Pure Bending

II. PURE Bending II. PURE BENDING

In this section we will discuss the analysis of structures that are under the action of pure bending. As such, there will be no transverse shear force along the beam section considered. The problems of beam bending considered here are based on the EulerBernoulli Beam Theory. In this section we will examine the problems in which the bending moment is applied either symmetrically or unsymmetrically on homogeneous or non-homogeneous beams. In addition, we will discuss the elastic and inelastic bending of beams having symmetric or unsymmetric cross sections. The determination of neutral axis location for elastic and inelastic beams will also be discussed. The variation of bending-induced normal stresses on the beam cross section will be shown in several example problems. Finally, in this section we will discuss the bending of curved beams including the determination of the neutral axis and distribution of normal stresses.  II.1 Introduction • •

II.1-1 Rectangular Moments of Inertia and Product of Inertia II.1-2 Parallel Axis Theorem

 II.2 Elastic Bending of Homogeneous Beams  II.3 Elastic Bending of Non-homogeneous Beams  II.4 Inelastic Bending of Homogeneous Beams  II.5 Elastic Bending of Curved Beams

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II. PURE Bending Section II.1 Introduction The straight beam section under consideration is in the state of pure bending (i.e., transverse shear force is zero along this section). Consequently, as the beam bends, plane sections remain plane but rotate relative to each other as shown in the figure below.

As illustrated in the figure above, the top surface of the beam is shortened due to compression, and the bottom surface is elongated due to tension - both as a result of bending moment M. By examining the figure below, it becomes apparent that at some location between the top and bottom surfaces of the beam there is a surface whose length is the same as the original length of the straight beam. This surface is neither in tension nor in compression, therefore, it is referred to as the "Neutral Surface". The intersection of the neutral surface with the plane of the beam cross section is called the "Neutral Axis".

Since the beam cross section only rotates without warping, the slope of the cross sectional plane is constant, indicating that axial deflection due to bending is linear. This finding implies that in originally-straight beams in pure bending, the axial strain must also vary linearly with zero value at the Neutral Axis.

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Note: The linearity of axial strain is a consequence of beam being (a) originally straight, and (b) in pure-bending state.

Therefore, no additional restriction such as material property or elasticity is imposed.

Now we impose additional restrictions: (c) bending stresses remain below material's elastic stress limit, (d) stress-strain relationship is linear.

As a result of restrictions (c) and (d), Hooke's law may be used.

In summary it can be said that in elastic and homogeneous beams, the Neutral Axis, NA always passes through the centroid of the cross section with its orientation determined according to the shape of the cross section and the orientation of the bending moment. The stress and strain variations for some elastic beams are given below.

EXAMPLES • • •

(a) Elastic, homogeneous beam with doubly symmetric cross section. (b) Elastic, homogeneous beam with symmetric cross section. (c) Elastic, non-homogeneous beam with doubly symmetric cross section.

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II. PURE Bending ection II.2 Elastic Bending of Homogeneous Beams

The general bending stress equation for elastic, homogeneous beams is given as

(II.1)

where Mx and My are the bending moments about the x and y centroidal axes, respectively. Ix and Iy are the second moments of area (also known as moments of inertia) about the x and y axes, respectively, and Ixy is the product of inertia. Using this equation it would be possible to calculate the bending stress at any point on the beam cross section regardless of moment orientation or cross-sectional shape. Note that Mx, My, Ix, Iy, and Ixy are all unique for a given section along the length of the beam. In other words, they will not change from one point to another on the cross section. However, the x and y variables shown in the equation correspond to the coordinates of a point on the cross section at which the stress is to be determined.

Sign Convention on Bending Moment Components Mx and My:

As far as the general bending stress equation is concerned, if a moment component puts the first quadrant of the beam cross section in compression, it is treated as positive (see the examples shown below). Notice that this is just a sign convention for the moment components and should not be confused with the sign associated with the bending stress.

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Neutral Axis: When a homogeneous beam is subjected to elastic bending, the neutral axis (NA) will pass through the centroid of its cross section, but the orientation of the NA depends on the orientation of the moment vector and the cross sectional shape of the beam.

