CEE 459

Test 2, 21 May 2004

1.  15%. Determine I_y, I_z, and I_yz with respect to centroidal axes (y, z), as shown.  Please write clearly the numbers and operations.

I_y1 = 1(4)³/12 + 4(2.5)² = 30.333

I_y2 = 10(1)³/12 = 0.833

I_y = 2I_y1 + I_y2 = 61.5 in⁴

I_z1 = 4(1)³/12 + 4(4.5)² = 81.333

I_z2 = 1(10)³/12 = 83.333

I_z = 2I_z1 + I_z2 = 246 in⁴

I_yz1 = 4(-4.5)(2.5) = - 45

I_yz2 = 4(4.5)(-2.5) = - 45

I_yz = I_yz1 + I_yz2 = -90 in⁴

2. 45%.  The simply supported beam subjected to load P at mid-span has the section shown below. The centroid C of the section is located as shown in the figure, and its inertia properties are as given below.

I_y =  91.2778  in⁴

I_z =  155.1111  in⁴

I_yz =  50.5556 in⁴

a) 10%.  Determine and show on a sketch the angle that the neutral axis makes with the y axis.

M_z = 0

s = (I_zz-I_yzy)M_y/D

D = I_yI_z -I_yz²

s = 0 at (I_zz-I_yzy)= 0.

tanb = z/y = I_yz/I_z = 0.3259

b = 18.05 deg.

b) 20%.  Determine the magnitude and location of the maximum compressive stress.

At midspan,

M_y = -(P/2)(4) = -40 k-ft

M_y' = component of M_y on neutral axis acts as shown on positive cross-section.  The compression side is thus above the neutral axis. 

B(y = - 3.222, z = 2.389)  is farthest from the neutral axis on the compression side.

D = I_yI_z -I_yz²  = 11,602.33 in⁴

s = (I_zz-I_yzy)M_y/D = [(155.1111)(2.389) - (50.5556)(- 3.222)](-40)(12)/11,602.33 = -22.069 ksi

c) 15%.  Assume that the load is applied such that there is bending without twisting. Determine the shear flow in AB at z = 0.

V_y = 0

V_z = 10 k

q = (I_zQ_y - I_yzQ_z)V_z/D

Q_y = - (1)(5.611)²/2 = -15.742 in³

Q_z =  - (1)(5.611)(3.222 - 0.5) = -15.273

q = [(155.1111)(-15.742 ) - (50.5556)(-15.273)]10/11,602.33 = -1.439  k/in

3.  40%.  The semi-circular cantilevered arch has an I cross-section, and is subjected to force P and moment C as shown.

a)  10%. Determine the curved beam factor Z of the I section.

Z = -1 + (R/A)Sb log(R_t/R_b)

R = 96

A = 12(2) + 18 = 42 in²

Z = -1 + (96/42)[12log(106/105) + 1log(105/87) + 12 log(87/86)] = 0.0069189

b)  20%.  Let P = 20k, C = 1213.91 k-in.

Determine the maximum normal stress in the cross-section at D, and indicate whether tensile or compressive.

s = (N + M/R)/A + My/I'(1+y/R)

I' = AR²Z = 2678.12 in⁴

At D

N = -P = -20 k

M = -C + PR =  706.09 k-in

N + M/R = -20 + 706.09/96 = -12.645

At y = - 10 in

s_D = -12.645/42 + 706.09(-10)/[2678.13(1 - 10/96)] = -0.301 - 2.943 = -3.244 ksi

Maximum stress is -3.244 ksi, at the inner fiber, and compressive.

c) 10%.  The flexibility relations of the arch for the displacements and forces shown in the figure are given as

Determine the reaction at support B for the arch supported and loaded as shown below.

F₁ = -P,  C = 0

v = 0

EI'v = 2R³(-P) + pR³(3/2 + Z)F₂ = 0

F₂ = 2P/p(3/2 + Z) = 0.4225P