Math Problem Statement

The image depicts a cantilever beam subjected to two loads: 1. A point load of 1,000 lb at a distance of 2 ft from the fixed end. 2. A uniformly distributed load of 150 lb/ft over the last 6 ft of the beam. How do we calculate the reactions at the fixed end and the bending moment?

Solution

The image depicts a cantilever beam subjected to two loads:

  1. A point load of 1,000lb1,000 \, \text{lb} at a distance of 2 ft from the fixed end.
  2. A uniformly distributed load of 150lb/ft150 \, \text{lb/ft} over the last 6 ft of the beam.

Let me analyze the reactions and deflections of this cantilever beam step by step.

Step 1: Break Down the Loads

  • Point load (P): P=1,000lbP = 1,000 \, \text{lb} located 2 ft from the fixed end.
  • Uniform load (w): w=150lb/ftw = 150 \, \text{lb/ft} distributed over a length of L=6ftL = 6 \, \text{ft}.

Step 2: Calculate Resultant Forces

  • The resultant force due to the uniform load is: Fw=w×L=150lb/ft×6ft=900lbF_w = w \times L = 150 \, \text{lb/ft} \times 6 \, \text{ft} = 900 \, \text{lb}
  • The location of this resultant force is at the centroid of the uniformly distributed load, i.e., 3 ft from the start of the distributed load, which is a total of 2ft+3ft=5ft2 \, \text{ft} + 3 \, \text{ft} = 5 \, \text{ft} from the fixed end.

Step 3: Reactions at the Fixed End

For a cantilever beam, the fixed end will have a reaction force (R) and a reaction moment (M).

  1. Sum of vertical forces: R=P+Fw=1,000lb+900lb=1,900lbR = P + F_w = 1,000 \, \text{lb} + 900 \, \text{lb} = 1,900 \, \text{lb}

  2. Sum of moments about the fixed end: The moment due to the point load is: MP=P×2ft=1,000lb×2ft=2,000lb-ftM_P = P \times 2 \, \text{ft} = 1,000 \, \text{lb} \times 2 \, \text{ft} = 2,000 \, \text{lb-ft} The moment due to the uniform load is: Mw=Fw×5ft=900lb×5ft=4,500lb-ftM_w = F_w \times 5 \, \text{ft} = 900 \, \text{lb} \times 5 \, \text{ft} = 4,500 \, \text{lb-ft} Therefore, the total moment at the fixed end is: M=MP+Mw=2,000lb-ft+4,500lb-ft=6,500lb-ftM = M_P + M_w = 2,000 \, \text{lb-ft} + 4,500 \, \text{lb-ft} = 6,500 \, \text{lb-ft}

The reactions at the fixed end are:

  • Vertical reaction force R=1,900lbR = 1,900 \, \text{lb}
  • Moment reaction M=6,500lb-ftM = 6,500 \, \text{lb-ft}

Would you like to see details on the deflection, shear, and bending moment diagrams, or have any specific questions?

Related Questions:

  1. How would the beam's deflection profile change under these loads?
  2. What is the equation for the bending moment along the beam?
  3. How can we calculate the maximum shear force in this beam?
  4. What is the significance of the location of maximum bending moment?
  5. How would increasing the uniform load affect the reaction forces?

Tip:

When analyzing cantilever beams, always resolve all forces and moments at the fixed support to ensure the structure remains in equilibrium.

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Math Problem Analysis

Mathematical Concepts

Statics
Beam Theory
Equilibrium of Forces and Moments

Formulas

R = P + F_w
M_P = P × Distance
M_w = F_w × Distance
F_w = w × L

Theorems

Principle of Superposition
Equilibrium of Forces
Moment Calculation in Beams

Suitable Grade Level

University Engineering Level