ENGG150 Mechanical Engineering Assignment 2: Detailed Solution

This document presents a detailed solution to a mechanical engineering assignment, likely for the ENGG150 course. The solution covers several key areas of mechanics, including the calculation of resultant forces and moments, analysis of static equilibrium, and truss analysis. The assignment involves calculating forces and moments using equilibrium conditions, determining forces in truss members, and applying concepts of statics. The solution includes step-by-step calculations, diagrams, and the application of relevant formulas. The document also includes a bibliography of resources used to solve the assignment.

Mechanical 1
ENGG150 ASSIGNMENT 2 (MECHANCIAL)
Student’s Full Name
Student Number
Tutor’s Name
Institution

Mechanical 2
Question 1
Let MA and FR be the moment at point A and resultant force1
Use equilibrium condition.
∑ Fx = -30sin 30o + (26 × 5
13 )
= - 15 + 10
= - 5 kN
Also total forces in y-direction is 0
∑Fy = - (30 cos 30o ) - (26 × 12
13 )
= - 25.9807 – 24
= - 49.9807kN
Calculating the resultant forces
FR =√ ¿ ¿
=√ ¿ ¿
= 50.2302
Getting the angle
tan θ = ∑ Fy
∑ Fx
= (−49.9807
−5 )
Θ = tan-1 (−49.9807
−5 )
1 Alys Holden et al, Structural Design For The Stage (CRC Press, 2nd ed, 2015)

Mechanical 3
θ = 84.28720
Getting moment at point A2
MA = (30 × sin 300 × 0.3) – (30 cos300 × 2) - 45 – (26 × 12
13 × 6) - (26 × 5
13 × 0.3)
= 4.5 – 51.96 – 45 - 144 – 3
= -239.46 kNm
Question 2
a)
b)
∑ Mc = 0 3
(50)( 3
5 ¿(4) + (50)( 4
5 ¿(0.2) – (1000×9.81×10-3 kN)(x) = 0
x = 13.04m
∑ Fx = 0
Cx – (50) ( 4
5 ¿ = 0
Cx = 40 kN
∑ Fy = 0
-Cy + (50) ( 3
5 ¿ – (1000×9.81×10-3) = 0
2 Alys Holden et al, Structural Design For The Stage (CRC Press, 2nd ed, 2015)
3 R. C. Hibbeler, Statics And Mechanics Of Materials (Pearson, 5th ed, 2018)

Mechanical 4
Cy = 20.19 kN
Question 3
a) FBD of the truss is
b)
Using the static equilibrium condition4
∑ Fy = 0
Ro + Rc = 2P ------------------ (1)
∑ M A = 0
9 ×P – RC ×6 – RB ×3 = 0 ----------------- (2)
Using equations 1 and 2;
RB = 6 kN RC = 6 kN Hence ‘-‘ P= 6kN
4 R. C. Hibbeler, Statics And Mechanics Of Materials (Pearson, 5th ed, 2018)

Mechanical 5
c)
Getting the FE, BC and BE
∑ M B = 0
FFE × 3 + Rc ×3 - P×6 = 0
FFE = ( 6 ×6−6 ×3
3 )
FFE = 6 kN Tensile
∑ M E = 0
FBC ×3 + P×3 = 0
FBC = -6 kN and ‘-‘ P = 6 kN
‘-‘represents an assumed opposite direction5
FBC = 6 kN Compressive
∑ FY = 0
FBE cos 0 – RC + P = 0
FBE cos 0 = 0 ‘-‘ RC = P = 6 kN
FBE = 0
5 R. C. Hibbeler, Statics And Mechanics Of Materials (Pearson, 5th ed, 2018)

Mechanical 6
Bibliography
Hibbeler, R. C., Statics And Mechanics Of Materials (Pearson, 5th ed, 2018)
Holden, Alys et al, Structural Design For The Stage (CRC Press, 2nd ed, 2015)