1 Class #2.1 Civil Engineering Materials – CIVE 2110 Strength of Materials Mechanical Properties of Ductile Materials Fall 2010 Dr. Gupta Dr. Pickett

2 Strain Energy External energy is required to deform specimen. Internal energy is stored in specimen during deformation process. during deformation process. Work = Energy = Force x Displacement Force starts from ZERO, Force starts from ZERO, Strain Energy = ΔU = “work done” Strain Energy = ΔU = “work done” ΔU = due to average force over load process. ΔU = due to average force over load process.

3 Strain Energy Strain Energy = ΔU = “work done” ΔU = due to average force over load process. ΔU = due to average force over load process.

4 Strain Energy – Modulus of Resilience Modulus of Resilience: - A measurement of a material’s ability to absorb energy absorb energy WITHOUT PERMANENT deformation. WITHOUT PERMANENT deformation. - Area under ELASTIC portion of stress-strain diagram. stress-strain diagram.

5 Strain Energy – Modulus of Toughness Modulus of Toughness: - A measurement of a material’s ability to absorb energy BEFORE FAILURE. absorb energy BEFORE FAILURE. - Area under ENTIRE stress-strain diagram. - Materials with high toughness will give warning before failure (GOOD) will give warning before failure (GOOD)

6 Modulus of Toughness - Concrete Concrete - need warning before failure - put in steel reinforcement bars - put in steel reinforcement bars High Toughness Low Toughness

7 Poisson’s Ratio Poisson’s Ratio: - For material that is: - Homogenous - Isotropic - in linear elastic range the material VOLUME must remain CONSTANT the material VOLUME must remain CONSTANT - deformations in the Longitudinal direction - deformations in the Longitudinal direction must be compensated for by deformations must be compensated for by deformations in the TWO directions PERPENDICULAR to the in the TWO directions PERPENDICULAR to the Longitudinal direction. Longitudinal direction.

8 Poisson’s Ratio Poisson’s Ratio: - Typical values: steel = aluminum = 0.35 aluminum = 0.35 cast iron = 0.28 copper alloys = concrete = 0.15 wood = 0.29 – 0.31

9 Shear Stress-Strain Diagrams Assumptions: - Material is: - loaded in pure shear - homogeneous - isotropic - ductile - in linear elastic range, has - proportional limit, - proportional limit, - Shear Modulus of Elasticity, G - Shear Modulus of Elasticity, G - Modulus of Rigidity, G - Modulus of Rigidity, G

10 Shear Stress-Strain Diagrams G Steel = 11x10 6 psi G Aluminum = 4x10 6 psi G Copper = 5.5x10 6 psi

11Creep Creep = time dependent permanent deformation permanent deformation Due to – Load - Temperature - Temperature Creep Strength = highest initial stress that the material can be subjected to material can be subjected to in order to avoid a specified in order to avoid a specified creep strain creep strain over a specified time over a specified time

12Fatigue Fatigue = BRITTLE fracture at stress < material’s Yield Stress stress < material’s Yield Stress due to repeated load cycles due to repeated load cycles Cause: Localized stress > average stress Fatigue Limit = stress below which NO failure occurs for s NO failure occurs for s specified number of cycles specified number of cycles Endurance Limit = Fatigue Limit

13Fatigue Fatigue Limit = stress below which NO failure occurs for s NO failure occurs for s specified number of cycles specified number of cycles Endurance Limit = Fatigue Limit

14 Shear Strain Example: Problem 3-30 Problem 3-30 Hibbeler 7 th edition, pg. 116

15 Shear Strain Example: Problem 3-32 Problem 3-32 Hibbeler 7 th edition, pg. 117