Analysis:
Find the reactions:

+\circlearrowright \sum{M_A} = 0 = 8.2  klf(180  ft)(180  ft/2)  –  B_y(180  ft)

B_y = 738  k (+↑)

+↑ΣF_y = 0 = A_y  –  8.2  klf  (180  ft) + 738  k

 

A_y = 738  k  (+↑)

 

\underrightarrow{+} ΣF_x = 0 = A_x

Note: The result for A_x confirms that the tie carries all the horizontal force.

Although the deck hangs from the arch by diagonal cables, we can use the equivalent downward load since we are trying to find only the tie force

Approximate the tied arch as a three-hinge arch:

FBD of AC:

\underrightarrow{+} \sum{F_x} = 0 = A_x + C_x

A_x = -C_x

+↑ΣF_y = 0 = A_y  –  8.2  klf  (90  ft)  –  C_y

A_y = 738  k + C_y

+\circlearrowright \sum{M_A} = 0 = 8.2 klf (90 ft) (90 ft/2) + C_x(36 ft) + C_y(90 ft)

C_x(36  ft) = -33,210  k  •  ft  –  C_y(90  ft)

C_x = -922  k  –  2.5C_y

FBD of CB:

\underleftarrow{+} \sum{F_x} = 0 = C_x + B_x

B_x = -C_x

+\uparrow \sum{F_y} = 0 = C_y  –  30  kN/m(34  m) + B_y

B_y = 1020  kN  –  C_y

+\circlearrowright \sum{M_B} = 0 = -C_x(36  ft) + C_y(90  ft)  –  8.2  klf(90  ft)(90  ft/2)

C_x  (36  ft) = -33,210  k  •  ft + C_y  (90  ft)

C_x = -922  k + 2.5C_y

Combine the C_x expression with the result from AC:

-922  k  –  2.5C_y = -922  k + 2.5C_y

5.0C_y = 0

C_y = 0

∴ A_y = 738  k  (+↑)

 

B_y = 738  k  (+↑)

Evaluation of Results:

Satisfied Fundamental Principles?

These reactions match what we calculated at the beginning of the problem. ✓

Find tie force:
Substitute C_y into the expression for C_x:

C_x = -922  k

 

∴ A_x = 922  k  (\underrightarrow{+})

 

B_x = 922  k  (\underleftarrow{+})

Note: This is the tie force.

Summary: