# Tractions on the surface of the block, and resultant forces and moment

• roldy
In summary, traction is the force per unit area acting on the surface of a block, typically measured in N/m² or Pa. It can be calculated by dividing the force by the surface area. Resultant forces are the net forces resulting from combining multiple forces on an object, taking into account magnitude and direction. They can be calculated using vector addition. The moment of a force is a measure of its tendency to cause rotation, calculated by multiplying the force by the perpendicular distance from the axis or point of rotation to the line of action of the force.
roldy

## Homework Statement

The plane strain solution for the stresses in the rectangular block 0 < x < a, 0 < y < b, 0 < z < c under certain loading conditions is given by.
$\sigma_{xx} = \frac{3Fxy}{2b^3}$
$\sigma_{xy} = \frac{3F(b^2-y^2)}{4b^3}$
$\sigma_{yy} = 0$
$\sigma_{zz} = \frac{3\nu Fxy}{2b^3}$
Find the tractions on all the six surfaces of the block. Determine the resultants (Forces and Moments) on the surface x = a.

## Homework Equations

$F = \int\int \sigma dA$
$M_x = Fx$
$M_y = Fy$
$M_z = Fz$

## The Attempt at a Solution

For the traction on each of the surfaces, I just plug the min and max values of the dimensions into the stress given.

x faces:
x = 0,
$\sigma_{xx} = 0$
$\sigma_{xy} = \frac{3F(b^2-y^2)}{4b^3}$
$\sigma_{xz} = 0$

x = a,
$\sigma_{xx} = \frac{3Fay}{2b^3}$
$\sigma_{xy} = \frac{3F(b^2-y^2)}{4b^3}$
$\sigma_{xz} = 0$

y faces:
y = 0,
$\sigma_{yy} = 0$
$\sigma_{yx} = \frac{3F}{4b}$
$\sigma_{yz} = 0$

y = b,
$\sigma_{xx} = 0$
$\sigma_{xy} = 0$
$\sigma_{xz} = 0$

z faces:
z = 0,
$\sigma_{zz} = \frac{3 \nu Fxy}{2b^3}$
$\sigma_{zy} = 0$
$\sigma_{xz} = 0$

z = c,
$\sigma_{zz} = \frac{3 \nu Fxy}{2b^3}$
$\sigma_{zy} = 0$
$\sigma_{xz} = 0$

The part I am confused about is finding the resultant forces on the face x = a.
I believe that all I do is use the tractions that I found for x = a. Thus,

$F_x = \int \limits_0^b \int \limits_0^c \sigma_{xx}dzdy = \int \limits_0^b \int \limits_0^c \frac{3Fay}{2b^3}dzdy$

$F_y = \int \limits_0^b \int \limits_0^c \sigma_{xy}dzdy = \int \limits_0^b \int \limits_0^c \frac{3F(b^2-y^2)}{4b^3}dzdy$

$F_z = \int \limits_0^b \int \limits_0^c \sigma_{xz}dzdy = \int \limits_0^b \int \limits_0^c (0)dzdy$

Then for the moments I just take these forces and multiply them by the distance to the x axis.

Is my work correct so far or am I missing something.

Also, is there any other approach to finding the resultant forces on the face x = a?

Your approach for finding the tractions on all six surfaces of the block is correct. To find the resultant forces on the face x = a, you can also use the equations for the resultant forces in the x, y and z directions:

F_x = \int \limits_0^b \int \limits_0^c \sigma_{xx}dzdy = \int \limits_0^b \int \limits_0^c \frac{3Fay}{2b^3}dzdy

F_y = \int \limits_0^b \int \limits_0^c \sigma_{xy}dzdy = \int \limits_0^b \int \limits_0^c \frac{3F(b^2-y^2)}{4b^3}dzdy

F_z = \int \limits_0^b \int \limits_0^c \sigma_{xz}dzdy = \int \limits_0^b \int \limits_0^c (0)dzdy

and then use the equations for the resultant moments in the x, y and z directions:

M_x = \int \limits_0^b \int \limits_0^c y\sigma_{xx}dzdy = \int \limits_0^b \int \limits_0^c \frac{3Fay^2}{2b^3}dzdy

M_y = \int \limits_0^b \int \limits_0^c y\sigma_{xy}dzdy = \int \limits_0^b \int \limits_0^c \frac{3F(b^2-y^2)y}{4b^3}dzdy

M_z = \int \limits_0^b \int \limits_0^c y\sigma_{xz}dzdy = \int \limits_0^b \int \limits_0^c (0)dzdy

These equations will give you the resultant forces and moments on the surface x = a. Your approach and this approach should give you the same results. Keep up the good work!

## 1. What is traction on the surface of a block?

Traction is the force per unit area that is acting on the surface of a block. It is typically measured in units of Newtons per square meter (N/m²) or Pascals (Pa).

## 2. How is traction calculated?

Traction is calculated by dividing the force acting on the surface by the area over which it is acting. It is represented by the symbol "T" and can be expressed as T = F/A, where F is the force and A is the surface area.

## 3. What are resultant forces?

Resultant forces are the net forces that result from combining multiple forces acting on an object. They take into account both the magnitude and direction of each individual force to determine the overall effect on the object.

## 4. How are resultant forces calculated?

Resultant forces can be calculated by using vector addition. This involves breaking down each force into its horizontal and vertical components, adding the components separately, and then combining them to find the overall magnitude and direction of the resultant force.

## 5. What is the moment of a force?

The moment of a force is a measure of its tendency to cause rotation around a particular axis or point. It is calculated by multiplying the force by the perpendicular distance from the axis or point of rotation to the line of action of the force.

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