CM 2500 EXAM 2 – PART 2 PROBLEM REVIEW – STRUCTURAL PRINCIPLES, SHEAR AND MOMENT ANALYSIS FOR PROCTORED EXAMITY ASSESSMENT

SHEAR AND MOMENT ANALYSIS FOR

CM 2500 – Exam 2 (Part 2: Problems) Review

Module 5: Shear and Moment



Key Concept 1: Maximum Bending Moment

Problem 1: Simply Supported Beam with Uniform Load

A simply supported beam of length 8 m carries a uniform distributed load of 5 kN/m across its entire
span.

Find:

1. Reaction forces

2. Maximum bending moment

3. Location of maximum moment



Solution:

Step 1: Reactions

Total load:

W=wL=5×8=40 kNW = wL = 5 \times 8 = 40 \text{ kN}W=wL=5×8=40 kN

Since loading is symmetric:

RA=RB=20 kNR_A = R_B = 20 \text{ kN}RA=RB=20 kN



Step 2: Shear Function

At distance xxx from left support:

V(x)=20−5xV(x) = 20 - 5xV(x)=20−5x

, Step 3: Location of Max Moment

Maximum moment occurs where:

V(x)=0V(x) = 0V(x)=0 20−5x=0⇒x=4 m20 - 5x = 0 \Rightarrow x = 4 \text{ m}20−5x=0⇒x=4 m



Step 4: Moment Function

M(x)=20x−5x22M(x) = 20x - \frac{5x^2}{2}M(x)=20x−25x2

At x=4x = 4x=4:

Mmax=20(4)−5(4)22=80−40=40 kN\cdotpmM_{max} = 20(4) - \frac{5(4)^2}{2} = 80 - 40 = 40 \text{
kN·m}Mmax=20(4)−25(4)2=80−40=40 kN\cdotpm



Final Answers:

• Maximum Moment = 40 kN·m

• Location = Midspan (4 m)



Key Concept 2: Point Load on Beam

Problem 2: Off-Center Load

A simply supported beam (length = 10 m) carries a 15 kN point load located 3 m from the left support.

Find:

1. Reactions

2. Maximum bending moment



Solution:

Step 1: Reactions

Take moments about A:

RB⋅10=15⋅3⇒RB=4.5 kNR_B \cdot 10 = 15 \cdot 3 \Rightarrow R_B = 4.5 \text{ kN}RB⋅10=15⋅3⇒RB
=4.5 kN RA=15−4.5=10.5 kNR_A = 15 - 4.5 = 10.5 \text{ kN}RA=15−4.5=10.5 kN



Step 2: Maximum Moment Location