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## Q. 3.6

Our firm is designing a bridge to cross a large creek. The bridge will be a tied arch.

In order to make a preliminary selection of member properties, the team needs to determine approximate internal forces. We have been tasked with finding the tie force and the vertical reactions at the abutments. A more experienced member of the team has estimated the combined dead and live load on the arch.

## Verified Solution

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: