Mechanics of Structures 2 Assignment

This article provides step-by-step solutions for Mechanics of Structures 2 Assignment. It includes finding reactions, maximum tensile bending stress, maximum shearing stresses, and analyzing the top half of the Universal Section. The article also includes a reference for further reading. Suitable for college students.

Mechanics of Structures 2 Assignment
Y
X
8 metres
3 metres
Uniform Load 8kN/m
2 metres 3 metres
40kN
Student Section Allocation Table
Student Onesteel Sections UB or UC
Shehryaar Saeed 1414631 200 UB 18.2
For the beam assigned to you solve:
a) Find the reactions (10 points)
Solution;
Consider the vertical equilibrium of at both ends say A & B
∑Fy=0, i.e the total sum of both upward and downward forces,
We take reaction at both ends to be RA & RB respectively.
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Upward force = downward
RA + RB = UDL + point load,
RA + RB = (8 x 3 + 40)
In order to get the reactions at both ends we have to take moments at one end.
Taking moments at point A we have,
∑MA=0,
Please note that the UDL acts at the centre of its span i.e 3/2=1.5
∑MA= (8 x 3x 1.5) + (40 x 5) – (8 x RB) =0
Rearranging we get,
8 x RB = 236KNM
Dividing we get, RB= 29.5KN
RA + 29.5KN= (8 x 3 + 40)
RA= 64-29.5
RA= 34.5KN,
Hence reactions at both ends are 34.5KN & 29.5KN respectively.
e) Determine the maximum tensile bending stress and indicate its location in the section (15
points)
Solution;
To find the maximum tensile bending stress, we divide maximum bending moment by the section
modulus.
6max=Mmax/S
Mmax= maximum bending moment.
S = section modulus
S = bh2/6 = 0.099 x (0.198)2/6=6.4687x 10-4m3
b=99 mm, h=198mm
At point x from point A we get the maximum moment = 5m,
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∑Mmax= (34.5 x 5) – (8 x 3 x3.5) = 88.5KNM
6max=88.5/ 6.4687x 10-4= 1.368 x 105MPa
The maximum tensile bending stress is 5m from point A. The maximum value is 1.368 x 105MPa
f) Calculate the maximum shearing stresses in the cross-section and draw a profile of these
shearing stresses showing their magnitudes at the critical points (20 points)
ﺡ=VQ/It
Where;
V= maximum shear force.
Q= 1st moment of area
I= centroid moment of inertia
T=thickness
ﺡ max=V(bh2-bhw2 +twhw2)/8Ictw
Where Ic= (bh3- bhw3+tw hw3)/12
Ic= ((99x 1983) - (99x 1843) + (4.5 x 1843))/12=1.4982x107mm4
Max shearforce=34.5KN,
ﺡ max= 34.5(99 x 1982- 99 x 1842 + 4.5 x 1842)/(8 x 1.4982x107x 184)
ﺡ max= 1.0666GPa
ﺡ min=Vb (h2-hw2)/8Ictw
ﺡ min=34.5 x99(1982-1842) /(8 x 1.498 x 107 x 184)
ﺡ min=0.8284GPa
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g) Analyse the top half of the Universal Section allocated to you, so it becomes a T- Section, and
use the 1st moment of area method, y=∑ y . dA
ATotal
to find the Centroid of the T- Section relative
to axis x-x. (10 points)
Itotal= ∑ (Ii +Ai di2)
Where;
Ii = moment of inertia of the individual segment about its own centroid axis.
Ai = area of individual segment,
d i= the vertical distance from the centroid.
Ii= bh3/12
Where;
b=base of rectangle,
h=height.
Therefore
Segment 1
I1= 99 x (7)3/12=2.8298x 103 mm4
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A1= 99 x 7= 693mm2
d1= 184+ 7/2 = 187.5mm
Segment 2
I2= 4.5x (184)3/12=2.336 x 106mm4
A2= 184 x 4.5= 828mm2
d2 = 184/2 =92mm
Reference
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Majid, T. A. (2014). Theory of Structures. Penerbit USM. Available from:
http://www.myilibrary.com?id=678949. (Date of Access: 7th May 2018).
Spofford, C. M. (2013). The theory of structures.
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