When the loading is unsymmetrical (at an angle) as seen in the figure below, the NA will also be at some angle - NOT necessarily the same angle as the bending moment.

Realizing that at any point on the neutral axis, the bending strain and stress are zero, we can use the general bending stress equation to find its orientation. Setting the stress to zero and solving for the slope y/x gives B-7

II. PURE Bending

(II.2)

A positive angle is defined as counter clockwise from the horizontal centroidal axis.

Notice that we can use the equation for orientation of NA to examine special cases. For example, if the cross section has an axis of symmetry, Ixy = 0. In addition if only Mx is applied, then NA will have angle of zero which is consistent with what we would expect from mechanics of materials. From this equation, we see that the orientation of NA is a function of both loading condition as well as cross sectional geometry.

EXAMPLE PROBLEMS • • • •

Example 1* Thin-walled beam with horizontally symmetric cross section under a horizontal bending moment Example 2* Thin-walled beam with horizontally symmetric cross section under an oblique bending moment Example 3* Thin-walled beam with unsymmetric cross section under an oblique bending moment Example 4 Skin-stringer beam with unsymmetric cross section under horizontal bending moment

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SECTION II.2 EXAMPLE 1

For the cross section and loading shown, determine • • •

(a) Neutral axis location and orientation, (b) Bending stress distribution, (c) Location and magnitude of the maximum bending stress.

Assume that the stresses due to the applied load do not exceed the elastic limit.

Use the Java screen shown below to modify this example and see the results.

EQUATION USED: Eq. A13.13

SOLUTION

(a) As seen in the figure above, the cross section is symmetric about the horizontal axis, therefore, the product of inertia is zero in this case. Furthermore, with the bending B-9

II. PURE Bending moment applied about the x axis, the y component of moment is zero. As a result of the previous two conditions, the NA orientation according to eqn. A13.15 will be horizontal passing through the centroid as expected. This problem is an example of symmetric bending. NOTE: There is no need to find the horizontal position of centroid because there is no need to calculate the moment of inertia about the y axis as y component of bending moment is zero. (b) Because of the conditions stated in part (a) of solution, eq. A13.13 reduces to

The bending stress distribution will be linear with a zero value at the NA.

(c) In this case, the maximum stress is at the farthest point from the NA. Because of horizontal symmetry about the NA, the stress at the top and bottom of the section will have equal magnitude with the one on top being compressive. To get the maximum value of stress, the reduced equation given previously will be used. The moment of inertia about the x axis is

This makes the maximum bending stress

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SECTION II.2 EXAMPLE 2

For the cross section and loading shown, determine • •

(a) Neutral axis location and orientation, (b) Location and magnitude of the maximum bending stress.

Assume that the stresses due to the applied load do not exceed the elastic limit. Also assume that each lenght shown is measured to the middle of the adjacent member.

Use the Java screen shown below to change the problem data and see the results.

EQUATIONS USED: Eq. A13.13 and Eq. A13.15

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SOLUTION

(a) As seen in the figure above, this is a symmetric cross section with unsymmetric loading. The response will be an unsymmetric bending.

With the product of inertia being zero due to cross-sectional symmetry, we need to calculate the components of the applied bending moment and the rectangular moments of inertia in order to determine the orientation of NA. The components of bending moment are:

By examinning the applied moment, it is clear that both of its components will produce compression on the first quadrant, hence, they are both positive. To calculate the rectangular moments of inertia, it is necessary to know the location of the centroid. Due to horizontal symmetry only the horizontal coordinate of centroid need to be calculated as its vertical coordinate is known due to symmetry.

The moments of inertia about the x and y axes are

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The neutral axis will pass through the centroid with an angle of

(b) The maximum axial stress is at the farthest point from the NA, either at point A, B, C, or D. To get the stress values, use equation A13.13 at all four points. With the product of inertia equal to zero A13.13 reduces to

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II. PURE Bending The bending stresses at point A, B, C, and D are

The results indicate that the maximum bending stress is at point B

SECTION II.2 EXAMPLE 3

For the cross section and loading shown, determine • •

(a) Neutral axis location and orientation, (b) Location and magnitude of the maximum bending stress.

Assume that the stresses due to the applied load do not exceed the elastic limit. Also assume that each length shown is measured to the middle of the adjacent member.

