Frequently a beam undergoes a deflectionwinz-direction as well as a deflectionviny-direction. In this case the bending is referred to asunsymmetric bending. This type of bending occurs if the be- am carries loads inz- as well as iny-direction or if the cross section is not symmetrical. These cases give rise to the shear forcesVy,Vz
and the bending moments My, Mz (see Volume 1, Section 7.4).
We will restrict ourselves to Bernoulli beams (i.e., we neglect the influence of shear) and we will distinguish between the following two cases.
Case 1: We assume that y and z are the principal axes of the cross section. Then we can use the results of the ordinary bending (Sections 4.4 and 4.5) if we consider the loads inz- andy-directions separately. The load in z-direction causes normal stresses σ and deflectionsw. They are given by (see (4.26) and (4.31))
σ= My
Iy z, w′′=− My
EIy .
Analogously, the load iny-direction leads to σ=−Mz
Iz
y, v′′= Mz
EIz
.
The different algebraic signs are due to the sign convention (see Volume 1, Section 7.4): positive moments at a positive face point into the direction of the positive coordinate axes (Fig. 4.43). Su- perposition yields the total normal stresses:
σ= My
Iy z−Mz
Iz y . (4.45)
top view side view
My x Mz
z Mz
y
My
y z
x z x
y Fig. 4.43
166 4 Bending of Beams
The deflectionswand v are independent of each other. They are obtained through integration of
w′′=− My
EIy, v′′= Mz
EIz. (4.46)
side view top view
Vz
Mz
My My+dMy
Vy
Vy+dVy
Vz+dVz
Mz+dMz
x x
z
z y
y dx
dx
qz qy
Fig. 4.44
Case 2:Now we assume thatyandzare not principal axes of the cross section. To derive the relevant equations we proceed as in the case of ordinary bending. Let us consider the forces and moments that act on an element (length dx) of the beam. They are shown in Fig. 4.44 (note the sign convention of the stress resultants). The conditions of equilibrium are
dVz
dx =−qz, dVy
dx =−qy, dMy
dx =Vz, dMz
dx =−Vy.
(4.47)
As in case 1, we assume that the deflectionsv andware inde- pendent of y and z: v = v(x), w =w(x). In addition, we apply the hypotheses of Bernoulli (see Section 4.5.1): the cross sections remain plane and stay perpendicular to the deformed axis of the beam. Now we introduce the angles of rotationψy andψz of the cross section about the y-axis and the z-axis, respectively (po- sitive sense of rotation according to the cork-screw rule). In the following, we first determine the displacementuin axial direction of an arbitrary pointP of the cross section with the coordinates y, z(Fig. 4.45). Due to asmallrotationψy only, this point is dis- placed by an amountz ψy in the positivex-direction. Similarly, a small rotationψz leads to the displacement−y ψz. Therefore, the
top view side view
v′ z
ψz
−yψz w
P′ P′
y
ψy
y
zψy
−w′
v y
z
z x
P P
x
Fig. 4.45
total displacement is obtained as u=z ψy−y ψz.
With the relations
ψy=−w′, ψz= +v′
(see Fig. 4.45 and note that the cross section is perpendicular to the deformed axis of the beam) we get
u=−(w′z+v′y).
The strainε=∂u/∂xis therefore given by
ε=−(w′′z+v′′y), (4.48)
and Hooke’s lawσ=E εfinally yields
σ=−E(w′′z+v′′y). (4.49)
The bending momentsMy andMz are the resultant moments of the normal stressesσ in the cross section (note the senses of rotation):
My =
z σdA, Mz=−
y σdA. (4.50)
168 4 Bending of Beams
With (4.49) we obtain My=−E
w′′
z2dA+v′′
y zdA , Mz=E
w′′
y zdA+v′′
y2dA .
We now introduce the moments of inertia Iy =
z2dA, Iz = y2dA and the product of inertia Iyz = −
y zdA (see (4.6)) and solve forw′′and v′′:
E w′′= 1
∆[−MyIz+MzIyz], E v′′= 1
∆[MzIy−MyIyz].
(4.51)
Here, ∆ = IyIz−Iyz2 . The deflections wand v can be determi- ned from (4.51) through integration if the bending moments are known.
If we insert (4.51) into (4.49) we obtain the normal stress:
σ= 1
∆[(MyIz−MzIyz)z−(MzIy−MyIyz)y]. (4.52) Thus, the distribution of the normal stress is linear in y and z ((4.52) is the equation of a plane). The normal stress is zero for
z
y =MzIy−MyIyz
MyIz−MzIyz
. (4.53a)
This equation defines the neutral axis in the cross section. The maximum normal stressσmax is located at the point that has the maximum distance from the neutral axis.
