Solid Mechanics

Md. Mohiuddin

Lecturer

Department of Mechanical Engineering

ME 2213

Beam

Beams are structural members subjected to lateral loads, that is, forces or moments having their vectors perpendicular to the axis of the bar.

Determinate Beam

• A beam may be determinate or indeterminate

Statically determinate beams are those beams in which the reactions of the supports may be determined by the use of the equations of static equilibrium

Indeterminate Beam

• If the number of reactions exerted upon a beam exceeds the number of equations in static equilibrium, the beam is said to be statically indeterminate.
• In order to solve the reactions of the beam, the static equations must be supplemented by equations based upon the elastic deformations of the beam.

• In the case of propped beam, there are three reactions R1, R2, and M and only two equations (ΣM = 0 and sum; F_y = 0) can be applied, thus the beam is indeterminate to the first degree (3 – 2 = 1).

Types of Loading

Types of Supports

Share Force and Bending Moment Diagram

• We usually need to know how the shear forces and bending moments vary throughout the length of the beam.
• The maximum and minimum values of these quantities are important in beam design.
• Information of this kind is usually provided by graphs in which the shear force and bending moment are plotted as ordinates and the distance x along the axis of the beam is plotted as the abscissa.
• Such graphs are called shear-force and bending-moment diagrams

Sign Convention

• Shear force is positive (+) when tends to rotate the material clockwise
• Bending moment is positive (+) when tends to compress the upper part of the beam and elongate the lower part.

Here,

V = Shear Force at section mn

M = Bending moment at section mn

Relation Between Share Force and Bending Moment

True only for distributed load

Discarding products of differentials (because they are negligible compared to the other terms)

Problem

Write shear and moment equations for the beams as shown in the figure below. Also, draw shear and moment diagrams, specifying values at all changes of loading positions and at points of zero shear. Neglect the mass of the beam.

Problem

Write shear and moment equations for the beams as shown in figure below. Also, draw shear and moment diagrams, specifying values at all change of loading positions and at points of zero shear. Neglect the mass of the beam.

Problem

Write shear and moment equations for the beams as shown in figure below. Also, draw shear and moment diagrams, specifying values at all change of loading positions and at points of zero shear. Neglect the mass of the beam.

Problem

Write shear and moment equations for the beams as shown in figure below. Also, draw shear and moment diagrams, specifying values at all change of loading positions and at points of zero shear. Neglect the mass of the beam.

Problem

Without writing shear and moment equations, draw the shear and moment diagrams for the cantilever beam acted upon by a uniformly distributed load and a couple as shown in the figure below.

Problem

Without writing shear and moment equations, draw the shear and moment diagrams for the cantilever beam acted upon by a uniformly distributed load and a couple as shown in the figure below.

Problem

The beam loaded as shown in the Figure consists of two segments joined by a frictionless hinge at which the bending moment is zero

Stresses in Beam

• Loads acting on a beam create internal actions (or stress resultants) in the form of shear forces and bending moments.
• The loads acting on a beam cause the beam to bend (or flex), thereby deforming its axis into a curve.
• The initially straight axis is bent into a curve called the deflection curve of the beam.

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Pure Bending & Non-Uniform Bending

• Pure bending refers to the flexure of a beam under a constant bending moment. Therefore, pure bending occurs only in regions of a beam where the shear force is zero (because V = dM/dx)

Pure Bending & Non-Uniform Bending

• Non-uniform bending refers to flexure in the presence of shear forces, which means that the bending moment changes as we move along the axis of the beam
• The figure shows an example of a beam that is partly in pure bending and partly in non-uniform bending.

Curvature of a Beam

Longitudinal Strains in Beam

• Because of the bending deformations cross sections mn and pq rotate with respect to each other
• Longitudinal lines on the lower part of the beam are elongated, whereas those on the upper part are shortened.
• The lower part of the beam is in tension and the upper part is in compression.
• Somewhere between the top and bottom of the beam is a surface in which longitudinal lines do not change in length. This surface, indicated by the dashed line ss is called the neutral surface of the beam.
• Its intersection with any cross-sectional plane is called the neutral axis

Longitudinal Strains in Beam

To evaluate the strain for an arbitrary line ef of a beam subjected to bending, which is not on the neutral surface

Length Before Bending:

Length After Bending:

Strain:

Normal Stresses in Beam

• Positive stress acting on that element dA produces an element of the moment
• This element of moment acts opposite in direction to the positive bending moment M. Therefore,

dA

Normal Stresses in Beam

dA

….....…………….………….(i)

………………………….(ii)

Combining equation (i) and (ii)

• For a symmetric beam, where will the stress be maximum?

