## Chapter 3

## Q. 3.7

We are designing a new pedestrian bridge to cross a major highway. The architect and senior engineer have decided on a thrust arch with fixed supports. The deck widens from one end to the other, so the dead load is not uniform.

Our team is still in the preliminary design phase, but the geotechnical engineers need order-of-magnitude estimates of the forces on the foundations. Each team member is performing an approximate analysis of the arch for different unfactored loads. A team member has calculated the seismic loads based on the assumed self-weight of the bridge. Although the static equivalent seismic load will be distributed along the arch, we can simplify that as five point loads for this preliminary analysis. Our task is to use approximate analysis to estimate the vertical and horizontal forces on the foundations due to seismic load in the plane of the arch.

## Step-by-Step

## Verified Solution

Analysis

Approximate the fix-supported arch as a three-hinge arch. This will not predict moment reactions, but it will predict all other reactions.

FBD of AC:

\underrightarrow{+} \sum{F_x} = 0 = A_x + 9 kN + 23 kN + C_xA_x = -C_x – 32 kN

+↑ΣF_y = 0 = A_y – C_yA_y = C_y

+\circlearrowright ΣM_A = 0 = 23 kN (9.6 m) + C_x (12 m) + C_y (34 m)C_x (12 m) = -C_y (34 m) – 221 kN • m

C_x = -2.83C_y – 18.41 kN

FBD of CB:

\underleftarrow{+} \sum{F_x} = 0 = -C_x + 28 kN + 33 kN + 19 kN – B_xB_x = -C_x + 80 kN

+↑ΣF_y = 0 = C_y + B_yB_y = -C_y

+\circlearrowright \sum{M_B} = 0 = -C_x (12 m) + C_y(34 m) + 28 kN (12 m) + 33 kN(9.6 m)C_x (12 m) = C_y (34 m) + 653 kN • m

C_x = 2.83C_y + 54.4 kN

Combine the C_x expression with the result from AC:

2.83C_y + 54.4 kN = -2.83C_y – 18.41 kN5.66C_y + 54.4 kN = -2.83C_y – 18.41 kN

C_y = -12.86 kN

∴ A_y = -12.86 kN (+↑)B_y = 12.86 kN (+↑)

Substitute C_y into the expression for C_x:

C_x = 2.83(-12.86 kN) + 54.4 kN = 18.01 kNUse C_x to find the horizontal reactions:

A_x = -18.01 kN – 32 kN B_x = -18.01 kN + 80 kN

A_x = -50.0 kN (\underrightarrow{+}) B_x = 62.0 kN (\underleftarrow{+})

Summary: