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

[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.

 

[KataKoniK]

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!