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Use the Java screen below to change problem data and see the results.

EQUATIONS USED: Eq. A13.13 and Eq. A13.15

SOLUTION

This problem requires more analysis as both the loading and cross-sectional shape are unsymmetric. The procedure is similar to the previous example. First need to find the centroid, moments of inertia about the x and y axes, and the product of inertia. B-15

II. PURE Bending The centroid is at

and the moments of inertia and product of inertia are

The components of the applied bending moment are determined as

The x component is negative because it causes tension in the first quadrant.

(a) Since this is a homogeneous section, and it is assumed to be within its elastic limits, the neutral axis will pass through the centroid. Its angle with respect to the x axis is (b) The maximum bending stress occurs at the farthest point from the NA, either at point A or B.

Therefore, point B is the location of maximum bending stress.

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II. PURE Bending Section II.3 Elastic Bending of NON-Homogeneous Beams Recall that in pure bending the normal strain variation is linear with a value of zero at the neutral axis. Before we discuss the relationship between stress and bending moment, let's first determine the location of NA. The equilibrium condition requires the sum of normal forces to go to zero. Mathematically, this is expressed as

Considering a section made of two different materials with Young's moduli identified by E1 and E2 and with stresses below the elastic limit of each material, we can write

Where yB is the distance from the NA to the bottom surface, yi is the distance from the NA to the material interface and yT is the distance to the top surface. Using the linear equation for strain variation gives

Factoring out the common terms and E1 gives B-18

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A close examination of this equation reveals that the bracketed term must be equal to zero. However, the ratio of E2 to E1 in front of the second term inside the brackets indicates that neutral axis will pass through the centroid of the modified homogeneous section, one with material 2 replaced with an equivalent material 1. This fact is captured in the figure below with the condition that E2 > E1.

Notice that only the width of the section is modified while its height is kept the same. This condition would have been reversed if the bending moment was applied about the vertical axis. The moment equation is written as

For the case of a section made of two different materials, as shown above, the integral is divided into two parts, one for each elastic material. Notice that y is measured from the neutral axis.

Using the linear normal strain variation we get

Factoring out the constant terms and normalizing with respect to E1 we get B-19

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The terms inside the brackets represent the moment of inertia of the modified homogeneous section about the neutral axis, and can be expressed as

Substituting this into the previous equation and solving for the stress gives

This equation can be used to calculate the bending stress only for the portion that is made of material 1. To calculate the bending stress in the portion that is made of material 2, it should be multiplied by the ratio of Young's moduli of the two materials as

The procedure described above is known as the Modified Section Method, and is used in the analysis of elastic non-homogeneous beam sections in bending.

EXAMPLE PROBLEMS •

Example 1 Design of an elastic, non-homogeneous beam under horizontal bending moment

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SECTION II.3 EXAMPLE 1

A bending moment of 74,734 N-m is applied to a uniform composite beam shown in the figure below. If the allowable stress for steel is 250 MPa and the allowable stress for aluminum is 100 MPa, what is the minimum width of the 5-mm thick steel plates attached to the aluminum I-beam. Also determine the neutral axis location and bending stress distribution.

EQUATIONS USED: SOLUTION

Since this is a doubly symmetric cross section, the centroid will be located at the center. This is true for both the original and the modified cross sections. With the applied moment acting about the horizontal axis, the NA will be horizontal in this case. From inspection, it is known that the maximum strain will occur in the steel plates. Also, with steel being the stiffest of the two materials it is very likely that it will carry more stress.

To calculate the modified moment of inertia, the entire section is modified to all aluminum. The modified moment of inertia in terms of w is B-21

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Let the steel plates be stressed to their limit. The top plate will have the same stress and strain as the bottom plate because of symmetry.

Check the stress at the farthest point in the aluminum part.

Use the stress equation and solve for the width

The bending stress distribution for this cross section is B-22

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ALTERNATE SOLUTION

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II. PURE Bending Section II.4 Inelastic Bending of Homogeneous Beams In this section we will first dicuss the inelastic behavior of beams in pure bending, and then elaborate on the method of analysis that we can use in such problems.

