Equilibrium of a Rigid Body Study Notes

Chapter 5: Equilibrium of a Rigid Body

• The primary objectives for this chapter include identifying support reactions and drawing comprehensive free-body diagrams (FBD).

Understanding Supports and Reactions

• Supports are components that specify how applied loads within a structure are transferred to the ground or to other structural members.
• Reactions are the forces and couple moments exerted on an object by its supports. These reactions hold the object in place; for example, a bridge is held up by reactions exerted by its supports at either end.

Support Types and Their Reactions

• Roller Support:

  • In real-life applications, a roller support allows a body to move horizontally and rotate.
  • It prevents translation in the vertical direction only.
  • Reaction: A single vertical force is exerted perpendicular to the surface of contact.

• Pin Support:

  • A pin support restricts movement in any translational direction but allows the body to rotate.
  • It prevents translation in both the horizontal and vertical directions.
  • Reaction: It involves two force components, generally represented as $\mathbf{F_x}$ and $\mathbf{F_y}$.

• Fixed Support (Cantilever):

  • A cantilever is a structure with a fixed support on one end while the other end remains free.
  • A fixed support resists all forms of translation and rotation.
  • Reaction: It must develop all three components in 2-D: vertical force ($\mathbf{F_y}$), horizontal force ($\mathbf{F_x}$), and a couple moment ($\mathbf{M}$).

General Rule for Support Reactions (2-D)

• As a general rule, if a support prevents the translation of a body in a given direction, a force is developed on the body in the opposite direction.
• Similarly, if rotation is prevented by the support, a couple moment is exerted on the body.
• Detailed tables for other support reactions can be found in reference materials such as Table 5-1 in the textbook.

Modeling and Static Indeterminacy

• Idealized Model: This is a simplified representation of a real-world object (e.g., a steel beam supporting roof joists). It is used to create a Free-Body Diagram to determine unknown support reactions at points such as A and B.
• Statically Indeterminate Situations:
  • A problem is considered statically indeterminate when it cannot be analyzed using only the Equations of Equilibrium (E-of-E).
  • This occurs when a structure has more supports than the minimum number necessary to maintain equilibrium.
  • For example, if a system has 5 unknown reactions but only 3 available equilibrium equations, it is statically indeterminate.
  • Another example shown involves a beam with 4 unknowns (e.g., $A_x, A_y, M_A, B_y$) and 3 equilibrium equations, which also qualifies as statically indeterminate.

Conditions for Rigid Body Equilibrium

• Unlike forces acting on a simple particle, the forces on a rigid body are usually not concurrent. This means they can create moments that cause the body to rotate.
• For a rigid body to be in equilibrium, two conditions must be met:
  • The net resultant force ($\sum \mathbf{F_R}$) must be equal to zero: $\sum \mathbf{F_R} = 0$
  • The net moment ($\sum \mathbf{M_O}$) about any arbitrary point O must be equal to zero: $\sum \mathbf{M_O} = 0$

Construction of a Free-Body Diagram (FBD)

1. Draw an Outlined Shape: Imagine the body is isolated or "cut free" from its environment and constraints. Draw only the perimeter of the body.
2. Show All External Forces and Couple Moments: This includes:
   • Applied loads (external weights or pressures).
   • Support reactions (forces and moments exerted by the supports on the body).
   • The weight of the body itself ($W$), usually acting through the center of gravity.

Example: Fixed Support Beam

• Problem Scenario: A beam of length $6\,m$ is fixed at point A. It carries an applied load of $1200\,N$ at a distance of $2\,m$ from support A. The beam's weight is $981\,N$, acting at the center of the beam ($3\,m$ from A).
• FBD Components:
  • Effect of fixed support at A: Reactions $A_x, A_y,$ and $M_A$.
  • Effect of applied force: $1200\,N$ acting downward.
  • Effect of gravity: $981\,N$ acting downward at the center of gravity ($3\,m$ mark).
• Calculations:
  • $\sum F_x = 0 \implies -A_x = 0 \implies A_x = 0$
  • $\sum F_y = 0 \implies A_y - 1200\,N - 981\,N = 0 \implies A_y = 2181\,N$
  • $\sum M_A = 0 \implies M_A - (1200\,N)(2\,m) - (981\,N)(3\,m) = 0$
  • $M_A - 2400\,N\cdot m - 2943\,N\cdot m = 0 \implies M_A = 5343\,N\cdot m$

Two-Force Members

• A two-force member is a structural component subjected to forces at only two points (e.g., points A and B).
• Properties:
  • The resultant forces at point A and point B must be equal in magnitude and act in opposite directions.
  • These forces must act along the line joining points A and B.
• Significance: Identifying two-force members simplifies equilibrium analysis because the direction of the resultant reaction is known immediately. This reduces the number of unknowns at a pin connection from two components to a single force with a known line of action.
• Example Application: On a platform supported at joint A (pin) and connected to link BC, link BC is recognized as a two-force member.

