me323 lecture 18 - purdue.edu
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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)