Inelastic behavior is possible in beams that are made of ductile materials, and as such can be loaded beyond the elastic limit or proportional limit of the material. This implies that the ultimate load carrying capability of a ductile beam is higher than its maximum elastic load. How much higher depends on mechanical properties of the beam material.

Naturally, the behavior of a beam in inelastic bending depends upon the shape of the material's stress-strain diagram. If the stress-strain diagram is known, it is possible to determine stress corresponding to a particular value of strain.

As in previous discussions we will assume that the material can be idealized as an elastoplastic material with maximum stress being the elastic limit stress, and the maximum strain being considerably higher than the elastic limit strain. It is possible for the elastoplastic material to have different characteristics in tension and in compression. For instance the corresponding elastic limit values and even the Young's moduli may be different. This tends to complicate the analysis to a certain degree. Assumptions: The analysis of an inelastic beam is based on the assumption that plane cross sections of a beam remain plane under pure bending, a condition that is valid for both nonlinear and linear materials. Therefore, normal strain in an inelastic beam varies linearly over the cross section of the beam. Restrictions: a. Beam has a symmetric cross section. It is not necessary for it to be doubly symmetric. B-24

II. PURE Bending b. Beam is loaded symmetrically, moment is acting about either the x or the y centroidal axis. Neutral Axis Location: The neutral axis of beams in inelastic bending may or may not pass through the centroid of the cross section. The following diagrams show the variations of bending strain and stress across a rectangular beam section ranging from fully elastic to fully plastic condition. Notice that the material is assumed to be elastoplastic with elastic limit in tension equal in magnitude to that in compression.

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Notice that the N.A. conicides with the horizontal centroidal axis even as the beam becomes fully plastic.

If in the previous example the stress-strain variation in compression was different from that in tension, then the position of N.A. would no longer coincide with the horizontal centroidal axis as beam is loaded beyond its elastic limit.

Notice that the resultant axial force is zero as the net compression force balances against the net tension force acting on the cross section.

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In such beam problems, the location of N.A. coincides with the horizontal centroidal axis when the beam is elastic. However, as it is loaded beyond the elastic limit, N.A. shift either up or down relative to the centroid depending on whether the material can carry more tension or compression. The farthest position of N.A. is determined by checking the cross-sectional stress variation for a fully plastic condition.

Helpful Observations in Inelastic Bending:

1. If the stress-strain variations in tension and compression are the same, then a. N.A. coincides with the centroidal axis (same as moment axis) if the cross section is symmetric about that axis. b. N.A. does not conicide with the centroidal axis if the cross section is unsymmetric about that axis.

2. If the stress-strain variations in tension and compression are different, then N.A. does not coincide with the centroidal axis regardless of cross-sectional symmetry about that axis. Determination of a Beam's Moment Capacity:

1. Check the stress-strain variations in compression and tension. Is there a difference between elastic limit stress in tension from that in compression? 2. Is the moment acting about the axis of symmetry or not? 3.

Case A. Moment is acting about the axis of symmetry and material properties in compression and tension are the same. • •

Maximum elastic moment is determined from the simplified form of Eq. (II.1) Inelastic moment for some given value of maximum strain less than fully-plastic strain is found from the moment equilibrium equation. First determine the strain variation B-27

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•

(remember it is linear in linearly elastic materials) across the beam. Then relate the strain variation to stress variation by checking the stress-strain diagram. Finally write the integral relating the bending moment to the stress distribution across the beam, and solve for the bending moment. Fully-plastic bending moment is obtained by drawing the stress pattern over the beam cross section. Keep in mind that in this case the location of N.A. is the same as centroidal axis or axis of symmetry in this case. Calculate the resultant force in compression, and resultant force in tension. Sum moments about the N.A. and find the total bending moment on the beam.

