Strength of Materials - Lecture Notes / Mehmet Zor

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(tvid - 5.a)

5.1

NORMAL STRESS DISTRIBUTION IN

SIMPLE (PURE) BENDING

Strength of Materials - Lecture Notes / Mehmet Zor

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Bending is a very common type of loading in applications.

For example,

• Beams used in construction,
• Bridges where vehicles pass,
• Cranes, cranes etc.

are subject to bending loading.,

5.1.1 Examples of Structures Subject to Bending Loading:

5.1 Simple Bending / Normal Stress Distribution

• Our first goal in this regard is to formulate the normal stress distribution occurring in any cross-section of an object under bending loading…>>

Note: Deformation in bending (elastic curve subject) is a more advanced subject and will be explained in a separate section later. (Explained in Strength 2 course.)

Figure 5.1.1.a

Figure 5.1.1.b

Figure 5.1.1.c

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General rule: If the moment vector is parallel to the plane, there is bending; if it is perpendicular to the plane, there is torsion.

bending in the z-axis direction

bending in the y-axis direction

If the moment vector is perpendicular to the plane, that is, in the direction of the plane normal (x axis in the figure), a torsional moment will occur. (The symbol T is used for torsional moment instead of M_x.)

Torsion

5.1.2 How Do We understand the Difference Between Bending and Torsion?

Note: Moment indices may vary depending on the placement of the axis set, but the general rule above does not change. We will always place the axes as in these figures. In other words, the vertical axis parallel to the section will be y, the horizontal axis parallel to the section will be z, and the axis perpendicular to the section will be x. (If the axis system changes, the sign of the stress formula may change, which will be given as information in the future.)

In bending loading, the moment vector in the section (according to the right-hand rule) is in the direction of one of the axes (y or z axes) on the section plane.

5.1 Simple Bending / Normal Stress Distribution

(Reminder: While the 4 fingers of our right hand are in the direction of moment rotation, our thumb shows the direction of the moment vector.)

Figure 5.1.2

Figure 5.1.3

Figure 5.1.4

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5.1.3 Bending Types:

1-) Simple (or Symetrical, or Pure) Bending:

• The section is symmetrical with respect to at least one of the y or z axes.
• There is 1 internal bending moment (M_int.) in the section in the y or z axes direction.

2-) Simple Bending with Shear:

• The section is symmetrical with respect to at least one of the y or z axes.
• In the section, there is 1 internal bending moment and 1 internal shear force in the y or z axes direction.

3-) Oblique (or Unsymmetrical) Bending: It is the type of bending that does not meet at least one of the conditions mentioned above for Simple Bending and Simple Bending with Shear. This type of bending is a more advanced topic and will be explained later in a different chapter.

Bending types and calculations according to the principal inertia axes will be explained later in a separate chapter.

We will deal with the first 2 types of bending within the scope of this topic.

5.1 Simple Bending / Normal Stress Distribution

x

y

z

Examples

Figure 5.1.6

Figure 5.1.5

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5.1.4 In bending loading, the type of stress occurring (due to moment):

As a result of the bending moment, does shear stress or normal stress occur in the section? We will decide this first. Later, we will formulate this stress distribution in the cross-section.

In the case of elastic loading, the assumptions made are:

Plane sections remain plane even after bending.

In other words, the sections retain their shape and there is no distortion.

Fibers (lines) perpendicular to each other remain perpendicular to each other after bending.

In other words, the horizontal fibers seen in the figure above are perpendicular to the vertical fibers before bending and continue to remain perpendicular to the vertical fibers after bending.

We haven't decided on the type of stress yet… we continue…>>

5.1 Simple Bending / Normal Stress Distribution

Let's imagine that the elastic rod in the figure is fiber by fiber. (For example, lines a-b, and c-d are fibers.)

Let's apply a bending moment (M_z) to this rod from both ends.

Note that since the cross section is symmetrical, simple bending occurs.

