# Strength of materials and theory of structures pdf

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- Structural Mechanics
- Structural engineering
- Strength of Materials and Structures Fourth edition JOHN CASE LORD CHILVER of Cranfield
- Strength of Materials

*Chilver — Strength of Materials and Structures: An Introduction to the Mechanics of Solids and Structures provides an introduction to the application of basic ideas in solid and structural mechanics to engineering problems. This book begins with a simple discussion of stresses and strains in materials, structural components, and forms they take in tension, compression, and shear.*

## Structural Mechanics

When a force is applied to a structural member, that member will develop both stress and strain as a result of the force. The applied force will cause the structural member to deform by some length, in proportion to its stiffness.

Strain is the ratio of the deformation to the original length of the part:. There are different types of loading which result in different types of stress, as outlined in the table below:. In the equations for axial stress and transverse shear stress , F is the force and A is the cross-sectional area of the member.

In the equation for bending stress , M is the bending moment, y is the distance between the centroidal axis and the outer surface, and I c is the centroidal moment of inertia of the cross section about the appropriate axis. In the equation for torsional stress, T is the torsion, r is the radius, and J is the polar moment of inertia of the cross section. In the case of axial stress over a straight section, the stress is distributed uniformly over the entire area.

More discussion can be found in the section on shear stresses in beams. In the case of bending stress and torsional stress, the maximum stress occurs at the outer surface.

More discussion can be found in the section on bending stresses in beams. We have a number of structural calculators to choose from. Here are just a few:. Just as the primary types of stress are normal and shear stress, the primary types of strain are normal strain and shear strain.

In the case of normal strain, the deformation is normal to the area carrying the force:. In the case of transverse shear strain, the deformation is parallel to the area carrying the force:.

The maximum shear strain occurs on the outer surface. In the case of a round bar, the maximum shear strain is given by:. The shear strains are proportional through the interior of the bar, and are related to the max shear strain at the surface by:. Stress is proportional to strain in the elastic region of the material's stress-strain curve below the proportionality limit, where the curve is linear.

The elastic modulus and the shear modulus are related by:. Essentially, everything can be treated as a spring. Hooke's Law can be rearranged to give the deformation elongation in the material:. When force is applied to a structural member, that member deforms and stores potential energy, just like a spring. The strain energy i. The total strain energy corresponds to the area under the load deflection curve, and has units of in-lbf in US Customary units and N-m in SI units.

The elastic strain energy can be recovered, so if the deformation remains within the elastic limit, then all of the strain energy can be recovered. Note that there are two equations for strain energy within the elastic limit. The first equation is based on the area under the load deflection curve. The second equation is based on the equation for the potential energy stored in a spring. Both equations give the same result, they are just derived somewhat differently.

More information on strain energy can be found here. Stiffness, commonly referred to as the spring constant, is the force required to deform a structural member by a unit length. All structures can be treated as collections of springs, and the forces and deformations in the structure are related by the spring equation:.

However, the maximum deflection is typically not known, and so the stiffness must be calculated by other means. Beam deflection tables can be used for common cases.

The two most useful stiffness equations to know are those for a beam with an axially applied load, and for a cantilever beam with an end load. Note that stiffness is a function of the material's elastic modulus, E , the geometry of the part, and the loading configuration. If there are multiple load paths in a structure i. To find the load carried by any individual member, first calculate the equivalent stiffness of the members in the load path, treating them as springs. Depending on their configuration, they will be treated as some combination of springs in series and springs in parallel.

If the members in the load path cannot be treated purely as springs in series or as springs in parallel, but are rather a combination of springs in series and in parallel, then the problem will need to be solved iteratively.

Fine a sub-grouping of the members that are either purely in series or in parallel, and use the equations provided to calculate the equivalent stiffness, force and deflection in the sub-group. The sub-group can then be considered a single spring with the calculated stiffness, force, and deflection, and that spring can then be considered as a part of another sub-group of springs. Continue grouping members and solving until the desired result is achieved.

Forces and stresses can be thought to flow through a material, as shown in the figure below. When the geometry of the material changes, the flow lines move closer together or farther apart to accommodate. If there is a discontinuity in the material such as a hole or a notch, the stress must flow around the discontinuity, and the flow lines will pack together in the vicinity of that discontinuity. This sudden packing together of the flow lines causes the stress to spike up -- this peak stress is called a stress concentration.

The feature that causes the stress concentration is called a stress riser. Check out our interactive plots for common stress concentration factors. Stress concentrations are accounted for by stress concentration factors.

To find the actual stress in the viscinity of a discontinuity, calculate the nominal stress in that area and then scale it up by the appropriate stress concentration factor:. When calculating the nominal stress, use the maximum value of stress in that area. For example, in the figure above, the smallest area at the base of the fillet should be used. Many reference handbooks contain tables and curves of stress concentration factors for various geometries.

Two of the most comprehensive collections of stress concentration factors are Peterson's Stress Concentration Factors and Roark's Formulas for Stress and Strain. MechaniCalc also provides a collection of interactive plots for common stress concentration factors.

The concentration of stress will dissipate as we move away from the stress riser. Saint-Venant's principle is a general rule of thumb stating that the distance over which the stress concentration dissipates is equal to the largest dimension of the cross section carrying the load. Calculation of stress concentration is particularly important when the materials are very brittle, or when there is only a single load path.

In ductile materials, local yielding will allow for stresses to be redistributed and will reduce the stress around the riser. For this reason, stress concentration factors are not typically applied to structural members made of ductile materials. Stress concentration factors are also not typically applied when there is a redundant load path, in which case yielding of one member will allow for redistribution of forces to the members on the other load paths.

