PRATT TRUSS AND EXAMPLE SOLUTION

        A Pratt truss includes vertical members and diagonals that slope down towards the center, Pratt Truss similar to the Howe Truss just the opposite of the Howe. The interior diagonals are under tension under balanced loading and vertical elements under compression. If pure tension elements are used in the diagonals then crossing elements may be needed near the center to accept concentrated live loads as they traverse the span. It can be subdivided, creating Y- and K-shaped patterns. The Pratt truss was invented in 1844 by Thomas and Caleb Pratt. This truss is practical for use with spans up to 250 feet (76 m) and was a common configuration for railroad bridges as truss bridges moved from wood to metal. They are statically determinate bridges, which lend themselves well to long spans. They were common in the United States between 1844 and the early 20th century.

 

 

This model is inspired by a classic bridge type called a Pratt truss bridge. You can identify a Pratt truss by its diagonal members, which (except for the very end ones) all slant down and in toward the center of the span. All the diagonal members are subject to tension forces only, while the shorter vertical members handle the compressive forces. Since the tension removes the buckling risk, this allows for thinner diagonal members resulting in a more economic design.

A truss structure supports only tension and compression forces in its members and you would normally model it using bars, but as this model uses 3D beams it also includes bending moments to some extent in a frame structure. In the model, shell elements represent the roadway and both the Shell and Beam interfaces are used.

 

 

Example: 

Analayse the truss system given below. 

solution:

Fundamentals

‎Write the equilibrium of forces along the x -axis.

∑Fx​=0

Write the equilibrium of forces along the y -axis.

∑Fy​=0

Write the equilibrium of moments about point A.

∑MA​=0

 

Sign Convention for force: Upward and right forces are positive.

Sign convention for moment: Anti clockwise moment is positive and clockwise moment is negative.

 

 

 

 

Take moment about point E:

∑ME​=0:

(Ey​×0.60) − (5kNx0.30) =0

0.6Ey​− 1.5kN=0

0.6Ey​=1.5kN

Ey​=1.5/0.6=2.5kN​

 

 

Apply force equilibrium in x and y directions.

Ay​+Ey​− 5kN=0

Ay​+2.5kN − 5kN=0

Ay​=2.5kN​

 

 

 

 

From the free body diagram of the entire truss:

θ=tan ^-1(20/15​)=53.13∘​

 

F_AH ​x sin53.13=2.5kN

F_AH ​= 2.5kN / sin53.13

F_AH = 3.125kN  (COMPRESSION)

 

From simetric condition:

 

F_FE ​x sin53.13=2.5kN

F_FE ​= 2.5kN / sin53.13

F_FE = 3.125kN  (COMPRESSION)

 

To calculate F_AB

F_AB = F_AH x cos53.13

F_AB = 3.125 x cos53.13  (TENSION)

F_AB = 1.875kN

From the simetric condition

F_DE = F_BC = F_DC = F_HG = F_GF = F_AB

F_DE = F_BC = F_DC = F_HG = F_GF = 1.875kN

F_DE , F_BC , F_DC ,F_AB (TENSION)

F_HG , F_GF (COMPRESSION)

 

 

 

From equilibrium conditions:

Ay​− F_HC ​ x sin53.13=0

2.5kN− F_HC ​ x sin53.13=0

F_HC ​x sin53.13=2.5kN

F_HC ​ = 2.5kN / sin53.13 = 3.125kN  (TENSION)

from the simetric condition:

F_FC = 3.125kN (TENSION)

 

 

F_GC, F_HB and F_FD = 0 (ZERO FORCE MEMBER)