me323 lecture 18 - purdue.edu

ME323 LECTURE 18

Alex Chortos

General Topics

▪ Chapter 2 – Normal stress and strain

▪ Chapter 3 – Shear stress and strain

▪ Chapter 4 – Design of deformable bodies

▪ Chapter 5 – Stress and strain – general definitions

▪ Chapter 6 – Axial members

▪ Chapter 7 – Thermal effects

▪ Chapter 8 – Torsion

▪ Chapter 9 – Beams – Shear and moment diagrams

Stress and Strain Summary

Definition of normal stress

Definition of normal strain

Poisson’s ratio 𝜀𝑦 = 𝜀𝑧 = -ν𝜀𝑥

𝜀𝑥 =𝐿 − 𝐿0𝐿0

=Δ𝐿

𝐿0

Shear Stress and Strain Summary

γ =δ𝑠𝐿𝑠

τ =𝑉

𝐴= 𝐺γ

σ =𝐹

𝐴= 𝐸𝜀

Normal stress can be decomposed into normal and shear forces

Summary: General Stress and Strain

σ𝑥 , σ𝑦 , σ𝑧

𝜏𝑥𝑦 , 𝜏𝑥𝑧 , 𝜏𝑦𝑧

Axial Deformation: Summary

ε 𝑥 =𝑑𝑢(𝑥)

𝑑𝑥

𝑒 = 𝑢 𝐿 − 𝑢 0 = න

0

𝐿

ε 𝑥 𝑑𝑥

𝑒 = න

0

𝐿𝐹 𝑥

𝐴 𝑥 𝐸 𝑥𝑑𝑥

𝑒 =𝐹𝐿

𝐴𝐸

Assumptions: constant force along the length, constant area, and constant modulus

Indeterminate Structures

1. Write the force balance equations (draw the FBD).

2. Write the elongation equations (relationship between

elongation and force for each member)

3. Write the compatibility equations (relationships between

elongations of different members) (draw the deformed

geometry).

4. Solve system of equations.

𝑒 =𝐹𝐿

𝐴𝐸𝑒 =

𝐹𝐿

𝐴𝐸+ α𝐿Δ𝑇

𝑒1 + 𝑒2 = 𝑐e.g.

𝐹1 + 𝐹2 = 𝑐e.g.

Types of Compatibility Conditions

𝑒1 + 𝑒2 = 𝑐

𝑒1𝑎=𝑒2𝑏

Total Elongation

If you’re ever unsure, draw the deformed geometry!

Similar Triangles

Indeterminate Planar Trusses

1. Solve for static equilibrium (force balance) assuming no changes in geometry.

2. Solve for the displacement of node C assuming no changes in the angles of the members.

• θ𝑗 is measured counterclockwise

from the x axis.• Origin of x axis is placed at the

point on the element that is pinned to ground.

Indeterminate Trusses

𝑒1 =𝐹1𝐿1𝐴1𝐸1

𝑒1 = 𝑢𝑐𝑜𝑠θ1 + 𝑣𝑠𝑖𝑛θ1

𝐹1 + 𝐹2 + 𝐹3 = 𝑐

𝑒2 =𝐹2𝐿2𝐴2𝐸2

𝑒3 =𝐹3𝐿3𝐴3𝐸3

𝑒2 = 𝑢

𝑒3 = 𝑢𝑐𝑜𝑠θ3 + 𝑣𝑠𝑖𝑛θ3

𝐹1 + 𝐹2 = 𝑑

Torsion

𝜏=𝑇ρ

𝐼𝑝

ΔΦ =𝑇𝐿

𝐺𝐼𝑝

𝐼𝑝 =π𝑟0

4

2

𝐼𝑝 =π

2(𝑟0

4 − 𝑟𝑖4)

Torsion Problems

ΔΦ1 = ΔΦ2

ΔΦ1 + ΔΦ2 = 0

Φ𝐶 = Φ𝐵 + ΔΦ2

Φ𝐷 = Φ𝐶 + ΔΦ3

Φ𝐷 = Φ𝐵 + ΔΦ1

ΔΦ1 = ΔΦ2 + ΔΦ3

𝑟𝐶

𝑟𝐵

𝑟𝐶

𝑟𝐵

𝑡𝑎𝑛Φ𝐶~Φ𝐶 =δ

𝑟𝐶

𝑡𝑎𝑛Φ𝐵~ − Φ𝐵 =δ

𝑟𝐵

Design for Strength

𝐹𝑆 =𝜎𝑓𝑎𝑖𝑙

𝜎𝑎𝑙𝑙𝑜𝑤𝑒𝑑

σfail can be σY or σU

τ𝑚𝑎𝑥 =𝑇ρ𝑚𝑎𝑥

𝐼𝑝=τ𝑓𝑎𝑖𝑙

𝐹𝑆σ =

𝐹

𝐴=σ𝑓𝑎𝑖𝑙

𝐹𝑆

Summary – Resultants in Beams

V x2 = V x1 + න

x1

x2

p x dx

M x2 = M x1 + න

x1

x2

V x dx

dV

dx= p(x)

dM

dx= V(x)

Example Question

BC

30 kN/m10 kN25 kN*m

0.5 m 0.5 m 0.5 m 0.5 m

V(x)

M(x)