A 2.0 inch long piece of 1/8 inch square balsa wood is loaded with a force of 25 pounds. If the piece stretches to 2.002 inches under load, what is the stress, strain, and Young's Modulus?

GIVEN: F = 25 *lbs.*, L_{0} = 2.0 *inches*, L = 2.002 *inches*, b=1/8" and h=1/8"

FIND: σ, ε, E

RELATIONSHIPS:

A = b h

σ = F/A

ε= *e*/L_{0}*e*=L-L_{0}

E = σ / ε

SOLUTION:

A = b h

A = 1/8 in. x 1/8 in.

A = 1/64 in^{2}

σ = F/A

σ= 25 *lbs.*/ (1/64 in^{2})__σ= 1600 lbs./in__^{2}

*e*=L-L_{0}*e*=2.002 in. - 2.0 in.*e*= 0.002 inches

ε= *e*/L_{0}

ε= 0.002 in./2.0 in.__ε= 0.001 in./in.__

E = σ / ε

E = (1,600 lbs./in^{2})/(0.001 in./in.)__E = 1,600,000 lbs./in__^{2}

σ= 1,600 p.s.i. |

Provide solutions to the following problems using the engineering problem solving format of Given, Find, Relationships, Solution as shown in the above example.

- A 2 inch long piece of 1/8" square balsa stretches 0.004 inches when 50 lbs. of force is applied. What is the stress? Strain? Young's Modulus (E)?
- A test sample of mild steel rod, 2 inches long and 0.25 inches in diameter is stretched to 2.0014 inches by a force of 1000 pounds. What are the elongation, strain, and stress? What is Young's Modulus (E) for the material?
- A 1/8" square balsa stick that is originally 24 inches long is used to lift an eight pound book. If the modulus of elasticity (Young's Modulus) of the balsa is 64 ksi (64,000 psi.), what is the length of the stick as it supports the book?
- A 44 (see figure below) configuration of 1/8" square and 24-inch long balsa sticks is used to lift you off of the floor. What is the stress on the material? If the stick stretches by 0.024 inches, what is the strain?
- For five of the symmetrical beams shown in the figure, find the cross-sectional area, overall height, and maximum y (y
_{max}). All answers should be expressed in inches (each block is 1/8" by 1/8"). Then find the Second Moment of Area around the neutral axis (I) in inches^{4}and the Section Modulus (Z) in inches^{3}. - An 8-inch long beam of 1/8" square balsa sticks in the "44" configuration shown above breaks with an applied force of 11 pounds. What is the applied moment, cross-sectional area, second moment of area, y
_{max}, and section modulus? What is the Yield Stress of the material? Bonus: If you stand the piece up into the "2222" configuration, how much force could it support? - For one of the asymmetrical beams used in the class, find the height of the centroid (neutral axis) above the base and the maximum y (y
_{max}). All answers should be expressed in inches (each block is 1/8" by 1/8"). Then find the Second Moment of Area around the neutral axis (I) and the Section Modulus (Z). - Using the most representative beams tested for your class estimate the yield stress of balsa wood.

A 0.1 inch diameter round aluminum 0.5 inches long test specimen is stretched 0.000125 inches as it breaks under a force of 20 pounds. Find the yield (or failure) stress, strain, and Young's Modulus.

GIVEN: F = 20 *lbs.*, L_{0} = 0.5 *inches*, e = 0.000125 *inches*, D=0.1 in.

FIND: σ_{yield}, ε, E

RELATIONSHIPS:

A = πr^{2}

σ = F/A

ε= *e*/L_{0}

E = σ / ε

SOLUTION:

A = πr^{2}

A = π(D/2)^{2}

A = π(D)^{2}/4

A = πx 0.1 in. x 0.1 in./4

__A = 0.008 in ^{2}__

σ = F/A

σ=20 *lbs.*/ 0.008 in^{2}__σ= 2,548 lbs./in__^{2}

ε= *e*/Lo

ε= 0.000125in./0.5in.__e= 0.00025 in./in.__

E = σ / ε

E = (2,548 lbs./in^{2})/(0.00025 in./in.)__E = 10,192,000 lbs./in__^{2}

σ |

A Crazy Eight beam in the 323Box configuration shown above breaks with an applied force of 24 pounds. What is the applied moment, cross-sectional area, second moment of area, maximum y, and section modulus? What is the Yield Stress of the material?

GIVEN: F = 24 *lbs.*, b=3/8" and h=3/8" (hollow)

FIND: M, A, I, y_{max}, Z, σ_{yield}

RELATIONSHIPS: (these are for beam bending)

A = b h

M = 2 in. x F

I = b h^{3} / 12

Z = I / y_{max} (this is the section modulus)

σ_{yield} = M / Z

SOLUTION:

For Crazy Eights the applied moment is half the force times half the distance or 2 inches times the force

M = 2 in. x F

M = 2 in. x 24 lbs.

__M = 48 in·lbs.__

The maximum allowed area is 1/8 square inches, what we have here

A = [b h]_{outside rectangle} - [b h]_{inside hole}

A = [(3/8in.) x (3/8in.)] - [(1/8 in.) x (1/8 in.)]

A = [9/64in^{2}] - [1/64in^{2}] = 8/64 in^{2}

__A = 1/8 in ^{2}__

I is the I of the outside rectangle minus the I of the inside square

I = [ b h^{3} / 12 ]_{outside rectangle} - [ b h^{3} / 12 ]_{inside hole}

I = [ (3/8in.) x (3/8in.)^{3} / 12 ] - [ (1/8 in.) x (1/8 in.)^{3} / 12 ]

I = 0.00165 in^{4} - 0.00002 in^{4}

__I = 0.00163 in ^{4}__

The maximum Y for a symmetrical beam is half the height

__y _{max} = 3/16 inches__ (by inspection)

The Section Modulus (Z) combines the geometric properties

Z = I / y_{max}

Z = 0.00163 in^{4} / (3/16 inches)

__Z = 0.0087 in ^{3}__

The Stress at Failure for the balsa wood is now easily calculated

σ_{yield} = M / Z

σ_{yield} = 48 in·lbs / 0.0087 in^{3}

__σ _{yield} = 5,517 lbs./in^{2}__

M = 48 in·lbs. |