Strategy for 2-D Equilibrium Problems

• Equations used:
  • $\sum F_x = 0$
  • $\sum F_y = 0$
  • $\sum M_O = 0$ (where O is any arbitrary point).
• Important Considerations:
  • If unknowns outnumber independent equations ($U > 3$), the system is statically indeterminate.
  • The order of equations matters. Solving $\sum M_O = 0$ first can often isolate one unknown if the point O is chosen at a support with multiple unknown forces.
  • If a calculated value for an unknown is negative, the actual direction of the force/moment is opposite to the direction originally assumed in the FBD.

Tutorial Problem #1: Beam Reactions

• Given: A beam with a $600\,N$ force at $45^\circ$ at one end, a $100\,N$ downward force $2\,m$ from the pivot, and a $200\,N$ downward force at the other end. Distance between supports is $7\,m$.
• Plan: Draw FBD and apply E-of-E.
• Calculations:
  • $\sum F_x = 0 \implies (600\,N)\cos(45^\circ) - B_x = 0 \implies B_x = 424.26\,N$
  • $\sum M_B = 0 \implies (100\,N)(2\,m) + (600\,N)\sin(45^\circ)(5\,m) - (600\,N)\cos(45^\circ)(0.2\,m) - A_y(7\,m) = 0$
  • $A_y = 319.50\,N$
  • $\sum F_y = 0 \implies A_y - (600\,N)\sin(45^\circ) - 100\,N - 200\,N + B_y = 0 \implies B_y = 404.76\,N$

Tutorial Problem #2: Lever and Link

• Given: Lever ABC pin-supported at A; connected to short link BD at point B. Load of $400\,N$ at point C.
• Link BD is a two-force member at an angle of $45^\circ$.
• Calculations:
  • $\sum M_A = 0 \implies -(400\,N)(0.7\,m) + F_{BD}\cos(45^\circ)(0.2\,m) + F_{BD}\sin(45^\circ)(0.1\,m) = 0$
  • $F_{BD} = 1319.9\,N \approx 1.32\,kN$
  • $\sum F_x = 0 \implies (400\,N) - F_{BD}\cos(45^\circ) + A_x = 0 \implies A_x = 533.33\,N$
  • $\sum F_y = 0 \implies -F_{BD}\sin(45^\circ) + A_y = 0 \implies A_y = 933.33\,N$

Tutorial Problem #3: Platform Support

• Given: Platform weight = $1962\,N$; cable at B at $70^\circ$ angle; pin support at A.
• Calculations:
  • $\sum M_A = 0 \implies T\cos(70^\circ)(1\,m) + T\sin(70^\circ)(2.2\,m) - (1962\,N)(1.4\,m) = 0$
  • $T = 1140.1\,N \approx 1.14\,kN$
  • $\sum F_x = 0 \implies A_x - T\cos(70^\circ) = 0 \implies A_x = 389.92\,N$
  • $\sum F_y = 0 \implies A_y + T\sin(70^\circ) - (1962\,N) = 0 \implies A_y = 890.69\,N$

Tutorial Problem #4: Frame and Frictionless Pulley

• Given: Frame ACD with cable ABD through a frictionless pulley at B. Load = $150\,N$.
• Geometry:
  • Length of cable segment AB: $\sqrt{0.3^2 + 0.125^2} = 0.325\,m$
• Calculations:
  • $\sum M_C = 0 \implies (150\,N)(0.225\,m) - (\frac{0.125}{0.325}T)(0.225\,m) + T(0.075\,m) - (\frac{0.3}{0.325}T)(0.175\,m) = 0$
  • Tension $T = 195.000\,N$
  • $\sum F_x = 0 \implies C_x + (\frac{0.3}{0.325})(195\,N) = 0 \implies C_x = -180.000\,N$
  • $\sum F_y = 0 \implies C_y - 150\,N + (\frac{0.125}{0.325})(195\,N) + 195\,N = 0 \implies C_y = -120.000\,N$