Applications to Deformations in StructuresAPLdExternal Work done by forces on structure = internal Strain EnergyConsider the truss shown at the right: External Work =FD where D = deflection at B.All 7 members have same AE (axial stiffness)

This assumes that the system is linear-elastic, and therefore the deflection D is a linear function of F.D, Deflection at BLoad, FThe total strain energy stored in the system is the sum of the individual strain energies in each of the truss members numbered i=1 to 7.

DBEquating External Work to Internal Strain Energy:Since, in this case, EA is constant for all members,

Try it!A lightweight aluminum truss, E=70000 MPa, has a height, Lo=1m, and is made of tubular stock with a cross sectional area of 250 mm2. Determine the deflection at B when the load, F=20kN.OK!..

Lets use an Energy method!

Solution:Solve the truss using statics to determine the forces in each member:Note: It doesnt matter if the axial load is compressive or tensile, both add to the total strain energy since the P term is squared.

Forces

Length of Members

Pi2Li

P1=-F

L1=L0

F2 L0

P2=

L2=

2

L0

P3=-F

L3=L0

F2 L0

P4=-2F

L4=L0

4F2 L0

P5=

L5=

2

L0

P6=F

L6=L0

F2 L0

P7=0

L7=L0

0

Sum

(7+4(2) F2 L0

_1097407079.unknown

_1097515264.unknown

_1097515386.unknown

_1097515444.unknown

_1097515361.unknown

_1097407100.unknown

_1097407047.unknown

OK, already find the deflection.What is the maximum stress in the truss?P4 =-2F=40kN (compression)S=P/A = 40000/250 = 160 MPa assumptions about linear elasticity are OK, if Sy>160MPa

Beam Elements:yxydxdAzyF(x)A beam that is symmetrical in x-section about the z-axis, is subjected to bending. Consider a infinitesimal volume element of length dx and area dA as shown. This element is subjected to a normal stress: sx=My/IThe Strain Energy Density on this element is:

For linear elastic material

Substituting, and multiplying by the Volume of the elementHence, the Strain Energy for a slice of the beam, of width dx, isyxdx

_1097407435.unknown

_1097407608.unknown

_1097407347.unknown

Assumptions.There is no axial force on the element (case of pure bending)The shear stress & strain on the element is relatively small, and hence the contribution of shear strain energy is negligible.These assumptions are generally valid for long slender beams.

Strain Energy in Entire BeamIFLdConsider the cantilever beam as shownxy

M=F(x-L)

_1097407676.unknown

_1097407765.unknown

Deflection ILdxyFExternal Work,Strain EnergyLinear-elastic,F dClassical Solution

=

_1097407826.unknown

_1097522656.unknown

_1097522689.unknown

_1097522704.unknown

_1097407857.unknown

_1097407765.unknown

Try it!yxPL/2L/2Shear ForcePL/4P/2-P/2MomentDetermine Elastic Strain Energy due to bending for simply supported 3-point bending member of constant X-section.For 0 xL/2: M=Px/2Note by symmetry we can find the total strain energy by doubling the strain energy of the LHS.

yL/2L/2PDBDetermine DB.

DB can be determined by an energy method:.

Find the Deflection for the x-section shown: L=2m, P=20 kN; h=50mm, b=20mm, SAE1045 Steel, E=200,000 MPa, Sy=400MPab=20 mmh=50mmzyIzz=bh3/12< Sy , OK

Elastic Strain Energy due to Transverse Shear Stressyx d = gxya txygxy

Shear Strain EnergyxydxdAzyF(x)f is called a form factor:Circle f=1.11Rectangle f=1.2Tube f=2.00I section f=A/Aweb

Try it.yxPL/2L/2Shear ForcePL/4P/2-P/2MomentDetermine Elastic Strain Energy due to shear strain for simply supported 3-point bending member of constant X-section.For 0 xL/2: T=P/2Note by symmetry we can find the total strain energy by doubling the strain energy of the LHS.

bmmh=2czyIzz=bh3/12A=bhfor rectangle x-sectionFor most metals G0.4E we find that:And for slender beam, L/h=10which may be neglected