Mechanics Of Deformable Bodies Ii

Flexural Members Engineering Science 10 Strength of Materials: Why Things Bend and Break?

Department of Engineering Science College of Engineering University of the Philippines-Diliman

AH-1Pitch Control Rod (Crash impact forces)

T-2 Wing Carry Through Beams (Inflight failure)

TH-57 Rotor Mast Note: Not a Helical (Torsional) failure, so it is not separated due to rotational forces.

Beams

Beams What is a beam? A slender (usually) horizontal structural member that is subjected to a load that tends to bend it.

Beams Examples: Floor joists and rafters Tree branches Vertebral column and neck

Beams Types: Fixed beam (simply supported beam)

Beams Types: Cantilever beam

Beams A cantilever beam can be thought of as half of a fixed beam turned upside down.

Beams See!

Beams Types: Beam with overhang

Beams Types: Continuous beam

Beams All materials and structures deflect, to greatly varying extents when they are loaded.

The science of elasticity is about the interactions between forces and deflections. The material of the bough is stretched near its upper surface and compressed or contracted near its lower surface by the weight of the monkey.

Bending

Bending Forces on a cantilever beam

Bending If the material near the neutral axis is removed, the beam collapses.

Bending Material in the middle provides shear resistance.

Geometry of deformation

Geometry of deformation We will consider the deformation of an ideal, isotropic prismatic beam. All parts of the beam that were originally aligned with the longitudinal axis bend into circular arcs.

Geometry of deformation The neutral plane/surface or axis:

Geometry of deformation Where is the neutral plane/surface or axis?  It is the centroid of the cross-section.  In simple circular or rectangular shapes, it is the middle.  For more complex shapes, it is the center of gravity of a cutout of uniform thickness with the object’s cross-sectional shape.

Geometry of deformation Bending stresses in a beam:

Stress

Stress Let’s consider a segment of a loaded beam. The segment may be deformed as shown below.

Stress Hence we can sketch of the stress normal to the axis of the beam …

Stress The radius of curvature:

1 M = − ρ EI Where M is the internal bending moment acting on a section, E is the modulus of elasticity and I is the moment of inertia of the section.  Positive ρ means the positive y-axis is on the concave side of the neutral axis.  EI is known as the flexural rigidity of the beam.

Stress The fiber stress:

My σ = I Where M is the internal bending moment acting on a section, y is the distance of a point (fiber) from the neutral axis and I is the moment of inertia of the section.

Cross section

Cross section Flexural rigidity: EI gives the resistance to bending of a structure. Several conditions must be met in order to simply calculate it. The material must be:  Homogeneous  Isotropic  Linearly elastic  Deform equally under tension and compression  Not change shape appreciably under load

Cross section What does the I, the moment of inertia mean? Essentially, this says that the beam is stiffer if the material in the beam is located further away from the centroid (neutral axis).  So any small area is more effective at stiffening the beam depending on the square of the distance.  Hence, if you want to make a strong beam with little material, make sure that the material is as far as possible from the centroid.

Cross section What does the moment of inertia mean?  Hence, we have ‘I’ beams.

Cross section I beams are very stiff in bending but not that resistant in torsion.

Cross section Box girders (hollow square tube)

Very resistant to torsion

But stresses tend to concentrate at the corners

Cross section Hollow tubes

Best if the direction of the load cannot be predicted

Resistant to bending and torsion

Cross section

Typical beam cross sections and the ratio of I to the value for a solid square beam of equal crosssectional area.

Cross section

A T-cross section can be used to decrease either the maximum tensile or the maximum compressive stress to which the beam is subjected.

Cross section Some organisms take advantage of I to localize bending.

Sea anemone, Metridium

Daffodil

Beam technology

Beam technology The I beam:

Beam technology Steel section terminology:

Beam technology While we are mostly assuming beams made of steel or other metals, many are made of concrete and concrete does not support a tensile stress.

For concrete beams, we assume that only the material on the compressive side of the neutral axis actually carries a load.

Beam technology One solution is pre-stressed concrete. Where metal bars set within the concrete are prestressed to provide an initial compression to the concrete beam. so it can withstand some tension, until the prestress is overcome

The yellow guidelines highlight the camber (upward curvature) of a prestressed double T. The pre-stressing strands can be seen protruding from the bottom of the beam, with the larger strands at the bottom edge. The tension is these strands produces the camber, the beam is straight when cast.`

Beam technology Arches

Beam technology How an arch works?

