Problem Type 5

A simply supported beam with a uniform shape (rectangular, circular) holds a weight (uniformly distributed or concentrated). A point X in the beam is of interest.

Given:

• Load: q (distributed, N/m) or P (concentrated, N)
• Point X coordinates (x, y)
• Beam dimensions (width, height, diameter…)
• No axial (horizontal) load

Find:

• Stresses at point X

Solution

1. Find internal force and moment in point X

Let internal force be \(V_x\), moment \(M_x\).

For concentrated load \(P\) at distance \(m\) from origin:

\[V_x = \frac{P(L - m)}{L}, \quad M_x = \frac{Px(L - m)}{L}, \quad 0 < x < m\] \[V_x = -\frac{Pm}{L}, \quad M_x = \frac{Pm}{L}(L - x), \quad m < x < L\]

For uniformly distributed load:

\[V_x = \frac{qL}{2} - qx, \quad M_x = \frac{qLx}{2} - \frac{qx^2}{2}\]

For cantilevers, concentrated load \(P\):

\[V_x = P,\quad M_x = P(L - x)\]

For cantilevers, uniform load \(q\):

\[V_x = q(L - x),\quad M_x = \frac{q(L - x)^2}{2}\]

2. Find moment of inertia

For rectangular cross section, width \(b\), height \(h\):

\[I = \frac{bh^3}{12}\]

For circular cross section, diameter \(d\):

\[I = \frac{\pi d^4}{64}\]

3. Calculate stresses

\[\sigma_x = -\frac{M_x y}{I}\]

For rectangular beam:

\[\tau_x = \frac{V_x}{2I} \left( \frac{h^2}{4} - y^2 \right)\]

For \(\sigma_x < 0\), the stress is compressive.

Example

%% Inputs
y = 25e-3; % m (distance from point to neutral axis)
q = 28e+3; % N/m (distributed load)
L = 1; % m (length of beam)
x = L - 0.2; % m (x coord of point)
b = 25e-3; % m (beam width)
h = 100e-3; % m (beam height)
% Here the beam is rectangular.

%% Calculations
Vc = (q*L)/2 - q*x;
Mc = (1/2)*q*L*x - (1/2)*q*(x^2);
I = (b*(h^3))/12;
omega = -(Mc*y)/I;
Tc = (Vc/(2*I)) * ((h^2)/4 - y^2);

%% Print
disp("Mc = " + Mc + " N*m")
disp("I = " + I + " m^4")
disp("omega_c = " + omega + " N/m^2")
disp("tau_c = " + Tc + " N/m^2")

Problem Type 10

An element is under plane stress. Given:

• Stress profile \(\sigma_x,\ \sigma_y,\ \tau_{xy}\)

Find:

• Principal stresses \(\sigma_1,\ \sigma_2\)
• Maximum shear stress \(\tau_{\text{max}}\)
• Draw the principal stresses on a properly oriented plane
• Draw the max shear stress on a properly oriented plane

Solution

1. Find \(\sigma_{\text{avg}},\ \sigma_{\text{diff}}\):

\[\sigma_{\text{avg}} = \frac{\sigma_x + \sigma_y}{2}\] \[\sigma_{\text{diff}} = \sigma_x - \sigma_y\]

2. Find \(\tau_{\text{max}}\):

\[\tau_{\text{max}} = \sqrt{\frac{\sigma_{\text{diff}}^2}{4} + \tau_{xy}^2}\]

3. Find the principal stresses \(\sigma_1,\ \sigma_2\)

\[\sigma_1 = \sigma_{\text{avg}} + \tau_{\text{max}}\] \[\sigma_2 = \sigma_{\text{avg}} - \tau_{\text{max}}\]

Note: \(\tau_{\text{max}} = \frac{\sigma_1 - \sigma_2}{2}\). AND \(\sigma_x + \sigma_y = \sigma_1 + \sigma_2\). Use these to verify your values.

4. Find the angles for the drawing \(\theta_p,\ \theta_s\)

\[\theta_p = \frac{1}{2} \arctan \frac{2\tau_{\text{xy}}}{\sigma_{\text{diff}}}\] \[\theta_s = \frac{1}{2} \arctan \frac{\sigma_{\text{diff}}}{2\tau_{\text{xy}}}\]

5. Assign principal angles

To verify which angle is for \(\sigma_1\) and \(\sigma_2\), the safe way is to replace in the typical transformation equation. For example:

\[\sigma_{x1} =\sigma_{\text{avg}} + \frac{\sigma_{\text{diff}}}{2} \cos{2\theta_p} + \tau_{\text{xy}} \sin{2\theta}\]

If \(\sigma_{x1} = \sigma_{1}\), then \(\theta_{p1} = \theta_p\) and \(\theta_{p2} = \theta_p + 90°\).

Otherwise, \(\theta_{p2} = \theta_p\) and \(\theta_{p1} = \theta_p + 90°\).

Verify that \(\theta_s = \theta_{p1} \pm 45°\).

Example

%% Inputs
sigma_x = 86; % MPa
sigma_y = -28; % MPa
tau_xy = -32; % MPa

% Positive => tensile
% negative => compressive

% For shear stress
% Positive => points to (+)(+) and (-)(-) quadrants
% negative => points to (-)(+) and (+)(-) quadrants
%% Calculations
sigma_diff = sigma_x - sigma_y;

sigma_avg = (sigma_x + sigma_y)/2;
tau_max = sqrt((sigma_diff^2)/4 + tau_xy^2);

sigma_1 = sigma_avg + tau_max;
sigma_2 = sigma_avg - tau_max;

theta_p = 0.5*atand(2*tau_xy/sigma_diff);
theta_s = 0.5*atand(-sigma_diff/(2*tau_xy));

% Assign principal angles
theta_1 = theta_p;
theta_2 = theta_p;

sigma_test = sigma_avg + (sigma_diff/2)*cosd(2*theta_p) + tau_xy*sind(2*theta_p);
if (abs(sigma_1 - sigma_test) < 1e-6)
    theta_2 = theta_2 + 90;
else
    theta_1 = theta_1 + 90;
end

%% Print
disp("\sigma_1 = " + sigma_1 + " MPa")
disp("\sigma_2 = " + sigma_2 + " MPa")
disp("\tau_{max} = " + tau_max + " MPa")

disp("\theta_p = " + theta_p + " deg")
disp("\theta_s = " + theta_s + " deg")

disp("\theta_1 = " + theta_1 + " deg")
disp("\theta_2 = " + theta_2 + " deg")