Solids And Structures Semester 1 PDF

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Solids And Structures Ԇ��Solids And StructuresShear CentreSymmetric Loading of Thin-WalledMembersSolids And Structures ԇThis causes bending, shear and twisting. In this case, the shear stress formula is not valid as twisting is occurring. But, the flexure formula is valid. The forces between AB and ...

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Solids And Structures Shear Centre Symmetric Loading of Thin-Walled Members

This causes bending, shear and twisting. In this case, the shear stress formula is not valid as twisting is occurring. But, the flexure formula is valid. The forces between AB and DE are equal and so produce a couple. But, these cannot be balanced and so V needs to be moved. This removes the problem of excess stress from no symmetric loading. The couple can be eliminated if V is moved to the left through a distance e such that the moment about B is equal to F x h.

To eliminate twisting a ‘plate’ can be attached to make the end symmetrical causing only bending.

Question Will the section twist under effect of oblique load? The section won’t twist due to there being two separate loads that can be split up into horizontal and vertical components

Example Determine the location, e, of the shear centre, O, of a channel section of uniform thickness when b = 100mm, h = 150mm and t = 4mm.

If V is in the final part, it is wrong.

Impact Loading Impact: When an object strikes another such that large forces are developed in a short period of time.

Deflection of a Spring Under Impact Loading Naive Method: Force balance A Find the Mass detached from spring

B Mass in contact with the spring

Smart Method: Energy Balance

Ue = Ui Ue - Initial Potential energy stored in the mass Final potential energy stored in the spring

mg/k = Static maximum deflection

This is Dynamic Load Factor DLF

Deflection of a beam under impact loading Work done = strain energy

A. Weight F=mg falling through height h to produce max deflection. B. Force applied gradually to maximum value F.

C. Force gradually applied to maximum value Feq or produce

Deflection of a Beam Under Impact Loading The elastic alloy deflected forms of A and C are identical so the strain energy U stored in A will be equal to that stored in C. By conservation of energy the potential energy PE of the weight F at heigh h, case A is fully converted into U in the beam when F falls to produce max deflection.

P E(A) = U (A) = U (C) = W (C)

Vertical impact (kinetic energy) KEMV^2/2

Plasticity in Ductile Materials

Deformation Conservation of energy for plastic deformation implies that

Parameters Affecting Failure Mode, Performance and Cost Length Thickness Material Section Dimensions

Bend Radius Hole Size Faster Type Flange Length Bolt Pre-Load Pitch Of Fasteners Tolerance Quantity

Crashworthiness Parameters Mean Load Maximum Load Energy Absorption Crush Efficiency Structural Effectiveness

Impact Duration Simplifying assumption: Constant deceleration

Inelastic Impact: Energy Loss

M2 ∗ V2 = (M1 + M2 )V32

This is the energy the crashbox needs to absorb.

Constitutive Strain Rate Equation The Cowper-Symonds constitutive equation is

Epsilon is the uniaxial plastic strain rate Sigma 0 is the static flow stress Sigma 0 dash is the dynamic flow stress D and q are constants for a particular material Mild steel: D40.4 s^1, q=5 Stainless Steel: D100 s^1, 1 10 Aluminium Alloy: D6500 s^1, q=4

Intro to Energy Methods Energy Methods

Energy methods have a very important role to play in engineering design and analysis. It will be shown that, by considering the balance between internal strain energy and the external work done on a structure, the deflections and rotations at a particular point in a structure can be readily computed.

Concepts of Stored Elastic Strain Energy When an elastic material is deformed, work is done by the applied load in producing the deformation The strain energy is the increase in energy associated with the deformation. This energy is stored in the material and may be recovered when the material resumes its initial dimensions upon release of the load. “Elastic” means that the loading and unloading load-displacement or stress-strain curves coincide.

Only elastic energy can be recovered.