Flexural Formula

Normal Stresses in Beam

Self Study

Review:

• What is the Area Moment of Inertia?
• How to determine the area moment of inertia of different shapes of areas about their centroids?

Area Moment of Inertia

Problem

Determine the minimum height h of the beam shown in Fig. P-508 if the flexural stress is not to exceed 20 MPa.

Problem

A beam with an S310 × 74 section (see Appendix B of the textbook) is used as a simply supported beam 6 m long. Find the maximum uniformly distributed load that can be applied over the entire length of the beam, in addition to the weight of the beam, if the flexural stress is not to exceed 120 MPa.

Problem

A 30-ft beam, simply supported at 6 ft from either end carries a uniformly distributed load of intensity w_o over its entire length. The beam is made by welding two S18 × 70 (see appendix B of text book) sections along their flanges to form the section shown. Calculate the maximum value of w_o if the flexural stress is limited to 20 ksi. Be sure to include the weight of the beam..

Unsymmetrical Beam

• The compressive and tensile stresses will be the same for a symmetrical section.
• This will be desirable if the material is equally strong in tension and compression.
• However, some materials, such as cast iron, are stronger in compression than tension.
• It is therefore desirable to use a beam with an unsymmetrical cross-section with the stronger fibers located at a greater distance from the neutral axis than the weaker fibers.

Unsymmetrical Beam-Problem

Find the maximum tensile and compressive flexure stresses for the cantilever beam shown in the figure below.

Economic Section

• In beams with rectangular or circular cross-sections, the fibers near the neutral axis experience lower stress compared to those located at the top or bottom.
• As a result a significant portion of the cross-section remains under-stressed.
• The presence of under-stressed regions in the cross-section of a beam reduces its efficiency in resisting flexure.
• For steel beams or composite beams, instead of adopting the rectangular shape, the area may be arranged so as to give more area on the outer fiber maintaining the same overall depth and saving a lot of weight.

Economic Section- Problem

A 10 m beam supported at the ends carries a uniformly distributed load of 16 kN/m over its entire length. What is the lightest W-shaped beam that will not exceed a flexural stress of 120 MPa?

Economic Section- Problem

A 10 m beam supported at the ends carriers a uniformly distributed load of 16 kN/m over its entire length. What is the lightest W shape beam that will not exceed a flexural stress of 120 MPa?

Shear Stresses

• Why Share Stress at a point is equal in horizontal and vertical plans?

Shear Stresses in Beams

Shear Stresses in Beams

• Shear stresses acting on the cross-section are parallel to the shear force, that is, parallel to the vertical sides of the cross-section.
• The shear stresses are uniformly distributed across the width of the beam, although they may vary over the height.
• There are horizontal shear stresses acting between horizontal layers of the beam as well as vertical shear stresses acting on the cross sections. These complementary shear stresses are equal in magnitude at any point in the beam.

Shear Stresses in Beams

• It is easier to evaluate the horizontal shear stresses acting between layers of the beam.
• Of course, the vertical shear stresses have the same magnitudes as the horizontal shear stresses.

Shear Stresses in Beams

Shear Stresses in Beams

….....…………….………….(i)

….....……………….………….(ii)

Combining equation (i) and (ii)

Shear Formula

First Moment, Q & Maximum Share Stress

Where,

Shear Stresses in the Web

Area of the upper flange

Area of the rectangle efcd

Problem

The T section shown in the figure is the cross-section of a beam formed by joining two rectangular pieces of wood together. The beam is subjected to a maximum shearing force of 60 kN. The neutral axis is 34 mm from the top and I_NA = 10.57 × 106 mm⁴. Using these values, determine the shearing stress (a) at the neutral axis and (b) at the junction between the two pieces of wood.

