## Chapter 3

## 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.

## Step-by-Step

## Verified Solution

Analysis:

Find the reactions:

B_y = 738 k (+↑)

+↑ΣF_y = 0 = A_y – 8.2 klf (180 ft) + 738 kA_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_xA_x = -C_x

+↑ΣF_y = 0 = A_y – 8.2 klf (90 ft) – C_yA_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_xB_x = -C_x

+\uparrow \sum{F_y} = 0 = C_y – 30 kN/m(34 m) + B_yB_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_y5.0C_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:

∴ A_x = 922 k (\underrightarrow{+})

B_x = 922 k (\underleftarrow{+})

Note: This is the tie force.

Summary: