1. Draw a stress-strain diagram to represent the behavior outlined below.  Before starting, read further down in order to choose an appropriate scale.

a) a linearly elastic part from the origin O to point A.

b) a perfectly plastic part AB whose plastic flow has a strain range equal to 10 times the elastic one.

c) a parabolic strain hardening part BC and a necking part CD characterized by the following:

    c1) an ultimate stress equal to twice the elastic limit.

    c2) a fracture stress equal to one and a half times the elastic limit.

    c3) a strain hardening range equal to twice the plastic flow range.

    c4) a necking range equal to one and a half times the plastic flow range.

2. Let the elastic limits for stress and strain in the graph of question 1 be s = 200 MPa and e = 0.001, respectively.
a) Determine the modulus of elasticity.

b) Is "Modulus of Elasticity" synonymous with "Young's Modulus"?  

E = s/e = 200/0.001 = 200,000 MPa or 200 GPa
c) Determine the modulus of resilience = Area under the elastic part of the curve.. 

u_r = s_ele_el/2  = 200MPa (0.001 m/m)/2 = 0.1 MJ/m³

d) Determine the modulus of toughness = Area under the curve up to the fracture point.  Note that the area under a parabolic segment symmetric about the axis of the parabola is equal to two thirds of the area of the rectangle constructed on the base and height of the segment. 

 u_t = s_ele_el/2 + s_el(10e_el) + s_el(20e_el) + (2/3)s_el(20e_el) + 1.5s_el(15e_el) + (2/3)(0.5s_el)(15e_el)

        = s_ele_el(0.5 + 10 + 20 + 40/3 + 22.5 + 5) = 71.33s_ele_el  = 142.67u_r = 14.27 MJ/m

e) What is approximately the energy expended to perform a tensile test up to fracture of a specimen having a gauge length of 30 cm and a diameter of 5 cm?  Recall that a Joule = 1 N.m.

Volume = V = 30p(25/4) = 589.05 cm³ = 589(10^-6) m³

Energy expended = u_tV = 14.27 MJ/m3 [589(10^-6) m³] = 0.0084 MJ or 8,400 J

f) If the specimen is unloaded from the point at the onset of strain hardening what will be the final gauge length?

Unloading path is parallel to the elastic path.  The permanent-set strain is e_ps = e_pl - e_el = 10e_el

 Final gauge length = 30(1+e_ps) = 30(1 + 0.01) = 30.3 cm or 303 mm.

Statically Determinate Problems

3.  In the figure, bar bd is rigid, and cable ac has the properties shown.

a) Name the equilibrium equation that yields the force in cable ac.

Moment equilibrium about point b.

b) Name the type of motion that bar bd can have.

Rotation about point b.

c) How does knowledge of the force in cable ac allow to determine the displacement at a?

From the force F in cable ac, get the cable elongation FL/EA, which is also the displacement at point a.

d) How does knowledge of the displacement of point a allow to determine the displacement of any other point of the bar?

The displacement v at a point at distance r from point b is vertical and equal to rq, where q is the bar rotation. So for any pair of points,

v₁/v₂ = r₁/r₂.

e) Describe the sequence of operations by which to determine the displacement at d caused by a given load P. 

1) Determine the force in the cable by the equation in a).

2) Determine the displacement at point a as in c).

3) Determine v_d from v_a as in d). 

4.  In the figure, bar ab is rigid, and cables ac and bd have the properties shown. How should the cable flexibilities be related if
a) bar ab is to remain level?

By statics, the forces in the cables are equal. The bar remains level if the cable elongations are equal.  This require equal flexibilities: L/EA = L'/E'A'.

b) bar ab is to tilt clockwise?

L'/E'A' > L/EA

c) bar ab is to remain remain level, and P is applied at the quarter point from a?

By statics, the force in cable ac is 3 times the force in cable bd.  For the cable elongations to be equal, the flexibility of ac needs to be 1/3 that of cable bd: L/EA = (1/3)L'/E'A'.