Figure 5.1.7

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• As a result of the bending moment, we see that the upper fibers shorten and the lower fibers lengthen. (For example, a-b fiber shortens and becomes a'-b' fiber, c-d fiber lengthens and becomes c'-d' fiber.)
• This indicates that there is a compressive force on the shortening upper fibers and a tensile force on the lengthening lower fibers.

5.1 Simple Bending / Normal Stress Distribution

• Therefore, compressive stresses (-σ_x) occur in the upper fibers and tensile stresses (+σ_x) occur in the lower fibers. Therefore, the type of stress that occurs in bending is normal stress. In terms of strength, unless otherwise stated, balance equations are applied and the process is carried out according to the undeformed state. This is called the first order principle. For this reason, the stresses in the fibers are shown for the undeformed state of the beam.
• The top fiber shortens the most and the bottom fiber lengthens the most. In this case, maximum and minimum stresses occur in these fibers.

Figure 5.1.8

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• Since there is no elongation or shortening in the fibers on the neutral plane, the stresses in these fibers are zero. In other words, normal stresses are zero at all points on the neutral axis of a cross-section.

• The maximum elongation and shortening will occur at the points furthest from the neutral axis, thus maximum tensile stress and minimum compressive stress will occur.
• Stress values decrease as you approach the neutral axis.
• All these items are valid in case of elastic loading in simple bending or simple bending with shear.

5.1 Simple Bending / Normal Stress Distribution

• As you move from bottom to top, the elongation in the fibers decreases. At a transition point, fiber elongation becomes zero. The plane where these non-extending fibers are located is called the neutral plane. (As will be proven later, the neutral plane is the horizontal plane parallel to the x-z plane passing through the center of gravity of the sections). The horizontal line where the neutral plane intersects the section is called the neutral axis. Try to understand these concepts thoroughly by examining the figure on the side.

Figure 5.1.9

Figure 5.1.10

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5.1.5 Derivation of the normal stress (σ_x) equation in simple bending:

For fiber JK

Final length:

Total elongation:

Elastic Strain:

We know that arc length = central angle x radius.

According to this, we can write:

5.1 Simple Bending / Normal Stress Distribution

Fiber J-K

Normal stress in elastic region :

(From Hooke's equation in uniaxial loading)

• As a result of bending, the shape of the rod takes the form of an arc with center C and radius r.

Figure 5.1.11

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In this case, the y coordinate of the center of gravity of the section is:

The internal force is actually the sum of the stresses in its own direction.

5.1 Simple Bending / Normal Stress Distribution

Remember: In bending, the coordinate system must be placed at the center of gravity of the section.

Accordingly, the normal internal force in the x direction is:

, Accordingly

If you pay attention to the operations on the previous page, the y distance was taken as the distance to the neutral axis, that is, the non-extending fiber. Fro this reason,

Figure 5.1.11

Figure 5.1.12

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The differential internal force acting on the element dA at a point at distance y in the section:

Moment of the force dF about the neutral axis :

Total internal moment in the section :

…. If we substitute :

Note: When M_z-int. >0, see that compressive stresses should occur at the points above the neutral axis (y>0). In this case, for sx to be negative, there must be - at the beginning of the equation. In some books, the axis set may be taken differently. In this case, there may be no minus sign in the formula.

Now we will understand the stress distribution in the section by interpreting this formula..>>.

As a result, we can find the normal stress at any point of any cross-section of a beam subjected to simple bending with equation 5.1.1:

( 5.1.1)

Equation 5.1.1 is valid for simple bending or simple bending with shear, in the elastic region, for isotropic materials with a symmetrical cross-section with respect to at least one axis.

5.1 Simple Bending / Normal Stress Distribution

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It was

Figure 5.1.13

5.1.6 Let's understand the normal stress distribution in simple bending thoroughly:

We will re-examine the stress formula (equation 5.1.1) at any point in cross-section:

The section must be symmetrical with respect to at least one axis.