Torsional Members Engineering Science 10 Strength of Materials: Why Things Bend and Break?

Department of Engineering Science College of Engineering University of the Philippines-Diliman

F-18 Engine Shaft Torsional Buckling

UH-1N Turbine: Helical shaft failure

Torsional loads

Torsional loads Torque – twisting couple 1) Turbine exerts torque T on the shaft. 2) Shaft transmits the torque to the generator. 3) Generator creates an equal and opposite torque T’.

Torsional loads Net of the internal shearing stresses is an internal torque, equal and opposite to the applied torque,

Torsional loads Torque applied to shaft produces shearing stresses on the faces perpendicular to the axis.

Torsional loads Conditions of equilibrium require the existence of equal stresses on the faces of the two planes containing the axis of the shaft.  The existence of the axial shear components is demonstrated by considering a shaft made up of axial slats. The slats slide with respect to each other when equal and opposite torques are applied to the ends of the shaft.

Shaft deformations

Shaft deformations From observation, the angle of twist of the shaft is proportional to the applied torque and to the shaft length.

φ ∝ T φ ∝ L

Shaft deformations When subjected to torsion, every cross-section of a circular shaft remains a plane and undistorted.

Cross-sections for hollow and solid circular shafts remain plain and undistorted because a circular shaft is axisymmetric.

Shaft deformations Cross-sections of noncircular (non-axisymmetric) shafts are distorted when subjected to torsion.

Shearing

Shearing Consider an interior section of the shaft. As a torsional load is applied, an element on the interior cylinder deforms into a rhombus. Since the ends of the element remain planar, the shear strain is equal to angle of twist. It follows that

ρφ Lγ = ρ φ or γ = L

Shear strain is proportional to twist and radius.

Shearing The shearing stress varies linearly with the radial position in the section.

The elastic torsion formula,

Tc Tρ τ max = and τ = J J

Torsional failure

Torsional failure Ductile materials generally fail in shear. Brittle materials are weaker in tension than shear.

Torsional failure When subjected to torsion, a ductile specimen breaks along a plane of maximum shear, i.e., a plane perpendicular to the shaft axis.

Torsional failure When subjected to torsion, a brittle specimen breaks along planes perpendicular to the direction in which tension is a maximum, i.e., along surfaces at 45o to the shaft axis.

Angle of twist

Angle of twist Recall that the angle of twist and maximum shearing strain are related,

cφ γ max = L In the elastic range, the shearing strain and shear are related by Hooke’s Law,

τ max Tc γ max = = G JG

Angle of twist Equating the expressions for shearing strain and solving for the angle of twist,

TL φ = JG

Angle of twist If the torsional loading or shaft cross-section changes along the length, the angle of rotation is found as the sum of segment rotations

Ti Li φ = ∑ i J i Gi

Transmission

Transmission Design of transmission shafts: Principal transmission shaft performance specifications are:  power  speed Designer must select shaft material and crosssection to meet performance specifications without exceeding allowable shearing stress.

Transmission Determine torque applied to shaft at specified power and speed,

P = Tω = 2π fT P P T= = ω 2π f Where P is the power, T is the torque and ω is the angular velocity.

Transmission Find shaft cross-section which will not exceed the maximum allowable shearing stress,

Tc τ max = J J π 3 T = c = c 2 τ max

(

( solid shafts )

)

J π T 4 4 = c2 − c1 = c2 2c2 τ max

( hollow shafts )

Transmission

The derivation of the torsion formula,

Tc τ max = J

assumed a circular shaft with uniform cross-section loaded through rigid end plates.

Transmission The use of flange couplings, gears and pulleys attached to shafts by keys in keyways, and crosssection discontinuities can cause stress concentrations. Experimental or numerically determined concentration factors are applied as

Tc τ max = K J

Transmission Shaft design:  Shaft must have adequate torsional strength to transmit torque and not be overstressed.  Shafts are mounted in bearings and transmit power through devices such as gears, pulleys, cams and clutches.  Components such as gears are mounted on shafts using keys.  Shaft must sustain a combination of bending and torsional loads.

Slender Columns Engineering Science 10 Strength of Materials: Why Things Bend and Break?

Department of Engineering Science College of Engineering University of the Philippines-Diliman

Buckling

Columns

Columns Columns – Straight structural members subjected to compressive axial loads.

