True or False:

a) A Failure stress is a material property.  T

b) An allowable stress is a combination of a material property and a factor of safety.  T

c) A factor of safety is a material property.  F

d) A stress cannot physically exceed the allowable stress.  F

e) The factor of safety is the ratio of the failure stress to the allowable stress.  T

f) In a  properly designed structural element, the maximum stress does not exceed the allowable stress.  T

1. What's wrong with this type of statement: "The strain at point A is 10^-3 mm".

2. Indicate the sign of the strains e_x, e_y, and g_xy for the deformations of  the rectangular figure shown in (a) and (b).

(a) e_x > 0,  e_y < 0,   g_xy > 0

(b) e_x > 0,  e_y < 0,   g_xy > 0

3. Line AB is 100 inch long. Determine its change in length DL and its average strain e in each of the following cases. Consider a decrease in length as negative.

a) A moves right by 0.1 inch.

b) B moves right by 0.1 inch.

c) Both A and B move right by 0.1 inch.

d) A moves left by 0.1 inch, and B moves right by 0.1 inch.

e) B moves right by 0.1 inch and up by 0.2 inch.

What is the answer by the linear theory?

4.  Bar OA of length L rotates about point O by angle r into position OA'.  The two adjoining figures show, respectively, the exact geometry and the approximate geometry consistent with the linear theory (samll r). Do each of the following questions by exact relations, and then by the linear theory.

a) Determine the displacements u and v in terms of r and L.

Exact:

u = -(OA - OA'') = -(L - Lcosr)
v = -(A'A'') = -Lsinr 

Linear Theory:

u = 0

v = -Lr

b) What is the elongation D of cable AB?

Exact:

D = BA' - a = sqrt[(a - v)² + u²] - a

Linear Theory:

D = AA' = Lr

c) What is the rotation r' of cable AB?

Exact:

r' = tan^-1[u/(a - v)]  = - tan^-1[(L - Lcosr)/(a + Lsinr)]

Linear Theory: 

r' = 0

 5.  Bar AB is inclined to the horizontal by angle a, and its ends displace horizontally to A' and B', respectively.  Show on the figure the axial displacements AA" and BB" at A and B,  respectively, then obtain by the linear theory the change in length of AB in terms of AA', BB', and q.

 DL = BB" - AA"  = BB'cosa - AA'cosa       

6.  A square element ABCD of side a deforms into parallelogram AB'C'D'.  The deformation may be defined by displacement u_b, deformed length a', and angle g, as shown.  All vertical line elements have the same elongation D_y = a' - a, and all horizontal line elements have the same elongation D_x = u_b.  Using the linear theory,

a) Write the expression of  D_x for line DC in terms of the displacements at D and C.

D_x = u_c - u_d

b)  Write the expression of  D_y and g for line AD in terms of the displacements at D.

D_y = v_d

g = u_d/a

c) Write the relations expressing D_y and g for line BC in terms of the displacements at B and C

D_y = v_c

g = (u_c - u_b)/a

d) Solve the preceding relations for the displacements in terms of D_x, D_y and g.  Verify that u_b = D_x .

From b): v_d = D_y,  u_d = ga

From a): u_c = u_d + D_x = ga + D_x

From c): v_c = D_y,  u_b = u_c - ga = ga + D_x - ga = D_x

e)  Determine the elongation of line AC in terms of (u_c, v_c).

D_AC = axial displacement at C = u_ccos45^o + v_ccos45^o = (u_c+ v_c)sqrt(2)/2

7.  Joint A of bars AB and AC displaces to position A' by small displacements u and v.  The elongation D_AB of bar AB is the axial displacement AP, and is negative in the case of the figure.  Similarly, AQ is the elongation of bar AC.

a)  Express D_AB and D_AC in terms of u and v and angles q_b and q_c. 

D_AB = AP  =  u cosq_b  - v sinq_b  
D_AC =  AQ = -u cosq_c -  v sinq_c

b)  Let BC = 10 ft, q_b = 60^o, and q_c = 45^o.  Re-write the expressions found as two simultaneous equations having (u, v) as unknowns, and right-hand sides in terms of D_AB and D_AC.  If a temperature change causes a  uniform strain e = 10^-3 in the bars, what will be the displacements of the ring?

cosq_b  = 0.5,   sinq_b  = sqrt(3)/2 = 0.866

cosq_c = sinq_c = sqrt(2)/2 = 0.707

The simultaneous equations for u and v are

 0.5u   - 0.866v  =  D_AB
0.707u  +  0.707v  = - D_AC

The initial lengths of the bars are found by the law of sines:

AB/sinq_c  = AC/sinq_b = BC/sin(p - q_b  - q_c) = 10/sin(75^o) = 10.353 ft
AB = 10sqrt(2)/2sin(75^o) = 7.3205 ft = 87.85 in.

AC = 10sqrt(3)/2sin(75^o) = 8.966 ft = 107.59 in.

The bar elongations caused by the temperature are

D_AB = e(AB) = 0.08785 in

D_AC =  e(AC) = 0.1076 in

Thus

 0.5u   - 0.866v  =  0.08785 in
0.707u  +  0.707v  = - 0.1076 in

The solution is found as
u = -0.0322 in.
v = -0.1200 in.

8.  Rectangular element ABCD deforms into AB'C'D'.  The corner displacements and the rotations of lines AB and AD are indicated in the figure.  Strains and rotations are assumed to be small.

i)  Using the linear theory,  obtain the expressions of the following quantities in terms of the displacements:

a) e_AB, e_DC, e_AD, e_BC 

e_AB = u_b/a

e_DC = (u_c - u_d)/a

e_AD = v_d/b

e_BC  = (v_c - v_b)/b

b) r_AB, r_DC, r_AD, r_BC

r_AB = v_b/a                 

r_DC = (v_c - v_d)/a

r_AD = u_d/b

r_BC = (u_c - u_b)/b

c) (g_xy)_A, (g_xy)_B, (g_xy)_C, (g_xy)_D

(g_xy)_A = r_AB + r_AD  = v_b/a + u_d/b

(g_xy)_B = r_AB + r_BC   = v_b/a + (u_c - u_b)/b

(g_xy)_C = r_DC + r_BC   = (v_c - v_d)/a + (u_c - u_b)/b

(g_xy)_D = r_DC +r_AD   = (v_c - v_d)/a + u_d/b

ii) Assume that e_x, e_y, and g_xy are constant, and, consequently, that the deformed shape is a parallelogram.  The deformed shape is determined by e_x, e_y, and g_xy, but not its rotational position about point A, which may be determined for example by r_AB.   Determine the six displacements in terms of  e_x, e_y, g_xy, and  r = r_AB.

From the deformation of AB,

u_b = ae_x

v_b= ar

From the definition of (g_xy)_A,

r_AD = g_xy - r

From the deformation od AD, 

u_d = br_AD = b(g_xy - r)

v_d = be_y

From  e_DC = (u_c - u_d)/a  = e_x , and r_DC =  (v_c - v_d)/a = r, obtain

u_c = u_d + ae_x =  b(g_xy - r) + ae_x

v_c = v_d + ar = be_y + ar