Part of the front view of the beam. Examine the stress distribution carefully.

Try to see that the moment is positive and that the upper fibers will be subjected to compression and the lower fibers will be subjected to tension according to this moment.

Maximum and minimum stresses occur at the points farthest from the neutral axis (N.A). Because M_z-int. and I_z are constant in a cross-section, as can be understood from equation 5.1, the stresses have the highest values at the points with the highest y value.

Stresses for any 2 points A and B

Why do we use M_z-int. and not Mz?

Because there may be different external moments M_z in different parts of the beam. However, we need to put the internal moment in the section we are examining into the formula. To avoid this confusion and clarify the situation, we use M_z-int. Even though M_z is used in other reference books, the internal moment is still meant.

5.1 Simple Bending / Normal Stress Distribution

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(5.1.1)

Figure 5.1.14

Figure 5.1.15

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5.1 Simple Bending / Normal Stress Distribution

5.1.7 How does the stress formula change in case of internal moment M_y-int.?

• By making the above comments before solving a problem, you can predict the sign of the stress at a point and verify the sign of your calculated result.
• You can also derive the stress formula and its sign according to different axis systems with similar comments.

If the internal moment in the section is M_y-int, this time the stresses will change depending on the z coordinate. In our stress formula, the y and z indices will be swapped and the formula will become like this:

Why didn't we put a «-» (minus) sign at the beginning of the formula this time?Try to understand the answer from the comments below:

( 5.1.2)

Top view

(a)

Figure 5.1.16

(b)

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• When there is M_z-int. moment, the stress distribution at all points of the section is actually as shown on the side.
• y koordinatı aynı olan noktaların gerilmeleri aynıdır.

​

• For this loading, compressive stresses occur at the points above the Neutral Axis (N.A) and tensile stresses occur at the points below it.
• In N.A, the stresses are zero.

5.1.8 Perspective view of stress distribution:

5.1 Simple Bending / Normal Stress Distribution

Figure 5.1.17

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Example 5.1.1

For the beam in the figure, find the bending moment that will initiate yielding.

Solution:

The lower fibers lengthen according to the loading in the I-I section. The maximum tensile stress occurs at the lowest points (for example, at point B). As a result of the solution, we predicted that there should be + stress at point B and – stress at point A. We can check the results according to this determination.

The same result can be found from the equation

Taking

(moment that will initiate yielding)

5.1 Simple Bending / Normal Stress Distribution

Figure 5.1.18

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5.1 Simple Bending / Normal Stress Distribution

Solution:

a-)

From Static Equilibrium we find the internal moment in a section:

Internal moments are the same in all sections

The yield strength of the T-section built-in (cantiliver) rod in tension and compression is 120MPa. To the free end of the rod,

a-) If a moment of M_z = 3kNm is applied,

b-) If a moment of M_y = 1kNm is applied,

c-) If a force of F= 1kN is applied in the -y direction from point A,

Check whether there will be any yield in the rod.

d-) How does the calculation method change when M_x = 3kNm torque is applied? Do your research.

Example 5.1.2

Figure 5.1.19

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To control yielding, we must calculate the maximum and minimum normal stresses from equation 5.1.1 and compare them with the yield stress.

But first we calculate the y coordinate of the center of gravity and the moment of inertia I_z:

5.1 Simple Bending / Normal Stress Distribution

Strength of Materials - Lecture Notes / Mehmet Zor

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Prediction: Since the M_z-int. moment is positive, minimum compressive stress should occur at point A and maximum tensile stress should occur at point B. We don't have to make this comment. However, if we can do it, our hesitations about the sign of the resulting stresses will disappear.

yielding occurs.

b-)Answer: 34.45 MPa < 120MPa (no yielding occurs),

d-) Answer: …..

c-) Answer : (no yielding occurs),

Solutions and explanations are given in Example 5a.2 in tvid- 5.a.

5.1 Simple Bending / Normal Stress Distribution