﻿ Both Ends Supported Bar under Distributed LoadExamples | Product | Murata Software Co., Ltd.

# Example53Both Ends Supported Bar under Distributed Load

General

• Distributed load is applied on a both ends supported bar.

• The top face is set with the distributed load boundary condition. See [How to Set Distributed Boundary Condition and Body Attribute] for more information.

• The deformation, the displacement and the mechanical stress are solved.

• Unless specified in the list below, the default conditions will be applied.

### Analysis Space

 Item Settings Analysis Space 3D Model unit mm

### Analysis Conditions

The default conditions are good enough for this exercise.

 Item Settings Solver Mechanical Stress Analysis [Galileo] Analysis Type Static analysis Options N/A

### Model

The bar is a rectangular solid body. The material is polycarbonate.
Distributed load is applied on the top face. Both edges of the bottom face is fixed in Z direction.

### Body Attributes and Materials

 Body Number/Type Body Attribute Name Material Name 0/Solid BEAM 002_Polycarbonate(PC) *

* Available from the Material DB

### Boundary Conditions

Tapered pressure is applied as shown below.

Pressure = 0 at X = 0, Pressure = 1 [MPa] at X = 50 [mm]

 Boundary Condition Name/Topology Tab Boundary Condition Type Settings FIX/Face Mechanical Displacement Select the Z component. UZ=0 LOAD/Face Mechanical Pressure Select “Use distribution data”.   [Coordinates-Pressure] Table

### How to view the distribution

Click “Run Mesher” on the pull-down menu of [Run Mesher/Solver] and see the meshing result.

Select “Distributed pressure” at [Field] to view the entered distribution.

You may click “Run Mesher/Solver” instead. In that case, select “Mesh Information” at [Mode] and then select “Distributed pressure” at [Field].

The linearly changing pressure can be observed.

### Results

The figure below is the gradation contour of Z component of Displacement.

It is a bottom view.

The lowest displacement is 27.584 [mm] at X = 26 [mm]

The formula

gives 27.297 [mm] at X = 25.967 [mm]

They are well matched.

Here,

y: Displacement, l, w, h: Dimensions of the bar, Pmax: Maximum pressure, x: Coordinate, E: Young’s modulus, I: Second moment of area

Second moment of area for a rectangle section is given by

The contour indicates the X normal stress.

It is a bottom view.

The maximum stress is 241.087 [MPa] at X = 28.938 [mm]

The formula

gives 240.563 [MPa] at X = 28.867 [mm]

They are well matched.

Here,

σ: Stress, l, w, h: Dimensions of the bar, Pmax: Maximum pressure, x: Coordinate, Z: Section modulus

Section modulus for a rectangle section is given by

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