Work Done by a Single Load Consider a bar subject to external tension load F

Work done during a small increment of extension du is f*du. Total work done to strain is the area under the load-displacement curve. Material must be linear-elastic and load-displacement relationship linear.

Work Done By A Moment Beams and shafts support moments in bending M and torsion T. Moment M can be replaced by equivalent force couple h/2d(twist)

Strain Energy in Tension Area under the load-displacement curve is the stored energy in the bar. Dividing AL then the stored energy per unit volume is given by:

Torsion angle is = TL/GJ

Strain Energy in Bending Stress = My/I

Problem Example

Summary

Castigliano’s Theorem and Applications Summary

Steps to find deflection and slope using Castigliano’s Theroem:

 If needed, add imaginary point loads at the point of interest: Point force if interested in deflection, point moment if interested in slope.  Compute internal strain energy  Use castigiano’s theorem:  Put virtual point forces and moments to zero in the final expression.

For Deflection:

For Rotation:

Discontinuous Structures

A beam ABC having flexural stiffness EI and torsional stiffness GJ, is bent in an L-shape in the horizontal plane, Determine the deflection under the vertical end load F at C.

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Statically Indeterminate (or Redundant) Structures

This propped cantilever has three unknown reactions: Ra, Rb and Mb. As there are only two equilibrium equations: The sum of FV = 0 and the sum of moments = 0, the problem is statistically indeterminate i.e. the reactions cannot be found by equilibrium alone.

Week 6 Lecture Stuff If you know U the deflection or slopes at the point of application of force and moments can be found with the following equations:

Pin-Jointed Frames Overview

Recap of Energy Methods Analysis of Frames Principle of Virtual Work Applied to Frames Worked Examples

Recap of Energy Methods When a structure is deformed by external loads it stores potential energy as internal strain energy By conservation of energy, such internal strain energy must be equal to the work done by the external loads on the structure Work done by external loads:

Uniaxial tension/compression

Bending

Torsion

Analysis of Pin-Jointed Frames https://s3-us-west-2.amazonaws.com/secure.notion-static.com/d9f682 8278e844859ec06b56d3625304/Pin-Jointed_Frames.pdf

Example 1 https://s3-us-west-2.amazonaws.com/secure.notion-static.com/6cb74d 5e-c091432a-841e-01345cae4ce7/Pin-Jointed_Frames_Example_1.pdf

Approach to Analysis of Pin-Jointed Frames  Find support reactions using equilibrium  Find Member forces using method of joints  Apply exterior work = interior work

Pin-Jointed Frames 2 Overview

Recap of previous lecture Virtual Work Method: Deflection at any generic joints in the structure, analysis of pin-jointed frames with multiple loads Examples

Recap In last lecture we have seen how we can use energy methods to analyse pinjointed frames.

The work done by external forces at the joints equals the internal work resulting from the tensions and compressions in the members.

Example

Vertical Displacement Since the displacement that we wish to find is in the direction of the applied load then energy conservation implies

Virtual Work The principle of virtual work is one of the most powerful tools in a structural analysis. Virtual work on a system is the work resulting from either virtual forces acting through a real displacement or real forces acting through a virtual displacement.

Imaginary Forces In determining the vertical displacement, the work terms comprised real (actual) forces and real (actual) displacements.

The displacement at any joint in any direction can be determined, whether loaded or not, by the use of an imaginary load of unit magnitude. This

Horizontal Displacement If an imaginary unit load is placed horizontally at D, then the equilibrium System of forces in the members due to the unit load only (the 10kN load removed) can be found. The values are given in the following table

The real horizontal displacement if joint D due to the actual load of 10kN is Uh. The external virtual work due to the unit load is 1 x Uh. Applying the work equation gives:

Multiple Loads

The method of using real forces and extensions cannot be used on a frame carrying several external loads, unless all the displacements at the loaded joints, except one, are known, since there would be several unknown u values on the left-hand side.

