Given four frames with loads shown below Assume the concrete

Given four frames with loads shown below. Assume the concrete columns are square measuring 16\" by 16\" with an E 5,200 ksi. Assume the flexural stiffness of the beam is very large compared to the columns and that the axial rigidity of both the beams and columns is large compared to their flexural stiffness. 1. Find: a. Calculate Stiffness, K for each frame. b. With lateral loads applied to the frames as shown, what is the lateral drift (deflection) of each frame? c. Assume that there is lumped weight of 125 kips at each of the beams. period (T), frequency (1), and circular frequency (o) for each frame? What is the 15 k 15 k 12\' 12\' 10\' 14 14\" 19 (ü) 15 k 15 k 12\' 12 10 14\' 14\' (iv) able

Solution

Solution:-

(a)

(i) stiffness for first frame

K₁ = 12 EI/L³ + 12 EI /L³     (FOR FIXED SUPPORT )

K₁ = 12 * 5200 *1000 * 5461.53/(14*12)³ + 12 *5200 * 1000 * 5461.53/( 14*12)³

K₁ = 143.74 * 10³ pound/inch

(ii) stiffness for second frame

K₂ = 12EI/L³ + 12EI/L³

K₂ = 12 * 5200 * 1000 *5461.33/(14*12)³ + 12 * 5200* 1000 * 5461.33/(10*12)³

K₂ = 269.086 * 10³ pound/inch

(iii) stiffness for third frame

K₃ = 3 EI/L³ + 3EI/L³     ( Simple support)

K₃ = 3 * 5200 *1000 * 5461.33/(14*12)³ + 3 * 5200 *1000 *5461.33/(14*12)³

K₃ = 35.93 * 10³ pound/inch

(iv) stiffness for fourth frame

K₄ = 12EI/L³ + 3EI/L³

K₄ = 12 * 5200 * 1000 * 5461.33/(14*12)³ + 3 * 5200* 1000* 5461.33/(10*12)³

K₄ = 121.174 * 10³ pound/inch

(b) solution:-

Lateral deflection for (i) frame

F =K₁ *

= 15 *1000/(143.74 * 1000)

= 0.104 inch

Lateral deflection for (ii) frame

= 15 *1000/(269.086 *10⁶) = 0.0557 inch Answer

Lateral deflection for (iii) frame

= 15 *1000/(35.93 *10³ ) = 0.417 inch    Answer

Lateral deflection for (iv) frame

= 15 *1000/(121.174 *10³) = 0.1237 inch  

(c) Solution:-

Givem

Lumped mass = 125 kips

For frame (i)

T = 2 (M/K₁) =2 (125*1000/(143.74 *10³))

                            = 5.85 seconds  

Frequency = 1/T = 0.17 second^-1

Angular frequency = (K/M)

                                  = (143.74 *10³/(125 *10³)) = 1.072 radian/second

For (ii) frame

T =2(M/K₂)

= 4.28 second Answer

f = 1/T = 0.23 second^-1

Angular frequency = (K₂/M) = 1.467 radian/second

For (III) FRAME

T= 2(M/K₃) = 11.71 SECOND

Frequency = 1/T = 0.085 second^-1

Angular frequency = (K₃/M) = 0.536 radian/second

For frame (iv)

T = 2(M/K₄)

= 2((125*10³)/(121.174*10³))

= 6.38 second

Frequency = 1/T =0.156 second^-1

Angular frequency = (K/M) = 0.984 radian/second