The normal stress each portion the rod elena pasternak slide

Handout for Students

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(biaxial loading)

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Lec notes 1, sl 41,42

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ν	=		strain		=	1	=	1	- dimensionless
ν	=		strain			1
Dr Elena Pasternak			strain			1

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Volumetric strain. Dilatation

e=[(1+ε11)(1+ε22)(1+ε33)-1]=ε11+ε22+ε33 +o(εij)

Since strains are small, we can neglect higher order terms:

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Beer et al (2015)

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To work on this problem you need to revise lec notes 2 sl 5, 7-8

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Deformations Under Axial Loading

From Hooke’s Law:

Fig. 2.17 Undeformed and	δ	=
	δ	=

		δ	=	∑	P i	L i
deformed axially-loaded rod.					P i	L i
deformed axially-loaded rod.					A i
Beer et al (2015)				i
				i

Example
Concept Application Problem 2.01 (Beer et al, 2015)

SOLUTION:

Beer et al (2015)

Apply a free-body analysis on each component to determine the
internal force

to determine internal forces

P 1=240 10 N 3

L 1	=	L 2				L 3	=
A 1	=	A 2	=	580 10	−6	A 3	=	190 10
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Knowing that the magnitude of P
is 4 kN, determine (a) the value of
Q so that the displacement at A is
zero, (b) the corresponding
displacement of B.

Beer et al (2015)

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Lec notes 2, sl 6

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δ = ∫L P x

0 EA x
( )

Denoting by E the Young’s modulus
(modulus of elasticity) of the material
and neglecting the effect of its
weight, determine the displacement
of point A.

Beer et al (2015)

structures

Force-displacement relations

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When statics (equilibrium eqns) is not sufficient to
determine either the reactions or internal forces, the system is statically indeterminate.

To analyse such structures we must supplement the
equilibrium equations with additional eqns pertaining to the displacements of the structure.

(equation of compatibility)

Compatibility eqn expresses the fact that the change in length of the bar must be compatible with the
conditions at the supports.

Redundant reactions are replaced with unknown
loads which along with the other loads must
produce compatible deformations.

(Superposition method)

bar and loading shown, assuming a close fit at

both supports before the loads are applied.

Solve for the displacement at B due to the

redundant reaction at RB.

and the reaction found at RB.

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A 1=A 2=400×10−6 m 2 A 3=A 4=250×10−6 m 2

L 1=L 2=L 3=L 4=.0 150 m

i
=.1 125

E
×10 9

P 1		=	P 2	=		−
A 1		=	400		×	10−		6	m		2	A 2		=	250
L 1		=	L 2	=		.0 300						×	10 3		)R B
L 1		=	L 2	P i		L i	=	−		(.1 95
δ	R		∑	P i		L i
			∑	A i		E i						E
			i	A i		E i						E

Find the reaction at A due to the loads and the reaction at B

∑	F y		=	=	kN		+
∑	F y		323	kN	R	A	=	323
R	A	=			R	A	=	323
R	A	=			R B		=	577

Problem 2.05

Determine the reactions at A and B
for the steel bar and loading (see
Figure), assuming that a 4.5mm
clearance exists between the bar and
the ground before the loads are
applied, E=200GPa.

5. Statically indeterminate structures

Force-displacement relations

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Problem 2.39
A polystyrene rod consisting of two
cylindrical portions AB and BC is
restrained at both ends and supports
two 30-kN loads as shown. Knowing
that E=3.1GPa, determine (a) the
reactions at A and C, (b) the normal
stress in each portion of the
rod.

7. Reflection/Feedback

Reflect on your learning in ENSC3004 Solid Mechanics this week and give us feedback on your learning experience.

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