Columns Types of columns:

Short

Long

Columns Failure of Modes of Columns: Crushing

Columns Failure of Modes of Columns: Buckling

Columns

Crushing failure governs in short columns while buckling failure governs in long columns.

Buckling

Buckling When a structure (subjected usually to compression) undergoes visibly large displacements transverse to the load then it is said to buckle. Buckling may be demonstrated by pressing the opposite edges of a flat sheet of cardboard towards one another. For small loads the process is elastic since buckling displacements disappear when the load is removed.

Buckling

Buckling Types: Euler buckling  Column is smoothly bent from end to end.  Compressive forces on the concave side and tension forces on the convex

Buckling Types: Euler buckling The critical Euler load:

2

π EI P= 2 (KL) Where E is the Young’s modulus, I is the moment of inertia, K is the effective length factor and L is the length of the column.

Buckling Types of column and the Euler buckling:

Pinned ends:

Buckling Types of column and the Euler buckling:

Fixed and free ends:

Buckling Types of column and the Euler buckling:

Fixed ends:

Buckling Types of column and the Euler buckling:

Pinned and Fixed ends:

Buckling

2

π EI P= (KL)2 Types of column and the Euler buckling:

Buckling Types: Local buckling  The way empty cans fail.  Indicated by growth of bulges, waves or ripples.

Buckling Buckling proceeds in manner which may be either : Stable - in which displacements increase in a controlled fashion as loads are increased, ie. the structure's ability to sustain loads is maintained, or Unstable - in which deformations increase instantaneously, the load carrying capacity nose-dives and the structure collapses catastrophically.

Buckling

Neutral equilibrium is also a theoretical possibility during buckling – this is characterized by deformation increase without change in load.

Compression structures

Compression structures Advantage: so long as they are put together correctly, gravity will do the work of keeping everything in its proper place (no specific fasteners needed). Traditionally, large compression structures have been built of brick/stone. Disadvantage: susceptible to failure by bending; stone has little resistance to tension.

Compression structures How high can we build a column/wall? Theoretical limit: the point at which the vertical load becomes so great that the material/s are crushed beneath the weight of the material/s above them. The crushing strength of stone and brick is > 40 MN/m2 We should be able to build a column/wall several miles high.

Compression structures How high can we build a column/wall? The wall will fail long before the theoretical limit is reached. Why? Important point: mortar between bricks does not glue the bricks together (it carries no tensile load). Mortar fills in the surface irregularities in the bricks’ surfaces, so that the compressive load is not carried by a few high spots (which would increase local stress).

Compression structures Analysis of column/wall stability? Assumptions:  Compressive stresses are so low that the material will not be broken by crushing.  Mortar is used, so the fit between the stones is so good that the compressive force is distributed over the whole area of the joint.  Friction in the joints is so high that failure will not occur because of the stones sliding over one another.  The joints (mortar) have no tensile strength.

Compression structures Symmetrical vertical force,P

Compressive stresses are spread evenly over the entire width of the joint

Compression structures What if P is eccentric?  Compressive stresses are no longer uniform.  Stress will vary linearly across the width of the column/wall if the material is Hookean.  Whole joint is still in compression.

Compression structures P is now at the edge of the middle third of the column/wall.  Stress at the outside edge of the wall is zero.  We still have compressive forces across the entire width of the column/wall.

Compression structures P is outside of the middle third of the column/wall.  The outside edge of the wall is now in tension.  But mortar cannot carry tension stresses!  What will happen?

Compression structures The column/wall cracks. We are still okay, because the remaining width of the wall can still carry the load.

Compression structures Disaster!  If P’s line of action moves outside of the wall, the whole joint is put into tension.  The wall tips up and falls over.

Compression structures What if the vertical force is applied to the wall obliquely?  We need to know the point at each successive joint at which we can consider the weight to be acting.  We can connect the points and plot a thrust line.

Compression structures Thrust line

Compression structures Thrust line and stability

If the thrust line comes in contact with the edge of the wall at any point, the wall is liable to fall over. How can one insure that the thrust line stays within the middle third of the wall?

Compression structures One of the most effective strategies is to simply add more weight to the top of the wall

Compression structures That is the practical reason behind the decorative statues adorning so many large stone buildings

Compression structures

cornice or gargoyle

Next Meeting:

Project Bridge building competition, 9/14. Bring toothpicks