# Solid Mechanics Spring 2007

Mechanics

Jul 18, 2012 (6 years and 4 days ago)

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Solid Mechanics Spring 2007

1/13
Chapter 2 & 3 - Concept of Stress, Strain and Deformation

A
F

A
F

Objectives

- Distinguish between normal, shear and bearing stresses
- Analyze simple frames and structures, which consist of members that are pin-connected
- Calculate normal and shear stresses in a general plane
- Define strain, ε.
- Determine modulus of elasticity (E), yield strength and ultimate strength from a stress vs.
strain plot (σ vs. ε)
- Understand fatigue and how to consider it when designing structures that will be subjected to
- Use factor of safety to determine the maximum allowable loads under design conditions
P P
P
P
F
P
P
cut
F
Solid Mechanics Spring 2007

2/13

Rod BC, length L, cross-sectional area A

B

C

Single Shear

Rivet CD connects plates A and B, which are subjected to tension forces with magnitude, F.

_________________________ develops on the _______________________ of the rivet.

FBD - section of rivet

Double Shear

Splice plates C and D and bolts connect plates A and B.

F
F
C
D
F
F
C
D
A
B
Solid Mechanics Spring 2007

3/13

FBD of bolt FBD of section

Bearing stresses in connections

Bolts, pins and rivets exert forces on the surfaces upon which they contact. The forces exerted
on the bearing surfaces result in _______________________.

Pin exerts a force P onto the bearing surface of the plate. P is equal and opposite to the force
exerted by the rivet onto the plate.
Bearing stress σ
B
is defined as

Analysis of simple structures

Analyze 2-D structures by considering the normal, shear and bearing stresses in the various parts
of the structure (members, bolts, pins, etc.)

1. Normal stress in 2-force members
a) Determine reaction forces by drawing the FBD and writing equilibrium equations of the
entire structure
b) Consider equilibrium at various joints
c) Sometime it is useful to consider the FBD of a section of the structure
2. Determine the shear stresses in connectors (pin, bolts, rivets, etc.)
3. Determine bearing stresses at contact surfaces

Solid Mechanics Spring 2007

4/13
Example - Link BD is a steel bar 40 mm wide and 12 mm think. Knowing that each pin has a
diameter of 10 mm, determine a) the maximum value of the average normal stress in the link
when
0=
α
degrees, b) the shear stress in pin D and c) the bearing stress in the link at D.
Solid Mechanics Spring 2007

5/13
Stress on Inclined Planes

Axial forces cause both normal and shear stresses on planes that are not perpendicular to the
axis.

For an axially loaded member, consider stresses on the cross-section at an angle θ

Draw FBD of left section

Resolve P into components

N

V

Average normal and shear stresses on the incline plane are

The area of the inclined cross-section A
θ
is related to the normal cross-section A
o

The average normal stress is maximum at

The average shear stress is maximum at

Solid Mechanics Spring 2007

6/13

BA
δ
δ
δ
+
=

ε
σ
E
=

EA
FL

So far, we've determined stresses that occur in axially loaded members and learned to design
them to avoid failure under specific loads.

Now, we consider deformation due to applied loads.
- deformations may be good or bad
- can use deformations to help solve statically indeterminate problems
F
A B
Solid Mechanics Spring 2007

7/13

Rod BC, length L, cross-sectional area A

B

C

Normalize deformation to determine normal strain

Stress-strain diagrams

Relation between σ and ε gives the mechanical properties of the material.
We can obtain a stress strain diagram by performing tensile tests on a material specimen.

For each pair of readings (P,δ) compute

Stress

Strain

σ

ε

Hooke's Law
Solid Mechanics Spring 2007

8/13

Recall that

Stress is

Strain is

So the relation between axial deformation and applied load is

If a rod has various cross sections, different materials, or loaded at places other than the end of
the rod, then you must sum over i sections

Solid Mechanics Spring 2007

9/13
Example

Given: E
Al
= 70 x 10
9
Pa, P
1
= 100 N, P
2
= 75 N, P
3
= 50 N
Find: Deflection at points K and M
Assumptions: Neglect weight of bars, linearly elastic

Diagram:

1.75 m
1.25 m
1.5 m
P1
P2
P3
A = 0.8 m
2

A = 0.5 m
2

J
K
L
M
Solid Mechanics Spring 2007

10/13
Example

Known: Rigid beam rests on posts
Given: AC is steel, d
S
= 20mm, E
S
= 200 Gpa, BD is aluminum d
Al
= 40mm, E
Al
= 70 Gpa, force
at point F = 90 kN
Find: Displacement at point F

A
B
C
D
90 kN
400 mm
200 mm
300 mm
F
Solid Mechanics Spring 2007

11/13
Statically Indeterminate Problems

So far, solved problems by
1) Determining internal forces produced in members using
a) FBD's
b) Equilbrium equations
1) Determining stresses and deformations due to these forces

In the analysis of MANY engineering structures and machines, internal forces and often-external
reaction forces can not be determined with these principals alone. These problems are statically

indeterminate
.

Example - Bar JK of length L, fixed supports at J and K, centric load P at a distance L
1
from pt J,
cross-section area A.

Method of Superposition

Break statically intedeterminate problem into 2 statically determinate problems.

= +

J
K
P
Solid Mechanics Spring 2007

12/13
Problems Involving Temperature Change

Consider bar AB of uniform cross-section and length L, which rests freely on horizontal surface

If the temperature of the bar is increased ____, then the rod elongates by ____

Normalize elongation by ___________________ to define thermal strain ___

Statically indeterminate problem - thermal stress/strain

Consider bar JK with fixed supports at both ends

What happens if the temperature of the bar is raised?

As before, use superposition to determine the force P and the normal stress σ

Total elongation is

Thermal stress is

A B
J
K
L
Solid Mechanics Spring 2007

13/13
Factor of Safety

Even after careful design and analysis of a structure of machine, we typically design structures to
be stronger to "be on the safe side". Design with a factor of safety.

From material testing, determine a maximum load called the ultimate load, Pu, which is the
maximum load a material can handle.

And ultimate shear stress is

Structures and machines will be designed so that the allowable load is considerably less than the
ultimate load. Thus only a fraction of the ultimate load carrying capacity is utilized when the

Factor of safety (F.S.) is defined as

How do we choose F.S.?