Solution First we determine the moments of inertia about the axesy andz. If we divide the area into several rectangular parts and neglect terms of higher order in the thicknesst, we obtain
Iy= 1
12 t(2a)3+ 2
"1
3t a3+a2(a t)
#
=10 3 t a3, Iz= 2
"1
3t a3+a2(a t)
#
=8 3t a3, Iyz = 2$
−%a 2a(a t)&
−% aa
2(a t)&'
=−2t a3. The orientation of the principal axes follows from (4.16)
tan 2ϕ∗= 2Iyz
Iy−Iz =−2·2·3 10−8 =−6 which leads to
ϕ∗1=−40.3◦, ϕ∗2=ϕ∗1+ 90◦= 49.7◦. (a) The principal moments of inertia are calculated from (4.17):
I1,2= t a3 3
⎡
⎣10 + 8
2 ±
10−8 2
2
+ (−6)2
⎤
⎦= (
3±
√37 3
) t a3
→ I1= 5.03t a3, I2= 0.97t a3. (b) The principal axes are shown in Fig. 4.11b. In order to decide which principal moment of inertia (b) corresponds to which direc- tion (a), we may insert (a) into (4.14). However, in this example one can see by inspection that the maximum moment of inertia I1belongs to the angleϕ∗1since the distances of the area elements from the corresponding axis are larger than those for the angle ϕ∗2.
4.3 Basic Equations of Ordinary Bending Theory
4.3120 4 Bending of Beams
c a b
∂u d
∂z
∂w∂x du=εdx
V+dV x+dx
M+dM MV
z x
w P
dx
z
τ σ dz
z
x
P′ z
dA
x z x
u=zψ ψ
dx
dx
dz dz
q
Fig. 4.12
i.e., we assume that the axes y and z are the principal axes of the cross sectional area (Iyz = 0). In addition, the applied loads are assumed to cause only a shear forceV in thez-direction and a bending moment M about the y-axis. These assumptions are satisfied, for example, if the z-axis is an axis of symmetry of the cross section and the applied forces act in thex, z-plane.
We have to use three different types of equations: equations from statics, geometrical (kinematic) relations and Hooke’s law.
The equations from statics are the equilibrium conditions formu- lated for a beam element as shown in Fig. 4.12a. They are taken from Volume 1, Section 7.2.2:
dV
dx =−q, dM
dx =V . (4.18)
The bending momentM and the shear forceV are the resultants of the normal stressesσ(acting in thex-direction) and the shear stressesτ(acting in thez-direction), respectively (see Fig. 4.12b):
M =
z σdA, V =
τdA . (4.19a,b)
The normal force N =
σdA (4.19c)
is zero due to the assumptions. Since only one normal stress and one shear stress are needed here, we have omitted the subscripts of the stress components: (σ=σx,τ=τxz).
The kinematic relations between the strain components and the displacements are taken from Section 3.1:
ε= ∂u
∂x, γ=∂w
∂x +∂u
∂z. (4.20)
Here,uandware the displacements in thex-direction (axis of the beam) and in thez-direction, respectively. Since no other strain components are needed, the subscripts are also omitted:ε=εx, γ =γxz. The strainε and the shear strainγ describe the defor- mation of an arbitrary element of the beam with length dxand height dz (Fig. 4.12b). This is illustrated in Fig. 4.12c.
We assume that the normal stressesσy andσz in the beam are small as compared with σ=σx and therefore may be neglected.
Then Hooke’s law is given by (compare Section 3.2)
σ=E ε, τ=G γ . (4.21)
It is not possible to uniquely determine the stresses and displa- cements with the aid of Equations (4.18) - (4.21). Therefore, we have to introduce additional assumptions. They concern the dis- placements of the points of a cross section at an arbitrary position x(Fig. 4.12d):
a) The displacementwis independent ofz:
w=w(x). (4.22a)
Hence, every point of a cross section undergoes the same deflection in the z-direction. This implies that the height of the beam does not change due to bending:εz=∂w/∂z= 0.
b) Plane cross sections of the beam remain plane during the ben- ding. In addition to the displacementw, a cross section undergoes
122 4 Bending of Beams
a rotation. The angle of rotation ψ=ψ(x) is a small angle; it is counted as positive if the rotation is counterclockwise. Thus, the displacementuof a pointP which is located at a distancezfrom thex-axis is given by
u(x, z) =ψ(x)z . (4.22b)
Experiments show that the assumptions a) and b) lead to results which are sufficiently accurate in the case of a slender beam with a constant cross section or with a slight taper.
Introducing (4.22a, b) and (4.20) into (4.21) yields σ=E∂u
∂x =E ψ′z , (4.23a)
τ =G ∂w
∂x +∂u
∂z
=G(w′+ψ) (4.23b)
where d( )/dx= ( )′ andw′ represents the slope of the deformed axis of the beam. Since |w′| ≪ 1, the slope is equal to the an- gle between the deformed axis of the beam and thex-axis. Using (4.23a), the Equations (4.19a) and (4.19c) yield the stress resul- tants
M =E ψ′
z2dA, N =E ψ′
zdA .
It was assumed that the normal force is zero: N = 0. Hence, according to the second equation, we obtain Sy =
zdA = 0, which implies that they-axis has to be a centroidal axis (see Vo- lume 1, Section 4.3). This is the reason for the particular choice of the coordinate system mentioned in Section 4.1. Introducing the second moment of areaI =Iy =
z2dA, the first equation can be written in the form
M =EI ψ′. (4.24)
Thus, the change dψof the angleψ in thex-direction is propor- tional to the bending momentM. The corresponding deformation
of a beam element of length dxis illustrated in Fig. 4.13a. Equa- tion (4.24) is aconstitutive equation for the bending moment. The quantityEI is referred to asflexural rigidityor bending stiffness.
Equation (4.23b) represents a shear stressτ which is constant in the cross section. This result is due to the simplifying assump- tions a) and b). In reality, however, the shear stress is not evenly distributed as will be shown in Section 4.6.1. In particular,τ = 0 at the outer fibers of the cross section. This can easily be shown with the fact that, according to (2.3), the shear stresses in two perpendicular planes are equal (complementary shear stresses).
Note that there are no shear stresses at the upper and the lower surface of the beam acting in the direction of the beam axis (no applied loads in this direction). Therefore, the shear stress also has to be zero at the extreme fibers of a cross section which is orthogonal to the surfaces of the beam. The actual distribution of the shear stress τ may approximately be taken into account by introducing a factorκ, called theshear correction factor, when (4.23b) is inserted into (4.19b):
V =κGA(w′+ψ). (4.25)
This is aconstitutive equation for the shear force. An element of the beam undergoes a shear deformationw′+ψ under the action of a shear force V (Fig. 4.13b). The quantity κGA = GAS is called theshear stiffnessandAS =κAis referred to as the “shear area ” (compare Sections 4.6.2 and 6.1).
a b
M V M
V dψ
x ψ
x
w′
dx dx Fig. 4.13
124 4 Bending of Beams