Question

Please help with this question.

4. For the truss shown below a-The truss shown is externally (circle correct answer) (2 point) a) determinate and unstable b) indeterminate and stable c) determinate and stable d) indeterminate and unstable Reason: (3 point) b- The truss shown is internally (circle correct answer) (2 point) a) determinate and unstable b) indeterminate and stable c) determinate and stable d) indeterminate and unstable Reason: (3 point) - Place an "O" on members with zero force on the truss (5 points) d-Solve for internal force in member BI = (5 points) CD = (5 points) DH (5 points) 20 K 60 K J H G 20 ft E B C D 25 K 4 @ 20 ft = 80 ft 3

Answer 1

a) External indeterminacy of truss,

D_SE = re - 3 = 3 - 3 = 0 (re is 3 that are V_A, H_A, V_E)

Stable truss because the support reactions line of action do not meet at single point.

Hence the truss is externally determinate and stable.

b) Internal indeterminacy of truss,

D_SI = m - ( 2j - 3 )

m = no. of members = 17

j = no. of joints = 10

D_SI = 17 - ( 2 x 10 - 3 ) = 0

Hence the truss is internally determinate and stable.

c) Zero force members,

Members GF and FE are orthogonal and no external force acting on them, hence they will be zero force member.

d) Find out support reactions,

Take sum of moment about hinge A,

V_E x 80 - 20 x 60 - 25 x 40 - 60 x 20 = 0 (ACW moment is considered +ve)

V_E x 80 = 3400

V_E = 42.5K

now from equilibrium of vertical forces we have,

V_A + V_E = 20 + 25 = 45K

V_A = 45 - 42.5

V_A = 2.5K

now from equilibrium of horizontal forces we have,

H_A = 60K

For member forces,

For BI, cut through a section passing HI, IB, BA and consider right section, now from equilibrium of vertical forces,

F_BI = 20 + 25 - V_E   (V_E = 42.5K)

F_BI = 2.5K (T)

For CD, cut through a section passing HG, HD, CD and consider right section, now take sum of moment about H,

F_DC x 20 + 20 x 20 - V_E x 40 = 0

F_DC x 20 = -400 + 65 x 40

F_DC = F_CD = 65K (T)

For DH, cut through a section passing HG, HD, CD and consider right section, now from equilibrium of vertical forces,

F_DH sin 45 + 42.5 + 20

F_DH = -22.5 / sin 45

F_DH = 31.82K (C)