CEE 220 - Test 1

55 minutes

1. 30%.  The two steel rods, AB and AC, support the vertical force P = 10 k.

a) 20%. Determine their required cross-sectional areas if the allowable tensile stress for the steel is s_a = 20 ksi.

SF_y = 0.8F_B - 10 = 0

F_B = 10/0.8  =  12.5 k

SF_x = -F_C + 0.6F_B = 0

F_C = 0.6F_B = 7.5 k

Required areas:

A_AC = F_C/s_a = 7.5/20 = 0.375 in²

A_AB = F_B/s_a = 12.5/20 = 0.625 in²

b) 10%. The pin at C acts in double shear, and has a diameter of 0.5 inch.  What is the average shear stress for that pin?

Cross-sectional area of pin = A = p(0.25)² = 0.1963 in²

t = 0.5F_C/A = 19.10 ksi

2. 30%. Disregard the cross sectional areas found in Question 1.  Instead, let the cross-sectional area be A = 0.5 in² for each bar.  Let the modulus of elasticity for steel be E = 29,000 ksi.

a) 10%.  What are the bar elongations?

d_AC = L_ACF_C/EA = (4)(12)(7.5)/(29000)(0.5) = 0.0248 in.

d_AB = L_ABF_B/EA = (5)(12)(12.5)/(29000)(0.5) = 0.0517 in.

b) What are the horizontal and vertical displacements at A?

d_AC = u,  u = 0.0248 in.

d_AB = v cosa - u sina = 0.8v - 0.6u

v = d_AB/0.8 + (0.6/0.8)u = 0.0833 in.

3. 40%.  The three bars have the same EA.  Determine the forces in the bars caused by the applied load.

Note: Partial credit where due will be based, respectively, on the three types of  governing equations for the structure, which must be clearly identified.

Equilibrium

SF_x = -F_C + 0.6F_B = 0

SF_y = F_D + 0.8F_B - 10 = 0

Force - Deformation Relations

d_AC = 4F_C/EA

d_AD = 4F_D/EA

d_AB = 5F_B/EA

Deformation - Displacement Relations

d_AC = u

d_AD = v

d_AB = 0.8v - 0.6u

Solution by Force Method

Compatibility: 

d_AB = 0.8d_AD  - 0.6d_AC

In terms of forces:

5F_B/EA = 0.8(4F_D/EA)  - 0.6(4F_C/EA)

5F_B = 3.2F_D - 2.4F_C

Solve with equilibrium equations:

F_C = 0.6F_B

F_D  =  10 - 0.8F_B

5F_B = 3.2(10 - 0.8F_B) - 2.4(0.6F_B)

9F_B = 32

F_B = 32/9 = 3.556 k

F_C = 0.6F_B = 2.133 k

F_D = 10 - 0.8F_B  = 7.156 k