Math Problem Statement

determine the support reactions from this image.

Determine the support reactions from this image.

Solution

To solve for the support reactions in this problem, we need to identify and analyze the given forces and supports. Here's a breakdown of the situation based on the provided image:

Known Data:

  • Uniformly distributed load (UDL) of 15 kN/m over a span of 4 meters starting from point A.
  • Concentrated load of 20 kN acting at point D.
  • The beam is supported at two points:
    • Pin support at point A (fixed support allowing rotation but no vertical/horizontal displacement).
    • Roller support at point B (allowing vertical displacement but no horizontal displacement).

Geometry:

  • Total span of the beam is 10 meters, divided as follows:
    • 4 meters from A to C (under the UDL).
    • 4 meters from C to B.
    • 2 meters from B to D.

Step-by-Step Solution:

1. Reactions at Supports

Let’s denote the following reactions:

  • RAR_A = Vertical reaction at support A.
  • RBR_B = Vertical reaction at support B.

Since point A is a pin support, it will also have a horizontal reaction HAH_A, but because there are no horizontal forces acting on the beam, HAH_A will be zero.

2. Equivalent Loads

  • UDL (15 kN/m over 4 m): This is equivalent to a single concentrated load of: WUDL=15kN/m×4m=60kNW_{\text{UDL}} = 15 \, \text{kN/m} \times 4 \, \text{m} = 60 \, \text{kN} acting at the center of the UDL, which is 2 meters from point A.

3. Moment Equation (Taking Moments about Point A)

To solve for RBR_B, we use the equilibrium condition that the sum of moments about point A is zero: MA=0\sum M_A = 0 Taking counterclockwise moments as positive: (60kN)×2m+(20kN)×10mRB×8m=0(60 \, \text{kN}) \times 2 \, \text{m} + (20 \, \text{kN}) \times 10 \, \text{m} - R_B \times 8 \, \text{m} = 0 Solving for RBR_B: 120+2008RB=0120 + 200 - 8R_B = 0 RB=3208=40kNR_B = \frac{320}{8} = 40 \, \text{kN}

4. Vertical Force Equilibrium

Now, using the vertical force equilibrium condition: Fy=0\sum F_y = 0 RA+RB=60kN+20kNR_A + R_B = 60 \, \text{kN} + 20 \, \text{kN} Substitute RB=40kNR_B = 40 \, \text{kN}: RA+40=80R_A + 40 = 80 RA=8040=40kNR_A = 80 - 40 = 40 \, \text{kN}

Final Reactions:

  • RA=40kNR_A = 40 \, \text{kN}
  • RB=40kNR_B = 40 \, \text{kN}

These are the vertical reactions at supports A and B.

Do you want more details or have any questions?

Follow-Up Questions:

  1. What is the horizontal reaction at point A?
  2. How would the solution change if the concentrated load were moved to the center of the beam?
  3. What are the internal shear forces at points C and B?
  4. How would you calculate the bending moment at point B?
  5. What would happen if an additional support were added at point D?

Tip:

When dealing with uniformly distributed loads, converting them to point loads makes moment calculations easier!

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Math Problem Analysis

Mathematical Concepts

Statics
Equilibrium of Forces
Moment Calculation

Formulas

Sum of moments about a point: ΣM = 0
Vertical force equilibrium: ΣFy = 0
Equivalent load for UDL: W = UDL × length

Theorems

Equilibrium condition: A system in static equilibrium must satisfy ΣF = 0 and ΣM = 0.

Suitable Grade Level

Undergraduate Civil/Mechanical Engineering