILT-UP BEAM (cont)
EXAMPLE 2
Nails having a total shear strength of 40 N are used in a beam that can be constructed either as in Case I or as in Case II, Fig. 7–18. If the nails are spaced at 90 mm, determine the largest vertical shear that can be supported in each case so that the fasteners will not fail.
EXAMPLE 2 (cont)
• Since the cross section is the same in both cases, the moment of inertia about the neutral axis is
Case I • For this design a single row of nails holds the top or bottom flange onto
the web. • For one of these flanges,
Solutions
433 mm 205833401012
125030
12
1
I
(Ans) N 1.27205833
3375
90
40
mm 33755305.22'' 3
V
VI
VQq
AyQ
EXAMPLE 2 (cont)
Case II• Here a single row of nails holds one of the side boards onto the web.• Thus,
Solutions
(Ans) N 3.81205833
1125
90
40
mm 11255105.22'' 3
V
VI
VQq
AyQ
SHEAR FLOW IN THIN-WALLED BEAM
• Approximation: only the shear-flow component that acts parallel to the walls of the member will be counted.
SHEAR FLOW IN THIN-WALLED BEAM (cont)
• In horizontal flanges, flow varies linearly,
• In vertical web(s), flow varies parabolically,
x
b
I
Vtd
I
txbdV
I
VQq
22
2/2/
2
2
42
1
2y
ddb
I
Vt
I
VQq
EXAMPLE 3
The thin-walled box beam in Fig. 7–22a is subjected to a shear of 10 kN. Determine the variation of the shear flow throughout the cross section.
EXAMPLE 3 (cont)
• The moment of inertia is
• For point B, the area thus q’B = 0. • Also,
• For point C, since there are 2 points of attachment:
• The shear flow at D, because there are 2 points of attachment :
Solutions
0'A
mkNmmkN
mm
mmkN
I
VQq cc /7.48/0487.0
10797.1
1750010
2
1
2
146
3
mkNmmkN
mm
mmkN
I
VQq DD /8.82/0828.0
10797.1
2975010
2
1
2
146
3
462 10797.135105027010212
1mmmmmmmmmmmmI
3
3
2975010503535102
352''
17500105035''
mmmmmmmmmmmmmm
AyQ
mmmmmmmmAyQ
D
C
SHEAR CENTRE
• Shear center is the point through which a force can be applied which will cause a beam to bend and yet not twist.
• The location of the shear center is only a function of geometry of the cross section and does not depend upon the applied load.
P
dFe f
PedFM fA
SHEAR CENTRE (cont)
P
dFe f
EXAMPLE 4
Determine the location of the shear center for the thin-walled channel section having the dimensions shown in Fig. 7–25a.
EXAMPLE 4 (cont)
• The cross-sectional area can be divided into three component rectangles—a web and two flanges.
• q at the arbitrary position x is
• Hence, the force is
Solutions
b
hthhbtthI
6222
12
1 223
bhh
xbV
bhth
txbhV
I
VQq
6/6/2/
2/2
bhh
Vbdxxb
bhh
VdxqF
bb
f
6/26/
2
00
EXAMPLE 4 (cont)
• Summing moments about point A, Fig. 7–25c, we require
• As stated previously, e depends only on the geometry of the cross section.
Solutions
(Ans) 23/
6/22
2
bh
be
bhh
hVbhFVe f