Assignment

A uniformly distributed load of 200 lb/ft is carried on a simply supported beam span. If the cross-section is as shown in the figure, determine the maximum length of the beam if the shearing stress is limited to 80 psi. Assume the load acts over the entire length of the beam.

Beam Deflection

• The calculation of deflections is an important part of structural analysis and design.
• Deflections are also important in dynamic analyses, as when investigating the vibrations of aircraft or the response of buildings to earthquakes.

Beam Deflection

• Under relatively small changes in shape, use an approximation

Beam Deflection: Problem

Beam Deflection: Problem

Beam Deflection: Area Moment Theorem

Beam Deflection: Area Moment Theorem

Beam Deflection: Area Moment Theorem

Beam Deflection: Area Moment Theorem

Problem: Area Moment Theorem

Rules of Sign: Area Moment Theorem

Problem: Area Moment Theorem

Find the maximum deflection for the cantilever beam loaded as shown if the cross section is 50 mm wide by 150 mm high. Use E = 69 GPa.

Problem: Area Moment Theorem

Compute the deflection and slope at a section 3 m from the wall for the beam shown in the Fig. Assume that E = 10 GPa and I = 30 × 106 mm⁴.

Problem: Area Moment Theorem

Determine the midspan value of EIδ for the beam shown in Fig.

Problem: Area Moment Theorem

Determine the value of EIδ at the right end of the overhanging beam shown in the figure. Is the deflection up or down?

Beam Deflection: Superposition

For only uniformly distributed load

For only concentrated load

Use tables of Appendix-G for deflection of common loads

Problem: Superposition

What would be the case if force P was upward?

Problem: Superposition

What would be the case if force P was upward?

Problem: Superposition

What would be the case if force P was upward?

Problem: Superposition

The beam shown in Fig. P-689 has a rectangular cross-section 4 inches wide by 8 inches deep. Compute the value of P that will limit the midspan deflection to 0.5 inch. Use E = 1.5 × 10⁶ psi.

Problem: Superposition

The beam shown in Fig. P-689 has a rectangular cross-section 4 inches wide by 8 inches deep. Compute the value of P that will limit the midspan deflection to 0.5 inch. Use E = 1.5 × 10⁶ psi.

Problem: Superposition

600 N/m

2 m

4 m

Problem: Superposition

Problem: Superposition

A simple beam AB of span length L has an overhang BC of length a. The beam supports a uniform load of intensity q throughout its length. Obtain deflection at the overhanging end

Break the beam into two parts and consider

• One part as simply supported
• Overhanging part as cantilever

Problem: Superposition

A simple beam AB of span length L has an overhang BC of length a. The beam supports a uniform load of intensity q throughout its length. Obtain deflection at the overhanging end

Problem: Superposition

A simple beam AB of span length L has an overhang BC of length a. The beam supports a uniform load of intensity q throughout its length. Obtain deflection at the overhanging end

Indeterminate Beam

• If the number of reactions exerted upon a beam exceeds the number of equations in static equilibrium, the beam is said to be statically indeterminate.
• In order to solve the reactions of the beam, the static equations must be supplemented by equations based upon the elastic deformations of the beam.

• In the case of propped beam, there are three reactions R1, R2, and M and only two equations (ΣM = 0 and sum; F_y = 0) can be applied, thus the beam is indeterminate to the first degree (3 – 2 = 1).

Analysis by Double Integration Method

• The most fundamental method for analyzing a statically indeterminate beam is to solve the differential equations of the deflection curve

For detail see section 9.2

First Integration provides the slope along the beam

Double Integration provides the deflection along the beam

Analysis by Double Integration Method- Problem

• Find the reactions at the supports

Analysis by Double Integration Method- Problem

Print Appendix G

Analysis by Method of Super Position

• The analysis begins by noting the degree of static indeterminacy and selecting the redundant reactions.

Analysis by Method of Super Position

Since the deflection of the original beam is zero at point B

Analysis by Method of Super Position

Analysis by Method of Super Position

Assignment

Find reactions to given problems by using the Double integration method and superposition method.