Case B. Moment is acting about the axis of symmetry but material properties in compression and tension are different. • •

•

Maximum elastic moment is determined from the simplified form of Eq. (II.1). Inelastic moment for some given value of maximum strain less than fully-plastic strain is found from the moment equilibrium equation. However, in this case the N.A. position is unknown. Therefore, an iterative solution based on a guessed position of N.A. is required. Guess a position for N.A. relative to the centroidal axis moment is acting about. From linearity of strain, determine the strain variation, then relate the strains to stresses and use the axial force equilibrium equation, and see whether the sum of forces goes to zero or not. If it goes to zero, then the location of N.A. is correct. Otherwise, guess again, and repeat the procedure. Once the location of N.A. is found, go to the moment-stress integral equation and solve for the value of moment. Fully-plastic bending moment is obtained by drawing the stress pattern over the beam cross section. Here once again the location of N.A. is unknown. However, we know the maxium stress in compression as well as in tension. With the stress being constant in the tension side and constant in the compression side. No iteration is necessary here as the location of N.A. can be determined by summing the axial forces to zero and determining the height of compression and tension portions of the cross section. Once N.A. position is known, then proceed with determining the moment summation about the N.A. to obtain the fully-plastic bending moment.

Case C. Moment is acting about a centroidal axis which is not an axis of symmetry, and material properties in compression and tension are different. Example 1 below deals with such a problem.

EXAMPLE PROBLEMS •

Example 1 Inelastic bending of a homogeneous beam section

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SECTION II.4 EXAMPLE 1

For the beam section and loading shown, determine • • •

(a) the maximum elastic bending moment that can be applied (b) the inelastic bending moment resulting in a strain of 0.003 at the bottom edge (c) the maximum (or fully-plastic) bending moment.

EQUATIONS USED

SOLUTION

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II. PURE Bending We must first find the location of centroid and the moment of inertia about the horizontal centroidal axis (x axis)

The moment of inertia about the x axis is all that is needed because with the product of inertia Ixy equal to zero and no moment about the y axis, the general elastic bending formula reduces to

(a) Because the elastic-limit stress in the tension side of the stress-strain diagram is less than that in the compression side, the tensile limit will be the maximum stress allowable. Since the bottom side is in tension, let the stress at that location be equal to the limit of 45,000 Pa. This gives a maximum moment of

Now, check the top to make sure the compressive stress doesn't exceed its limit (60,000 Pa).

Therefore, the maximum elastic moment is

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II. PURE Bending The strain and stress diagrams corresponding to the maximum elastic moment are shown below. Notice that the NA in this case coincides with the horizontal centroidal axis (x axis).

(b) We know the elastic-limit strain in tension and compression to be

With the strain at the bottom now at 0.003, the elastic-inelastic interface, so to speak, is at some location between the NA and the bottom surface. Since stress-strain variations in tension and compression are different in the inelastic region, the NA will not pass through the centroid. Therefore, NA will not coincide with the x axis in this case. However, the NA will be parallel to the x axis. So we know its orientation but not its location. The location of NA is found through an iterative process. Assume NA is @ 68.333 mm from the bottom. Through similar triangles, the elasticinelastic interface can be found

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If the location of NA is correct, then the summation of forces in the normal direction must be zero.

Based on the stress diagram shown above, the force equation is written as

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II. PURE Bending Since the sum of axial force components is not zero, the assumed location of NA is incorrect. The negative value indicates that the NA must be moved in the compression side as to reduce the area which is in compression or to increase the area in tension. For the second guess, assume NA to be @ 80 mm from the bottom surface and repeat the procedure.

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The force sum is now positive indicating that the correct location of NA is somewhere between the two locations assumed previously. Using linear interpolation, we come up with the third guess

Using linear interpolation again and finding the summation of forces yields B-35

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The result is satisfactory. Now with the location of the NA known, the moment corresponding to the strain of 0.003 at the bottom side can be calculated.

(c) To find the fully-plastic moment, we must first determine the location of NA resulting in the total axial force to go to zero. In this case, there is no linear stress region. The compression and tension sides are under the state of constant stress equal to the corresponding maximum stress values.

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Using the stress variation shown above and the corresponding force summation, we find the location of NA as follows

The fully-plastic moment is found to be

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II. PURE Bending The results indicate that in this case the fully-plastic moment is 57% larger than the maximum elastic bending moment. We also saw in this problem that the location of neutral axis changed depending on the magnitude of M. The two factors having the most influence on the results are: (a) Different material elastic limits in tension and compression, and (b) Unsymmetric cross section with respect to the moment axis.

To Section II.4

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