To determine the stresses we first choose an imaginary section c−c perpendicular to the axis of the beam. Determine the normal stress σ in any section perpendicular to the axis of the beam.
Strain
If the bar is elongated, the cross-sections undergo displacements in the x-direction denoted by u. So the displacements are on the left side of the element and on the right side.
Constitutive Law
The change ∆l of the lengthl of the test piece can be measured and the strain ε= ∆l/l can be calculated. This behavior of the material is called elastic; the behavior in the areaσ ≤σP is called linear elastic.
Single Bar under Tension or Compression
The displacement of the cross-section at the top of the bar is equal to zero:u(0) = 0. So we have to take into account the deformation of the bar. the temperature in region② is not changed).
Statically Determinate Systems of Bars
Therefore, within a good approximation, the arcs of the circles can be replaced by their tangents. Determine the angle of inclination of the beam caused by its weight after the structure is assembled.
Statically Indeterminate Systems of Bars
The total elongation of the rods is obtained by superposition of the "0" and "1" systems:. f) The compatibility condition (c) is again taken from the shift diagram (Figure 1.16c). In order to determine the forces in the rods of such a system, in addition to equilibrium conditions, incompatibility conditions are also necessary.
Supplementary Examples
Determine the stresses σ(x) and σ(l) and the elongation. l bars due to its own weight. A force F acts on it at point B. a) Calculate the forces Si. in bars and extensions ∆li bars.
Summary
However, the stress state at a point P of the section is not yet sufficiently determined by t for the following reason. In this case the directions of the normal to the section and of the stress component coincide.
Plane Stress
- Coordinate Transformation
- Principal Stresses
- Mohr’s Circle
- The Thin-Walled Pressure Vessel
In the special case that the stresses are site independent, the stress state is called homogeneous. By inserting ϕ∗∗ into the third equation of (2.6) the shear stress τξη =τmax is obtained with the correct sign. Using the transformation relations (2.6), the stresses σξ, ση and τξη can be calculated from the stresses σx, σy for aξ,η system.
The normal stresses for all cross-section directions have the same valueσξ=ση=σ0 and no shear stresses appear (cf.. a) principal stresses and principal directions, .. b) normal and shear stresses acting in the cross-section whose normal forms with the x-axis angle ϕ= 30◦.
Equilibrium Conditions
Therefore, the stress on the wall of a thin-walled spherical vessel has the value p r/(2t) in any arbitrary direction. In the considered case of plane stress, they consist of two coupled partial differential equations for the three components σx, σy and τxy = τyx of the stress tensor. The state of stress cannot be uniquely determined by these equations: the problem is statically indeterminate.
These are three coupled partial differential equations for the six components of the stress tensor.
Supplementary Examples
The "1" system consists of the beam and the strut, both subjected to the unknown force X = S. The bending moments cause the deflections w(x) and v(x), whereas the normal force only causes a displacement(x) in the direction of the beam axis. TdA (average change in cross-section) only causes a change in the length of the beam.
Summary
Hooke’s Law
Strength Hypotheses
In order to prevent such a failure, an allowable stress σallow is introduced and it is assumed that the stress σ in the bar cannot exceed σallow, i.e.: σ≤σallow (cf. Chapter 1). The stress σe applied to the uniaxial bar case is assumed to have the same effect on failure due to plastic flow or fracture as the given spatial stress state in the body under consideration. It is assumed that the material begins to fail when the maximum principal stress reaches a critical value.
Here it is assumed that the material state becomes critical when the energy required for the "deformation" of a material element (volume remains unchanged) reaches a critical value.
Supplementary Examples
Due to the symmetry of the problem, the moments of inertia about each axis through C are equal.
Summary
Second Moments of Area
Definitions
The magnitude of a moment of inertia depends on the location of the origin and on the orientation of the axes. Often an area A consists of several parts Ai whose moments of inertia are known (Fig. 4.3b). In a second example, we calculate the moments of inertia and the tower radius of a circular area (radius R), with the origin of the coordinate system at the centroid C (Fig. 4.5a).
The results for the circular region can be used to find the moments of inertia for a circular ring with outer radius Ra and inner radius Ri (Fig. 4.5c).
Parallel-Axis Theorem
As an example, we determine the moments of inertia with respect to the axes ¯y,z¯ for the rectangle shown in fig. Since the moments of inertia with respect to centroid axes are given by (4.8), we get from (4.13). Therefore, the sum of the moments of inertia Iy and Iz (= polar moment of inertia) is independent of the angle ϕ.
Therefore, the angle ϕ=ϕ∗ that makes the moments of inertia a maximum or a minimum is given by.
Basic Equations of Ordinary Bending Theory
They are related to the displacements of the points of a cross section in an arbitrary position x (Fig. 4.12d): . a). Thus, the change dψ of the angle ψ in the x direction is proportional to the bending momentM. Therefore, the shear stress must also be zero at the extreme strands of a cross section that is orthogonal to the beam surfaces.
An element of the beam undergoes a shear deformation w′+ψ under the action of a shear force V (Fig. 4.13b).
Normal Stresses
For example, if the bending moment M, the section modulus W and the permissible stress σ are known, it must be checked whether the maximum stress σmax meets the requirement. On the other hand, if M and σallow are given, the required section modulus can be calculated. Solution The maximum bending moment is located at the coupling; has size.
