11-10-2016, 06:36 AM

Simple beams: Reactions in Supports. [ R2SP ]

From the author’s Engineering Collection, included in the ETSII4 module.

This program calculates the reactions in the supports of a simple beam subjected to any combination of the following efforts: point, uniform, triangular loads and external moments. Trapezoidal loads can be expressed as a combination of a uniform load plus a triangular one.

The supports can be placed at any two points along the beam’s distance - xa, xb - taking the left end as origin of coordinates. The program will prompt for the load type to enter next, with the following message “LOAD? P:U:T:M:R”. Press the corresponding key for each load type and use it as many times as loads exist, then press “R” to calculate the reactions.

The expressions used by the program are a straight application of statics. Let Ra and Rb be the reaction in the supports; Pj and Mj the different point loads and external moments, applied at a distance xj.(j=1,2…n). Let q be the load per unit length of uniform and triangular loads applied between distances (x1,x2), and “m” the slope of the triangular load. Then we have:

(1) SUM Fj = Ra + Rb + SUM qj (x2-x1)j + SUM qj/2 (x2 - x1)j^2

(2) SUM Mj = xa Ra + xb Rb + SUM x Fj + SUM qj/2 (x2^2 – x1^2)j + SUM {[ m (x2^3 – x1^3)j/3 – (m x1 /2) (x2^2 – x1^2)j ] }

Example.

Calculate the reactions in the supports of a simple bean with the following configuration:

Xa = 2 m, Xb = 6m;

External moment M1 = 200 N.m (counter-clockwise is positive)

Uniform loads q1=120 N/m between x1=1 and x2=3

Uniform load q2 = 300 N/m between x1=3 and x2=4

Triangular load between x1=6 and x2=7, with final value of q3=100 N/m

The results are:

R1 = 485, 84 N

R2 = 4,1667 N

From the author’s Engineering Collection, included in the ETSII4 module.

This program calculates the reactions in the supports of a simple beam subjected to any combination of the following efforts: point, uniform, triangular loads and external moments. Trapezoidal loads can be expressed as a combination of a uniform load plus a triangular one.

The supports can be placed at any two points along the beam’s distance - xa, xb - taking the left end as origin of coordinates. The program will prompt for the load type to enter next, with the following message “LOAD? P:U:T:M:R”. Press the corresponding key for each load type and use it as many times as loads exist, then press “R” to calculate the reactions.

The expressions used by the program are a straight application of statics. Let Ra and Rb be the reaction in the supports; Pj and Mj the different point loads and external moments, applied at a distance xj.(j=1,2…n). Let q be the load per unit length of uniform and triangular loads applied between distances (x1,x2), and “m” the slope of the triangular load. Then we have:

(1) SUM Fj = Ra + Rb + SUM qj (x2-x1)j + SUM qj/2 (x2 - x1)j^2

(2) SUM Mj = xa Ra + xb Rb + SUM x Fj + SUM qj/2 (x2^2 – x1^2)j + SUM {[ m (x2^3 – x1^3)j/3 – (m x1 /2) (x2^2 – x1^2)j ] }

Example.

Calculate the reactions in the supports of a simple bean with the following configuration:

Xa = 2 m, Xb = 6m;

External moment M1 = 200 N.m (counter-clockwise is positive)

Uniform loads q1=120 N/m between x1=1 and x2=3

Uniform load q2 = 300 N/m between x1=3 and x2=4

Triangular load between x1=6 and x2=7, with final value of q3=100 N/m

The results are:

R1 = 485, 84 N

R2 = 4,1667 N