unsymmetrical bending
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General Solutions for Unsymmetrical Bending of Beams with Arbitrary Cross Sections
Figure 1 A Beam with An Arbitrary Cross Section
Consider a cantilever beam subjected to an end force P acting in the plane inclined at an angle θ to the y-‐‑axis as shown. The bending moment M, produced by the force P, is oriented in the direction making an angle θ to the z-‐‑axis as shown. The components of M in the y and z-‐‑directions, denoted by yM and zM ,
respectively, are given by θθ cossin MMMM zy == (a)
Consider the special case where 90=θ . From (a) we have MMy = and 0=zM , then the beam would
deflect in the xz plane, and the neutral axis would coincide with the y-‐‑aixs. With the “cross sections remain plane” assumption, we have
z
zxzzρ
κε == (b)
where zκ and zρ are the curvature and radius of curvature, respectively, of the beam in the xz plane. On the other hand, if 0=θ , then MMz = and 0=yM , the beam would deflect in the xy plane and the
neutral axis would coincide with the z-‐‑aixs. Again, with the “cross sections remain plane” assumption, we have
y
yxyyρ
κε −=−= (c)
where yκ and yρ are the curvature and radius of curvature, respectively, of the beam in the yz plane. It
is noted that the negative sign in (c) indicates that a line element located above the z-‐‑axis would be under compression when a positive zM is applied. Now consider the general cases in which the value of angle θ is arbitrary. Since the “cross sections remain plane” assumption is still valid, we can express the strain in the x direction in the following linear equation in y and z: zcybax ʹ′+ʹ′+ʹ′=ε (d) where aʹ′ , bʹ′ , and cʹ′ are constants. The stress is then given by czbyaE xx ++== εσ (e) where E is the Young’s modulus and aEa ʹ′= , bEb ʹ′= , and cEc ʹ′= . For a beam subjected to pure bending, the stress resultant over the cross section must vanish, i.e., 0=++= ∫∫∫∫ zdAcydAbdAadAxσ (f)
If the origin of the coordinate system coincides with the centroid of the cross section, then
0,0 ====∫∫
∫∫
dAzdA
zdAydA
y
x
y
z
P
θθ
α
y
z
MyM
zM
P
yz
dA
2
Consequently, 0=a and (e) becomes czbyx +=σ (g) The resultant moments on the cross section are given by yyzxy cIbIdAzcyzdAbzdAM +=+== ∫∫∫ 2σ (h)
yzzxz cIbIyzdAcdAybydAM −−=−−=−= ∫∫∫ 2σ (i)
where ∫∫∫ === yzdAIdAyIdAzI yzzy ,, 22
are the moments of inertia of the cross section. Solving (h) and (i) simultaneously yields
22 ,yzzy
yzzzy
yzzy
yzyyz
IIIIMIM
cIIIIMIM
b−
+=
−
+−= (j)
Substituting (j) into (g) gives
zIIIIMIM
yIIIIMIM
yzzy
yzzzy
yzzy
yzyyzx 22 −
++
−
+−=σ (k)
The neutral axis, by definition, is the line along which stress vanishes. Thus, by setting 0=xσ in (k), we have
θ
θ
θ
θα
cotcot
tantan
tanyyz
yzz
yzy
zyz
yzyyz
yzzzy
IIII
IIII
IMIMIMIM
zy
+
+=
+
+=
+
+== (l)
in which α (see Fig. 1) is the angle measured clockwise from the z-‐‑axis to the neutral axis, and
y
z
z
y
MM
MM
== θθ cot or ,tan (m)
Combining (k) and (m) and eliminating yM , we obtain the following simplified equation in terms of zM
and angle α for calculating the stresses:
( )αα
σtantan
yzz
zx II
zyM−
−−= (n)
If the y-‐‑ and z-‐‑axes coincide with the principal axes, then 0=yzI and (l) reduces to
θα tantany
z
II
=
which is the same as (6-‐‑19) in the textbook. Furthermore, if 0=θ (i.e., MMz = and 0=yM ), then 0=α
and (n) becomes
z
zx I
yM−=σ
Example: A 5.0 m long simply-‐‑supported beam is subjected to a force kN 4=P at a point that is 2.0 m away from the far end, as shown in the figure. The dimensions of the L-‐‑shaped cross section are shown in the figure. If the force P is applied in an inclined plane making a 10° angle with the y-‐‑axis, determine the locations and magnitudes of the maximum tensile and compressive stresses in the beam. Solution: (a) Locate the centroid of the cross section (in terms of distances 0y and 0z measured from point A):
