Solution for ENGG150 Assignment 2: Mechanical Engineering Problems

This document presents a detailed solution to ENGG150 Assignment 2 in Mechanical Engineering. The assignment covers three key problems: the first involves replacing a force and couple moment system acting on an overhang beam with a resultant force and couple moment at a point. The second problem focuses on analyzing a jib crane supported by a pin and rod, calculating reactions at the supports. The third problem requires the analysis of a truss structure, determining the forces in specific members (FE, BC, and BE) using static equilibrium conditions. The solution includes free body diagrams, calculations of forces and moments, and application of equilibrium equations to find the unknown quantities. The document also provides a bibliography of the resources used to solve the assignment problems.

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)