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In civil engineering, structures such as buildings, bridges, and towers are designed to remain stable and safe under various loads. The fundamental principle that ensures a structure does not move or collapse under these loads is equilibrium. Equilibrium refers to the state where all the forces and moments (rotational effects) acting on a structure balance out perfectly, resulting in no movement.
Imagine a seesaw perfectly balanced with equal weights on both ends; it stays still because the forces are balanced. Similarly, structures must satisfy equilibrium conditions to remain stationary and safe. Understanding equilibrium is crucial for analyzing forces within structural elements and designing them to withstand loads without failure.
In this chapter, we will explore the concept of equilibrium from first principles, learn how to represent forces acting on structures, and apply mathematical conditions to solve real-world engineering problems.
For a body (or structure) to be in equilibrium, two fundamental conditions must be satisfied:
These conditions apply to both two-dimensional (2D) and three-dimensional (3D) problems, though the number of equations differs.
In a 2D plane (X-Y), the equilibrium conditions are:
In 3D, an additional force equilibrium along the Z-axis and moments about X, Y, and Z axes are considered:
A moment is the turning effect produced by a force acting at a distance from a pivot point. It is calculated as:
A Free Body Diagram (FBD) is a simplified representation of a structure or its part, isolated from its surroundings, showing all external forces and moments acting on it. Drawing an accurate FBD is the first and most crucial step in solving equilibrium problems.
Why use an FBD? It helps visualize the problem clearly, identify unknown forces, and apply equilibrium equations systematically.
Step 1: Draw the Free Body Diagram (FBD) showing the beam, supports at ends A and B, and the uniform load.
Step 2: Calculate the total load on the beam:
Total load, \( W = w \times L = 2 \, \text{kN/m} \times 6 \, \text{m} = 12 \, \text{kN} \)
This load acts at the midpoint of the beam, i.e., 3 m from either support.
Step 3: Apply equilibrium equations.
Sum of vertical forces:
\[ R_A + R_B - 12 = 0 \quad \Rightarrow \quad R_A + R_B = 12 \, \text{kN} \]Sum of moments about point A (taking counterclockwise as positive):
\[ \sum M_A = 0 \Rightarrow -12 \times 3 + R_B \times 6 = 0 \] \[ -36 + 6 R_B = 0 \quad \Rightarrow \quad R_B = \frac{36}{6} = 6 \, \text{kN} \]Step 4: Find \( R_A \):
\[ R_A = 12 - R_B = 12 - 6 = 6 \, \text{kN} \]Answer: Reaction forces at supports A and B are both 6 kN upward.
Step 1: Draw the Free Body Diagram of joint A showing the load and forces in members AB and AC.
Let the force in AB be \( F_{AB} \) (tension positive away from joint) and force in AC be \( F_{AC} \).
Step 2: Resolve forces into components.
Step 3: Apply equilibrium equations at the joint.
Sum of forces in X-direction:
\[ F_{AB} + F_{AC} \cos 45^\circ = 0 \quad \Rightarrow \quad F_{AB} = -F_{AC} \times 0.707 \]Sum of forces in Y-direction:
\[ F_{AC} \sin 45^\circ - 10 = 0 \quad \Rightarrow \quad F_{AC} \times 0.707 = 10 \] \[ F_{AC} = \frac{10}{0.707} = 14.14 \, \text{kN} \]Step 4: Find \( F_{AB} \):
\[ F_{AB} = -14.14 \times 0.707 = -10 \, \text{kN} \]The negative sign indicates \( F_{AB} \) acts opposite to assumed direction, i.e., compression.
Answer: Force in member AC is 14.14 kN tension, and force in member AB is 10 kN compression.
Step 1: Draw the Free Body Diagram showing supports A (left) and B (right), point load at 3 m from A.
Step 2: Apply equilibrium equations.
Sum of vertical forces:
\[ R_A + R_B - 20 = 0 \quad \Rightarrow \quad R_A + R_B = 20 \, \text{kN} \]Sum of moments about A:
\[ \sum M_A = 0 \Rightarrow -20 \times 3 + R_B \times 8 = 0 \] \[ -60 + 8 R_B = 0 \quad \Rightarrow \quad R_B = \frac{60}{8} = 7.5 \, \text{kN} \]Step 3: Calculate \( R_A \):
\[ R_A = 20 - 7.5 = 12.5 \, \text{kN} \]Step 4: Calculate moment at the load point due to reactions (eccentric load effect):
Moment about point of load (3 m from A):
\[ M = R_A \times 3 = 12.5 \times 3 = 37.5 \, \text{kNm} \]Answer: Reaction at A is 12.5 kN, at B is 7.5 kN. The eccentric load causes a moment of 37.5 kNm at the load point.
Step 1: Draw the Free Body Diagram showing the ladder, weight at midpoint, normal reaction at base, friction force at base, and normal reaction at wall (horizontal).
Step 2: Calculate the vertical and horizontal components.
Weight \( W = 100 \, \text{N} \) acts at 2.5 m from base.
Step 3: Apply equilibrium equations.
Sum of vertical forces:
\[ N - W = 0 \quad \Rightarrow \quad N = 100 \, \text{N} \]Sum of horizontal forces:
\[ F_f - R = 0 \quad \Rightarrow \quad F_f = R \]Sum of moments about base:
\[ R \times 5 \sin 60^\circ - W \times 2.5 \cos 60^\circ = 0 \] \[ R \times 4.33 - 100 \times 1.25 = 0 \] \[ R = \frac{125}{4.33} = 28.87 \, \text{N} \]Step 4: Frictional force at base:
\[ F_f = R = 28.87 \, \text{N} \]Answer: The frictional force preventing slipping is approximately 28.9 N.
Step 1: Draw the Free Body Diagram showing the sign weight acting downward and tensions \( T_1 \) and \( T_2 \) in cables at 30° and 45° respectively.
Step 2: Resolve tensions into horizontal and vertical components.
Horizontal equilibrium:
\[ T_1 \cos 30^\circ = T_2 \cos 45^\circ \]Vertical equilibrium:
\[ T_1 \sin 30^\circ + T_2 \sin 45^\circ = 200 \]Step 3: From horizontal equilibrium:
\[ T_1 = T_2 \frac{\cos 45^\circ}{\cos 30^\circ} = T_2 \frac{0.707}{0.866} = 0.816 T_2 \]Step 4: Substitute in vertical equilibrium:
\[ 0.816 T_2 \times 0.5 + T_2 \times 0.707 = 200 \] \[ 0.408 T_2 + 0.707 T_2 = 200 \quad \Rightarrow \quad 1.115 T_2 = 200 \] \[ T_2 = \frac{200}{1.115} = 179.37 \, \text{N} \]Step 5: Calculate \( T_1 \):
\[ T_1 = 0.816 \times 179.37 = 146.4 \, \text{N} \]Answer: Tension in cable 1 is 146.4 N, and in cable 2 is 179.4 N.
When to use: Before applying equilibrium equations to any problem
When to use: When multiple unknowns are present to simplify calculations
When to use: For symmetric beams, trusses, or loading conditions
When to use: Throughout calculations to avoid errors
When to use: While summing moments to simplify equations
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