Mechanics of Structures 2 Assignment

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.
tmpg9lbzi2p2326845_573939401_ORDER744551.docx

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,
tmpg9lbzi2p2326845_573939401_ORDER744551.docx

∑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
tmpg9lbzi2p2326845_573939401_ORDER744551.docx