In the special case that y and z are the principal axes of the cross section, the product of inertia vanishes:Iyz = 0. Then (4.51) and (4.52) reduce to (4.46) and (4.45), respectively, and the neu- tral axis is given by
z
y =MzIy
MyIz
. (4.53b)
We may use the equations of either case 1 or those of case 2 to treat unsymmetric bending. In the first case we have to de- termine the principal axes of the cross section and then resolve the applied loads and the bending moments in the corresponding coordinate system. The normal stress and the displacements are given by (4.45) and (4.46) with respect to the principal axes. In case 2, the stress and the displacements are determined by (4.52) and (4.51) in an arbitrary coordinate system.
E4.15 Example 4.15 The beam in Fig. 4.46a is supported by ball-and-
socket joints and subjected to a force F that acts at an angle α= 30◦ to the vertical. It has a rectangular cross section (width b, height h= 2b).
Determine the normal stresses and the deflections at the center of the beam.
neutral axis c b
F
l/2 l/2
h
f a
F α
v
w A
B
b
z
z
y y
y
z Fig. 4.46
Solution The normal stresses can be determined from (4.45) since yandzare the principal axes of the cross section. We resolve the forceF into its components iny- andz-direction:
Fy =F sin α= F
2, Fz=F cosα=
√3 2 F .
170 4 Bending of Beams
This yields the bending moments My= l
2 Fz
2 =
√3l F
8 , Mz=− l 2
Fy
2 =−l F
8 (a)
at the center of the beam (note the algebraic signs). The moments of inertia are given by (compare Table 4.1)
Iy= b h3 12 =2
3b4, Iz= h b3 12 = 1
6b4. (b)
Inserting (a) and (b) into (4.45) we obtain σ=
√3l F3
8·2b4 z+l F6
8b4 y=3l F 4b4
(√ 3 4 z+y
) . The neutral axis follows from the conditionσ= 0 as
y=−
√3 4 z.
This axis is shown in Fig. 4.46b. As can be seen by inspection, the pointsA andB are at maximum distance from the neutral axis.
With the coordinatesyA=b/2, zA=b of pointAwe obtain the maximum normal stress (tension)
σmax=3l F 4b4
(√ 3 4 b+b
2 )
= 3l F 8b3
(√ 3 2 + 1
) .
The normal stress at pointB has the same magnitude but it is a compressive stress.
The displacementswandvat the center of the beam are taken from Table 4.3. The force componentFz causes the displacement
w= Fzl3 48EIy =
√3F l3 64E b4.
Analogously, the force componentFy leads to the displacement v= Fyl3
48EIz
= 4F l3 64E b4 .
The total displacementf (see Fig. 4.46c) is obtained as f =
w2+v2=
√19F l3 64E b4 .
E4.16 Example 4.16 A cantilever beam with a thin-walled cross section
(t≪a) carries a forceF that acts inz-direction (Fig. 4.47a).
Determine the deflection at the free end B. Locate and deter- mine the maximum bending stress.
a
b c
neutral axis
00 0 11
A1 B
wB B
fB vB
a at
a a a
y
z z
y
z
a
C
y
D l
F
F
x
Fig. 4.47
Solution The bending moments are given by
My =−F(l−x), Mz= 0. (a)
The moments of inertia of the cross section are taken from Exam- ple 4.3:
Iy =10
3 t a3, Iz= 8
3t a3, Iyz =− 6
3t a3. (b)
172 4 Bending of Beams
SinceIyz = 0, the axesy and z are not the principal axes of the cross section. Therefore, we determine the displacementswandv with the aid of (4.51). With (a), (b) and ∆ =IyIz−Iyz2 = 449 t2a6, the differential equations for the deflections are
E w′′=−MyIz
∆ = 6F
11t a3(l−x), E v′′=−MyIyz
∆ =− 9F
22t a3(l−x).
We integrate twice, use the boundary conditions w(0) = 0, w′(0) = 0, v(0) = 0, v′(0) = 0 and obtain the deflections (re- call Section 4.5.2)
w(x) = F l3 11E t a3
3x
l 2
−x l
3 ,
v(x) =− 3F l3 44E t a3
3x
l 2
−x l
3 . This yields
wB= 2F l3
11E t a3, vB =− 3F l3 22E t a3
at the free endB(x=l), and the total deflectionfB is found to be (Fig. 4.47b)
fB =
w2B+vB2 = 5F l3 22E t a3 .
If we insert (a), (b) and ∆ into (4.52) we obtain the normal stress σ= My
∆ (Izz+Iyzy) =− 3F(l−x)
22t a3 (4z−3y).
The maximum stress is located at the clamping (x= 0) at those points of the cross section that have the maximum distance from the neutral axis. The neutral axis is determined from the condition σ = 0 which yields y = 43z. Fig. 4.47c shows that the points C andD are at the maximum distance from the neutral axis. With the coordinatesyC= 0, zC=−aof pointC we get
σmax= 6F l 11t a2.
The normal stress atDhas the same magnitude but is a compres- sive stress.