Find the required side length c of the thin-walled square cross-section so that the permissible stress σallow = 200 MPa is not exceeded.
Deflection Curve
Differential Equation of the Deflection Curve
In this case, an element of the beam does not undergo a shear deformation due to the shear force: γ = 0. The curvature of the deformed axis of the beam is therefore proportional to the bending moment (see (4.31)). If we use the first and second equation of (4.30), we get another form of the differential equation of the elastic line.
If the loadq and bending stiffnessEI are given, the deflection w can be determined by integrating the above differential equation.
Beams with one Region of Integration
Equation (4.31) can only be applied to the static determination of problems, since only in this case the bending moment can be calculated in advance from the equilibrium conditions. Since in the latter case the bending moment cannot be calculated from the equilibrium conditions, we will use the differential equation (4.34b) in all three cases. Having determined the constants of integration, we also know the slope′, the bending moment M(x) and the shear force V(x).
Solution The maximum bending stress at any cross-section is equal toσ0 as according to (4.28).
Beams with several Regions of Integration
The deflection at the free end (ξ= 0) can be calculated using the limit expression lim. Note that it is four times the deformation of the cantilever with a constant moment of inertia I0. The constants of integration can be calculated from both the boundary conditions and the matching conditions, also called continuity conditions (cf. Vol. 1, Section 7.2.4).
Solution In this example, the shear force has a jump (discontinuity) at x= and the bending moment is given by.
Method of Superposition
If the support is removed, we get the "0" system, which corresponds to the one shown in the picture. Since the beam and strut are connected in the original system, their displacements at B must coincide. In the case of a statically indeterminate system up to the first degree, we remove one constraint to obtain a statically determinate one.
The auxiliary "1" system is then subjected to a load according to the constraint removed.
Influence of Shear
Shear Stresses
As an example, we determine the distribution of the shear stress due to a shear force in a rectangular cross-section (Fig. 4.33a). We assume that the shear stressesτ at a position s of the cross-section are constant over the thickness, that they are parallel to the boundary (Fig. 4.36a). The magnitude and direction of the shear stresses and the thickness may depend on the coordinate (arc length) s.
E4.12 Example 4.12 Determine the distribution of shear stresses due to the shear force in a solid circular section with radius r (Figure 4.39).
Deflection due to Shear
In section 6.1 it will be shown that the displacement correction factor for a rectangular cross-section has the value κ = 5/6. In this example, the deflection due to displacement is only approx. 3% of the deflection due to the bending moment. In general, the deflection due to shear can be neglected if the length of the beam is greater than five times the height of the cross-section.
Unsymmetric Bending
In addition, we apply the hypotheses of Bernoulli (see Section 4.5.1): the cross sections remain flat and remain perpendicular to the deformed axis of the beam. In the special case that y and z are the principal axes of the cross-section, the product of inertia vanishes:Iyz = 0. Solution The normal stresses can be determined from (4.45) since yandzare the principal axes of the cross-section.
The maximum stress is located at the clamping (x= 0) at those points of the cross-section which are the greatest distance from the neutral axis.
Bending and Tension/Compression
Note that the deformation due to the normal force is usually much smaller than the deformation caused by the bending moments. Consequently, the change in length of the beam can be neglected and the beam can be considered stiff in tension/compression. This is statically equivalent to the system with the same force F with the line of action equal to the x-axis and the moment MB =r F (Fig. 4.48b).
According to (1.18), the change in the length ∆l of the beam due to the compressive force F is given by.
Core of the Cross Section
Now we want to determine the allowed region of application points F, zF of the force so that only compressive stresses are induced in the cross section. To find the core of the cross section we assume that the neutral axis g−g is a tangent to the cross section (Fig. 4.50c). If the point of application of force F lies within the core, the neutral axis lies outside the cross section and the column is subjected to compression only.
In addition, it can be shown that the neutral axes at the angles of the cross-section correspond to the points of action of the force along the straight lines between P1 and P2, between P2 and P3, etc.
Thermal Bending
Here TlandTuare the temperature changes at the extreme lower fiber and the extreme upper fiber respectively. It can be seen that the difference Tl−Tu of the temperature causes a curvature of the beam axis, as does the bending moment M. Since there is no applied force, the bending moment M and the normal stress σ are zero.
Elastic curve(1) of system “1” (cantilever subjected to redundant force X = B) is taken from table 4.3.
Supplementary Examples
Torsion spring stiffness priA (kA = EI/a) and linear spring pri. Assume that the top surface of the pontoon remains above the surface of the water. If the beam is subjected to bending and tension/compression, the stresses and strains are obtained by superposition of solutions for different loads.
The non-uniform temperature change across the cross section results in a temperature moment that causes the beam axis to bend.
Summary
Circular Shaft
Thin-Walled Tubes with Closed Cross Sections
Thin-Walled Shafts with Open Cross Sections
Supplementary Examples
Summary
Strain Energy and Conservation of Energy
Principle of Virtual Forces and Unit Load Method
Influence Coefficients and Reciprocal
Statically Indeterminate Systems
Supplementary Examples
Summary
Bifurcation of an Equilibrium State
Critical Loads of Bars, Euler’s Column
Supplementary Examples
Summary
Stress
Introduction