( )( ) ( )( )( ) ( )
( )( ) ( )( )( ) ( )
mm 74.19107010120
451070510120
mm 74.39107010120
510706010120
0
0
=×+×
×+×=
=×+×
×+×=
z
y
3
Figure 2 Umsymmetric Bending of A Simply-Supported Beam
(b) Determine the moments of inertia:
( )( ) ( )( )( ) ( )( ) ( )( )( ) 4623
23
mm 10003.174.19457010127010574.1910120
1210120
×=−++−+=yI
( )( ) ( )( )( ) ( )( ) ( )( )( ) 4623
23
mm 10783.2574.39107012107074.396012010
1212010
×=−++−+=zI
( )( )( )( ) ( )( )( )( ) 46 mm 10973.04574.1974.3951070574.1974.3960112010 ×=−−+−−=yzI
(c) Determine the neutral axis: Since 10−=θ (i.e., positive θ measured clockwise from the z axis), we have, from (l),
( )( ) 580.010tan10973.010003.110tan10783.210973.0
tantan
tan 66
66
=−××+×
−××+×=
+
+=
θ
θα
yzy
zyz
IIII
The neutral axis thus is oriented at the angle ( ) 12.30rad 526.0580.0tan 1 === −α
(d) Determine the maximum stresses in the beam: The maximum bending moment along the beam occurs at the point where the load P is applied, i.e.,
( )( )( ) mmN 104.8mkN 8.40.5
0.30.20.4 6max ⋅×=⋅===
LPabM
where m 0.5=L , m 0.2=a , and m 0.3=b . The y and z components of this moment are mmN 10834.0sin 6
max ⋅×−== θMMy
mmN 104.727cos 6max ⋅×== θMMz
From Fig.2 it is easy to see that the two locations denoted by A and B are the farthest from the neutral axis, hence would experience the highest tensile or compressive stresses. The y and z coordinates of those two points are ( )74.19 ,74.39−A and ( )74.9 ,26.80B , respectively. The maximum tensile stress, occurs at point A, is given by
m 2
m 3
P mm 120
mm 80
mm 01
mm 01
P
α
10
10
M
0y
0z
y
z
A
B
4
( )
( ) ( ) ( )[ ]( ) ( )
MPa 0.109mmN 100.109
12.30tan10973.010783.212.30tan74.1974.3910727.4
tantan
26
66
6
=×=
×−×
−−×−=
−
−−=
αα
σyzz
zx II
zyM
The maximum compressive stress, occurs at point B, is given by
( )
( ) ( ) ( )[ ]( ) ( )
MPa 0.159mmN 100.159
12.30tan10973.010783.212.30tan74.926.8010727.4
tantan
26
66
6
−=×−=
×−×
−×−=
−
−−=
αα
σyzz
zx II
zyM
(e) Solve the same problem by using the method outlined in Section 6-‐‑5 of the textbook: The principal axes are denoted as Y and Z, respectively, in Fig.3. The directions of the principal axes and the principal moments of inertia of the cross section can be obtained by using (A-‐‑11) and (A-‐‑12) in the textbook. Let β be the angle between the principal axis and y-‐‑axis (see Fig.2), then from (A-‐‑11)
( ) 093.110783.210003.1
10973.0222tan2tan 66
6
=×−×
×−=
−−==
zy
yzp II
Iβθ
and
( ) 8.23093.1tan21 1 == −β
The principal moments of inertia can be obtained by substituting the value of β in (A-‐‑10), or using (A-‐‑12):
Figure 3 Neutral Axis and Principal Axes
P
α
10
M
y
z
A
B
Y
Z
ββ
π−
2
10β
ψ
5
,mm 10.5740 ,mm 10212.322
464622
minmax, ××=+⎟⎟⎠
⎞⎜⎜⎝
⎛ −±
+= yz
zyzy IIIII
I
Thus, 4646 mm 10212.3mm 10.5740 ×=×= ZY II
The components of M in the Y and Z directions, respectively, are now given by
( )( ) mmN 10662.410cos
mmN 10143.110sin6
6
⋅×=−=
⋅×=−=
β
β
MMMM
Z
Y
The angle ψ , which determines the orientation of neutral axis and is measured clockwise from the Z-‐‑axis as shown in Fig.3, is given by
( ) 374.110tantan =−=== βψY
Z
Y
Z
Z
Y
II
II
MM
ZY
from which ( ) 9.53374.1tan 1 == −ψ It can be seen from Fig.3 that 1.30=−= βψα which matches the value obtained previously. With respect to the principal axes, the stresses are given by
Z
Z
Y
Yx I
YMIZM−=σ
At point A, the coordinates are
mm 1.34cossinmm 4.28sincos
=+−=
−=+=
ββ
ββ
AAA
AAA
zyZzyY
and the stress is given by
( )( ) ( )( ) MPa 0.10910212.3
4.2810662.410574.0
1.3410143.16
3
6
3
=×
−×−
×
×=−=
Z
Z
Y
Yx I
YMIZM
σ
At point B, the coordinates are
mm 5.23cossin
mm 4.77sincos−=+−=
=+=
ββ
ββ
BBB
BBB
zyZzyY
and the stress is given by
( )( ) ( )( ) MPa 0.15910212.3
4.7710662.410574.0
5.2310143.16
3
6
3
−=×
×−
×
−×=−=
Z
Z
Y
Yx I
YMIZM
σ
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