An example of this is a pattern of bolts. If one bolt starts to give, then the other bolts in the pattern will take more of the load. At any point in a loaded material, a general state of stress can be described by three normal stresses one in each direction and six shear stresses two in each direction :. The first indicates the direction of the surface normal, and the second indicates the direction of the shear stress itself. Commonly, the stresses along one direction are zero so that the full state of stress occurs on a single plane, as shown in the figure below.

This is called plane stress. Plane stress occurs in thin plates, but it also occurs on the surface of any loaded structure. Surface stresses are commonly the most critical stresses since bending stress and torsional stress are maximized at the surface. The stresses balance so that the point is in static equilibrium.

Because the shear stresses are all equal in magnitude, the subscripts are dropped for simplicity. Note however that the sign of the stresses on the x face will be opposite to those on the y face.

The proper sign conventions are as shown in the figure. For normal stress, tensile stress is positive and compressive stress is negative. For shear stress, clockwise is positive and counterclockwise is negative. The transformation equations below give the values of the normal stress and shear stress on this rotated plane. At any point in the material, it is possible to find the angles of the plane at which the normal stresses and the shear stresses are maximized and minimized.

The maximium and minimum normal stresses are called principal stresses. The maximum and minimum shear stresses are called the extreme shear stresses. The angles above can be substituted back into the transformation equations to find the values of the principal stresses and the extreme shear stresses:.

Principal stresses are always accompanied by zero shear stress. Mohr's circle is a way of visualizing the state of stress at a point on a loaded material. From Mohr's circle, it also becomes clear what are the principal stresses, the extreme shear stresses, and the angles at which thoses stresses occur.

An example of Mohr's circle is shown in the figure below:. Check out our Mohr's Circle calculator based on the methodology described here. To construct Mohr's circle, first locate the center of the circle by taking the average of the normal stresses:. Place points on the circle representing the stresses on the x and y faces of the stress element.

Place points on the circle for the principal stresses. Place points on the circle for the extreme shear stresses. All of the points will lie on the perimeter of the circle.

The circle has a radius equal to the magnitude of the extreme shear stresses:. This line in Mohr's circle corresponds to the unrotated element in the figure below. If this line is rotated by some angle, then the values of the points at the end of the rotated line will give the values of stress on the x and y faces of the rotated element. It is important to note that the degrees of Mohr's circle are equivalent to degrees on the stress element. For instance, the points for x face and the y face are degrees apart on Mohr's circle, but they are only 90 degrees apart on the stress element.

To get a more intuitive feel for how Mohr's circle relates the stresses on a stress element and how the stress state changes as a function of rotation angle, see the accompanying Mohr's circle calculator.

## Structural engineering

It seems that you're in Germany. We have a dedicated site for Germany. This text book covers the principles and methods of load effect calculations that are necessary for engineers and designers to evaluate the strength and stability of structural systems. It contains the mathematical development from basic assumptions to final equations ready for practical use. It starts at a basic level and step by step it brings the reader up to a level where the necessary design safety considerations to static load effects can be performed, i.

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## Strength of Materials and Structures Fourth edition JOHN CASE LORD CHILVER of Cranfield

When a force is applied to a structural member, that member will develop both stress and strain as a result of the force. The applied force will cause the structural member to deform by some length, in proportion to its stiffness. Strain is the ratio of the deformation to the original length of the part:. There are different types of loading which result in different types of stress, as outlined in the table below:. In the equations for axial stress and transverse shear stress , F is the force and A is the cross-sectional area of the member.

Solved examples with detailed answer description, explanation are given and it would be easy to understand. Here you can find objective type Civil Engineering Strength of Materials questions and answers for interview and entrance examination. Multiple choice and true or false type questions are also provided. You can easily solve all kind of Civil Engineering questions based on Strength of Materials by practicing the objective type exercises given below, also get shortcut methods to solve Civil Engineering Strength of Materials problems.

This page is the portal of the Reviewer in Strength of Materials. You can find here some basic theories and principles. Most of the content however for this online reviewer is solution to problems.

*The strength of a material, whatever its nature, is defined largely by the internal stresses, or intensities of force, in the material.*

### Strength of Materials

Structural engineering is a sub-discipline of civil engineering in which structural engineers are trained to design the 'bones and muscles' that create the form and shape of man-made structures. Structural engineers also must understand and calculate the stability, strength, rigidity and earthquake-susceptibility of built structures for buildings [1] and nonbuilding structures. The structural designs are integrated with those of other designers such as architects and building services engineer and often supervise the construction of projects by contractors on site. See glossary of structural engineering.

The 1 ehtion was begun by Mr. John Case and Lord Chlver but, because of the death of Mr. John Case, it was completed by Lord Chlver.

Провели первый реальный тест. Несмотря на сомнения относительно быстродействия машины, в одном инженеры проявили единодушие: если все процессоры станут действовать параллельно, ТРАНСТЕКСТ будет очень мощным. Вопрос был лишь в том, насколько мощным. Ответ получили через двенадцать минут. Все десять присутствовавших при этом человек в напряженном ожидании молчали, когда вдруг заработавший принтер выдал им открытый текст: шифр был взломан. ТРАНСТЕКСТ вскрыл ключ, состоявший из шестидесяти четырех знаков, за десять с небольшим минут, в два миллиона раз быстрее, чем если бы для этого использовался второй по мощности компьютер АНБ. Тогда бы время, необходимое для дешифровки, составило двадцать лет.