Example

Lecture Example 1

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https://s3-us-west-2.amazonaws.com/secure.notion-static.com/4a7d89 5063f64fb886d9-a7b10cb87768/Pin-Jointed_Frames_2.pdf

Mohr Circle and Yield Criteria Overview Principal Stress and Strain Graphical Methods: Mohr’s Circle Yield Criteria: Tresca and Von Mises Worked Examples

Principal Stress State Complex stress and strain exists in most components and structures It isn’t sufficient just to determine these stresses in order to size the component or select a suitable material More severe stresses and strains may exist on certain planes Identification of the maximum stresses is essential for designing a component The Mohr circle represents all possible states of normal and shear stress on any plane through a stressed point.

Plane Stress Transformation Normal Stress: positive when along the outward normal from a side of the element. Shear Stress: positive when pointing in the y direction Angles: positive if counter clockwise

Mohr’s Circle Constuction

Principal Stresses

Mohr’s Circle Key Features

Maximum principal stress sigma1 acts on a plane at angle theta from the plane on which sigma x acts, where:

The minimum principle stress, sigma2, acts On a plane at 90 degrees from the sigma1 plane There is no shear stress on the principal stress planes Maximum shear stress:

And acts on planes at 45 degrees from the principal planes

Tension

The rightmost point that crosses the x axis is P/A. Everything else is just 0 and so is at the origin.

Compression

Pressure

Torison

Example Consider a cylindrical bar subjected to combined tension and torsion such the the stress state is as follows:

First, construct mohr’s circle:

To find radius, we can look at cd and use Pythagoras to determine Tmax. Since ch and cd are both radii:

Note on 3D Case For a 3D case:

We have 3 separate parts to look at:

Which gives us 3 Mohr’s Circles:

Yield Criteria Once we know principal stress we can find out if a material is yielding

There are two yield criteria based on concepts of shear which are now accepted and used for ductile materials

Maximum Shear Stress (Tresca) Criterion Yielding is dependent on the maximum shear stress in the material reaching a critical value This is taken as the maximum shear stress at yielding in a uniaxial tensile test Ductile metals tend to fail in tension by shearing on planes approximately 45 degrees to the direction of the tensile stress. This usually leads to a local contraction or ‘necking’ of the material.

Uniaxial Tension

3D Principal Stress State

Maximum Shear Strain Energy (Von Mises) Criterion The total elastic strain energy stored in a material could be considered as consisting of energy stored due to change in volume (distortion) and change in shape (shear) Proposed that stored strain energy could provide a viable criterion for complex yield conditions Equating the 3D principal stress state with the uniaxial stress state gives:

Week 10 Lecture Notes https://s3-us-west-2.amazonaws.com/secure.notion-static.com/89d7c5 bc-c281407e-a1118dd6108b55ac/Week_10_Examples.pdf

Revision For Semester 1 Stress In a Transverse Plane

2013 Exam Question

V = 20kN

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Energy Methods

Castigliano’s Theroem:  If needed, add imaginary point loads at the point of interest  Compute internal strain energy  Use Castigliano’s Theroem:  Put Virtual Point forces and moments to zero in the final expression

Step By Step For Each Topic

Thin-Walled Members Very common in engineering applications: Narrow beams

I section Wing Box Under this condition it is safe to assume the distribution of shear (flow) is constant across the thickness.

Longitudinal Shear Stresses in a Beam

If top and bottom surfaces are free of stress then the vertical shear flow is zero (because it is zero at the surfaces and must be constant across the thickness). Only shear flow parallel to the walls of the member is important.

Questions How does the shear flow vary in segments that are perpendicular to V? I section) V and I are given In horizontal segments A’ varies linearly from zero (at extremities to its maximum on the axis of symmetry. Q is linear making q linear. In vertical segments A’ varies linearly from zero (at top) to its maximum on neutral axis ad y’ varies linearly as well Q is quadratic and q is quadratic. q is shear flow per unit length Q is the first moment of area (y’ x Area)...