Statically Indeterminate Problems

5.  The steel rod and aluminum sleeve assembly may be modeled as two springs in parallel whose axial stiffness k is the sum of the constituent stiffnesses k_st and k_al.  Assuming that the cross sectional areas and the moduli of elasticity are related as follows, 

A_st = 1.5A_al

E_st = 3E_al

a) What is the ratio of  each spring stiffness to the stiffness of the assembly ?

k_st = E_stA_st/L,  k_al = E_alA_al/L,  k = E_stA_st/L + E_alA_al/L

k_st/k = E_stA_st/(E_stA_st + E_alA_al) = 4.5/(4.5 + 1) = 0.818

k_al/k = E_alA_al/(E_stA_st + E_alA_al) = 1/(4.5 + 1) = 0.182

b) If an axial load P = 10 k. is applied at the ends, what will be the axial forces in the steel tube and the aluminum sleeve?

N_st = (k_st/k)P = 8.18 k

N_al = (k_al/k)P =1.82 k

6.  For the three-bar truss,

a) Write the equilibrium equations governing the forces in the three bars.

F₁sina - F₃sinb = 0.6F₁ - 0.8F₃ = 0

F₁cosa + F₂ + F₃cosb = 0.8F₁ + F₂ + 0.6F₃ = P

b) Write the deformation-displacement relations expressing the bar elongations in terms of the displacements (u, v) of joint A.

d₁ = usina - vcosa  =  0.6u - 0.8v

d₂ = -v

d₃ = -usinb - v cosb  =  - 0.8u - 0.6v

c) Write the force deformation relations, assuming the bars have the same EA.

d₁ = 30F₁/EA

d₂ = 24F₂/EA

d₃ = 40F₃/EA

d) The equations written above should form a system of as many equations as unknowns.

 How does the force method proceed?

1) Obtain the compatibility equation by Gauss-elimination of u and v from the three equations in (b)

2) Express the compatibility equation in terms of the forces by means of equations (c).

3) Solve the system of 3 equations: two equations in (a) and one compatibility equations for the three unknown forces.

4) Compute the deformations using the equations (c)

5) Determine the displacements by solving two of the equations (b).

How does the displacement method proceed?

1) Rewrite the equations in (c) as stiffness equations: forces in terms of elongations.

2) Substitute for the elongations their expressions in (b).

3) Substitute for the forces in the equilibrium equations (a).

4) Solve the two equations thus obtained for the two displacements.

5) Obtain the forces from their expressions in step 2).

Thermal Stresses

7. The two bar system is maintained between two rigid walls.  If the temperature of the left bar is increased,

a) Will the left bar be in tension or compression?   Will its length increase or decrease?

The left bar is restrained from elongating, so it will be in compression. 
Its length will increase by less than the free thermal deformation.

b) Answer the same questions for the right bar.

By statics, the right bar has the same compression force as the left bar.

Its length will decrease since it is in compression and it has no thermal strain.  This decrease in length must be equal to the increase in length of the other bar, since the walls maintain a constant total length.

c) If the temperature of both bars is increased by the same amount, and assuming A₁ > A₂, will the joint of  the two bars move to the left or to the right?

If the bars are cut at their joint, they would elongate each by the free thermal deformation d_T = aLDT.  That would create an overlap of 2d_T at the joint.  To maintain geometric continuity, equal compressive forces are developed n the bars.  The elastic shortening these forces cause is smaller in the stiffer bar, which is the bar on the left.  Thus the net joint displacement is to the right. 

Here's another way of saying this: Each bar wants to push the joint outward. Bar 1 is the bully, and wins out.   

8.   The sleeve-rod assembly is maintained between two rigid plates. If the temperature of the system is increased by a uniform amount, and given that the coefficient of thermal expansion of aluminum is larger than that of steel, will the aluminum sleeve be in tension or in compression? What will be the state of the steel rod?

By statics, the total force on a cross section must be zero. By geometry, the length of the sleeve and of the rod must remain the same.  The aluminum would like to expand more than the steel, which has to restrain it.  Thus the aluminum sleeve will be in compression, and the steel rod will have an equal tension.