Test: Structural Analysis- 3 - Civil Engineering (CE) MCQ

At joint B, F_BC → Compressive

At joitnt C, F_CA → Tensile

Stiffness (k) is defined as the force per unit deflection.

K = P/δ 

⇒ δ = P/K

If stiffness matrix is doubled

i.e. K’ = 2K

then,

δ = P/K'

δ = P/2K

δ = (1/2)P/K

δ’ = δ/2.  

Therefore, if the stiffness matrix is doubled with respect to the existing stiffness matrix, the deflection of the resulting frame will be half the existing value.

At joint C, there is no horizontal reaction,

So, F_BC=0,
Due to Loading Symmetricity, 
V_a = V_c = P
∴ V_c = P/2
∴ F_CD = P/2
Now at joint E, 

∑F_{along the line BE} = 0

F_BE = P KN

For k₃₃ Apply unit displacement along 3 and restraining the displacement in 2 and 1
Δ → unity 

To find the vertical deflection at joint E, a unit vertical load is applied at joint E and the forces in member AB and CD are to be found out.

Consider joint A,

F_AB sin 60° = 0.5  

F_AB = 0.577 (compressive)

Similarly, F_CD = 0.577 (compressive)

due to increase in temperature, the increase in length of AB & CD = L ∝ t

= 3 × 12 × 10^-6 × 40

= 0.00144 m

So the deflection of joint E,

δ = Σ kδ’

where, k = Force produced in each member when a unit downward load is applied at joint E.

           δ’ = Deflection produced in the concerned member (AB & CD)  due to temparature rise.

= (- 0.577) × (0.00144) + (- 0.577) × (0.00144)

= - 0.00166 m

= 1.66 mm (upward)

[‘–’ means our assumed downward deflection is wrong, it will deflect upward]

Taking moment about C,

I. Member LK, LA and LB are meeting at joint ‘L’ and AL and LK re co-linear so force in the member LB is zero.

F_LB = 0

II. Member FG, FE, and FD are meeting at joint F, So force in the member FD = 0

F_FD = 0

III. Removing Member LB, DF, thus member BK, BC and BA meets at point B out of which BA and BC are co-linear and vice versa for member DG, DC and DE.

So, F_BK = 0

F_DG = 0

IV. Similarly, Member HM, HI and HG are meeting at joint H

F_HM = 0

Minimum tension occurs at deepest point and is equal to horizontal thrust.

T_min = H = 2000 kN