8.3 Internal Forces in Rigid Bodies: Trusses, Frames, Machines, and Friction

8.3 Internal Forces in Rigid Bodies: Trusses, Frames, Machines, and Friction

Engineering Application: Structural analysis relies on evaluating internal axial forces in trusses, multi-force pin interactions in frames, geometric properties (centroids and moments of inertia) for stress analysis, and contact friction thresholds governing mechanical stability.

Analysis of Planar Trusses

A truss is a structure composed of slender members joined together at their end points by smooth pins. Standard FE exam truss analysis assumes:

1. All members are weightless two-force members carrying purely axial tension ($T > 0$) or compression ($C < 0$).
2. External loads and support reactions are applied exclusively at pin joints.

Method of Joints

The Method of Joints analyzes the static equilibrium of individual pin joints ($\sum F_x = 0, \sum F_y = 0$). It is ideal when finding forces in all members sequentially starting from a joint with at most two unknown member forces.

Method of Sections

The Method of Sections cuts through a portion of the truss to expose internal member forces. By treating the isolated section as a rigid body in coplanar equilibrium, you can solve for up to three unknown member forces simultaneously using:

$\sum F_x = 0, \quad \sum F_y = 0, \quad \sum M_O = 0$

Strategy: To isolate a specific member force, sum moments about the intersection point of the remaining cut members.

Identification of Zero-Force Members

Inspecting joints for zero-force members simplifies truss analysis prior to writing equilibrium equations:

• Case 1 (Two non-collinear members): If a joint connects only two non-collinear members and has no external load or support reaction, both members are zero-force members.
• Case 2 (Three members, two collinear): If a joint connects three members with no external load or support reaction, and two of the members are collinear, the third non-collinear member is a zero-force member.

Frames, Machines, and Dry Coulomb Friction

Frames and Machines

Unlike trusses, frames and machines contain at least one multi-force member—a structural component subjected to three or more forces or moments (causing bending). To analyze frames:

1. Isolate the entire frame to find external support reactions.
2. Disassemble the frame into individual components.
3. Apply Newton's Third Law ($F_{A/B} = -F_{B/A}$) at all connecting pins and solve member equilibrium equations.

Dry Coulomb Friction Laws

Dry friction occurs between non-lubricated contacting surfaces. The maximum static friction force $F_{s,\max}$ before sliding occurs is proportional to the normal force $N$:

$F_f \le F_{s,\max} = \mu_s N$

where $\mu_s$ is the coefficient of static friction. Once motion impends, the angle of static friction $\theta_s$ relates to $\mu_s$ by:

$\tan \theta_s = \mu_s$

When sliding occurs, the kinetic friction force acts opposing motion with magnitude:

$F_k = \mu_k N$

where $\mu_k$ is the coefficient of kinetic friction (typically $\mu_k < \mu_s$).

Tipping vs. Sliding Criteria

For a rigid block of width $w$ and height $h$ subjected to a horizontal force $P$ applied at height $y$:

• Sliding occurs first if $P > \mu_s W$.
• Tipping occurs first if the line of action of the normal force moves beyond the footprint boundary ($x > w/2$), requiring $P > W \left( \frac{w}{2h} \right)$.

Worked Engineering Problems

Problem 1: Truss Method of Sections with Zero-Force Member Identification

Scenario: A planar Pratt bridge truss features panels of width $b = 3.0\text{ m}$ and vertical post height $h = 4.0\text{ m}$. Joints $A, B, C, D$ lie along the bottom chord, while $E, F, G$ lie along the top chord. A vertical downward load of $P = 60\text{ kN}$ is applied at bottom joint $B$. Determine if vertical post $FB$ is a zero-force member, and calculate the force in top chord member $EF$.

Solution:

1. Inspect joint $F$ for zero-force members: At top chord joint $F$, horizontal members $EF$ and $FG$ are collinear. No external load is applied at joint $F$. Vertical post $FB$ connects to joint $F$ perpendicular to the collinear line. By Case 2 zero-force member rules, $\sum F_y = 0 \implies F_{FB} = 0\text{ kN}$. Post $FB$ is a zero-force member.

2. Calculate global support reactions: Assume symmetric span $L = 12.0\text{ m}$ with pin at $A$ and roller at $D$. Reaction at $A$: $A_y = 45\text{ kN}$, reaction at $D$: $D_y = 15\text{ kN}$.

3. Method of Sections cut through members $EF$, $EB$, and $AB$: Pass a vertical section cut through panel 1. Isolate the left portion of the truss. Sum moments about bottom joint $B$ ($x = 3.0\text{ m}$): $\sum M_B = 0 \implies A_y (3.0\text{ m}) + F_{EF} (4.0\text{ m}) = 0$ $(45\text{ kN})(3.0\text{ m}) + F_{EF}(4.0\text{ m}) = 0$ $135 + 4.0 F_{EF} = 0 \implies F_{EF} = -33.75\text{ kN}$ The negative sign confirms that member $EF$ is in compression with magnitude $33.75\text{ kN}$.

Problem 2: Composite Area Centroid and Parallel Axis Moment of Inertia

Scenario: A structural T-section beam cross section consists of a top flange ($200\text{ mm}$ wide by $20\text{ mm}$ thick) welded to a vertical web ($20\text{ mm}$ wide by $180\text{ mm}$ high). Determine the centroid location $\bar{y}$ measured from the bottom edge of the web and compute the total centroidal area moment of inertia $I_x$.

Solution:

1. Partition into two rectangles:

• Segment 1 (Web): $A_1 = 20 \times 180 = 3600\text{ mm}^2$, centroid $\bar{y}_1 = 90\text{ mm}$.
• Segment 2 (Flange): $A_2 = 200 \times 20 = 4000\text{ mm}^2$, centroid $\bar{y}_2 = 180 + 10 = 190\text{ mm}$.
• Total Area $A = 3600 + 4000 = 7600\text{ mm}^2$.

1. Calculate composite centroid $\bar{y}$: $\bar{y} = \frac{A_1 \bar{y}_1 + A_2 \bar{y}_2}{A} = \frac{(3600)(90) + (4000)(190)}{7600} = \frac{324,000 + 760,000}{7600} = \frac{1,084,000}{7600} \approx 142.63\text{ mm}$

2. Apply Parallel Axis Theorem $I_x = \sum (I_{cx,i} + A_i d_i^2)$:

• Web offset $d_1 = |\bar{y} - \bar{y}_1| = |142.63 - 90| = 52.63\text{ mm}$. $I_{cx1} = \frac{1}{12}(20)(180)^3 = 9.720 \times 10^6\text{ mm}^4$ $I_{x1} = 9.720 \times 10^6 + (3600)(52.63)^2 = 9.720 \times 10^6 + 9.972 \times 10^6 = 19.692 \times 10^6\text{ mm}^4$
• Flange offset $d_2 = |\bar{y} - \bar{y}_2| = |142.63 - 190| = 47.37\text{ mm}$. $I_{cx2} = \frac{1}{12}(200)(20)^3 = 0.1333 \times 10^6\text{ mm}^4$ $I_{x2} = 0.1333 \times 10^6 + (4000)(47.37)^2 = 0.1333 \times 10^6 + 8.976 \times 10^6 = 9.109 \times 10^6\text{ mm}^4$

1. Sum total centroidal moment of inertia $I_x$: $I_x = I_{x1} + I_{x2} = 19.692 \times 10^6 + 9.109 \times 10^6 = 28.801 \times 10^6\text{ mm}^4 \approx 2.88 \times 10^7\text{ mm}^4$