mechanics of materials

Category: Education

Presentation Description

No description available.


Presentation Transcript

Final Review ME306: Mechanics of Materials: 

Final Review ME306: Mechanics of Materials By Prof. Shkel Spring 2007

Forces and Stresses: 

Forces and Stresses Bearing stress: Single and double shear:

Multiaxial Stress: 

Multiaxial Stress

Axial Load: 

Axial Load Superposition:

Midterm Fall03: Example 1: 

Midterm Fall03: Example 1 A square concrete post 12-in on a side is axially reinforced with four symmetrically placed cylindrical steel bars, each of cross-sectional area 1.40 in2. Calculate the maximum safe load P the post can carry. The allowable stress for the steel is 20,000 psi and the allowable stress for the concrete is 1000 psi. The modulus of elasticity for the steel is 30×106 psi and the modulus of elasticity for the concrete is 2×106 psi.

Example 1: Solution: 

Example 1: Solution

Thermal Stress: 

Thermal Stress

Poisson’s Ratio: 

Poisson’s Ratio Generalized Hooke’s Law

Shear Strain: Hooke’s Law: 

Shear Strain: Hooke’s Law

Torsional/Shear Stress in Shafts : 

Torsional/Shear Stress in Shafts

Torsional/Shear Stress in Shafts : 

Torsional/Shear Stress in Shafts

Midterm Fall03: Example 2: 

Midterm Fall03: Example 2 The two shafts AC and DF, each of ¾-in. diameter are connected by the two gears shown. Knowing that the shear modulus is 11.2 ×106 psi, and the shaft F is fixed, determine the following, when a 500-lbin torque is applied at A. a) The angle of twist in EF (in degrees). b) The maximum shear stress in EF. c) The rotation at A (in degrees).

Example 2: Solution: 

Example 2: Solution

Bending Strain: 

Bending Strain

Bending Stress: 

Bending Stress Eccentric Load

Midterm Fall03: Problem 3: 

Midterm Fall03: Problem 3 A steel band saw blade, 20 mm wide and 0.8 mm thick, runs over pulleys of diameter d. (a) Find the maximum bending stress in the saw blade if d=600 mm. (b) What is the smallest value of d for which the bending stress in the saw blade does not exceed 400 MPs? Use E=200GPa for steel.

Transverse Loading of Beams : 

Transverse Loading of Beams

Thin-Walled Beams: 

Thin-Walled Beams

Midterm Fall03: Problem 4: 

Midterm Fall03: Problem 4 The 12ft-long walkway of a scaffold is made by screwing two 12-in by ½-in. sheets of plywood to 1.5-in. by 3.5-in timbers as shown. The screws have a 5-in. spacing along the length of the walkway. The working stress in bending is w=850 psi for the plywood and the timbers; and the allowable shear force in each screw is Fw=250 lb. Determine: Maximum shear load supported by the walkway; Maximum bending moment supported by the walkway; What limit should be placed on the weight W of a person who walks across the plank? Hints: Consider a free body diagram and find the shear load and the bending moment produced by a person having weight W.

Example 4: Solution: 

Example 4: Solution

Stress Transformation: 

Stress Transformation

Free-body Analysis: 

Free-body Analysis

Principal Stresses : 

Principal Stresses

Mohr’s Circle for Plane Stress : 

Mohr’s Circle for Plane Stress

3-D Stresses & Mohr’s Circles : 

3-D Stresses & Mohr’s Circles Principal stresses

Thin-Walled Pressure Vessels : 

Thin-Walled Pressure Vessels

Thin-Walled Pressure Vessels: 

Thin-Walled Pressure Vessels

Transformation of Plane Strain : 

Transformation of Plane Strain

Strains Measurement : 

Strains Measurement

Stresses Under Combined Loads : 

Stresses Under Combined Loads

Summary of Combined Loading: 

Summary of Combined Loading

Euler’s Formula for Pin-Ended Columns : 

Euler’s Formula for Pin-Ended Columns The boundary conditions must be satisfied at ends A: x=0, y=0 than B=0 and B: x=L, y=0 1) A=0, the column is straight

Euler’s Formula: 

Euler’s Formula The boundary conditions must be satisfied at ends A: x=0, y=0 than B=0 and B: x=L, y=0 2) A0, Euler’s formula r is the radios of gyration L/r is the slenderness ratio

Extension of Euler’s formula: I: 

Extension of Euler’s formula: I For these boundary conditions

Extension of Euler’s formula: II: 

Extension of Euler’s formula: II For these boundary conditions

Extension of Euler’s formula: III: 

Extension of Euler’s formula: III A particular solution of this equation is The general solution is


Boundary conditions: @A: x=0, y=0 => B=0 @B: x=L; y=0: x=L; dy/dx=0: Solve The general solution is pL=4.4934 : For these boundary conditions

Extensions of Euler’s formula: 

Extensions of Euler’s formula

Final Spring04: Problem 1 : 

Final Spring04: Problem 1 A solid circular bar of diameter d = 1.5 in. is subjected to an axial force P and a torque T. Strain gages A and B mounted on the surface of the bar give readings a = 100  10-6 and b =-55 10-6. The bar is made of steel having E = 30  106 psi and  = 0.29. (a) Determine the axial force P and the torque T. (b) Determine the maximum shear strain max and the maximum shear stress max in the bar.

Final Spring04: Problem 2: 

Final Spring04: Problem 2 A flying buttress transmits a load P = 25 kN, acting at an angle of 60 to the horizontal, to the top of a vertical buttress AB (see figure). The vertical buttress has height h = 5.0 m and rectangular cross section of thickness t = 1.5 m and width b = 1.0 m (perpendicular to the plane of the figure). The stone used in the construction weighs = 26 kN/m3 What is the required weight W of the pedestal and statue above the vertical buttress (that is, above section A) to avoid any normal tensile stresses in the vertical buttress?

Final Spring04: Problem 3: 

Final Spring04: Problem 3 When the beam ABC is unloaded, there is a gap of length 0 between the beam and the support at B. Determine 0 for which all three support reactions are equal when the uniformly distributed load of intensity w0 is applied.

Final Spring04: Problem 4 : 

Final Spring04: Problem 4 An aluminum tube AB of circular cross section is fixed at the base and pinned at the top to a horizontal beam supporting a load Q = 200 kN. Determine the required thickness t of the tube if its outside diameter d is 100 mm and the desired factor of safety with respect to Euler buckling is n=3.0. (Assume E = 72 GPa.)