Structural Members using Bow’s Notation and Graphical Methods

TASK 1
1.0 Structural members using Bow’s notation and 1.1 graphical methods

(a) Magnitude of reactions x and y(nature is shown on the graph)
Rx = 12KN which is compressive
Ry = 12KN which is tensile
(b) Magnitude and nature of each force acting in each member of the structure
A2 = 1m × 12 = 12KN (R.Hibbeler, 2016)
B2 = 0.72m × 12 = 8.64KN
C1 = 0.75m × 12 = 9KN
D1 = 0.5m × 12 = 6KN
E1 = 0.88m × 12 = 10.56KN
1.2 Analytical method
(a) Nature and magnitude of reaction at x and y
(b) Magnitude and nature of forces acting in each member of structure
We consider the member to be in equilibrium
So FMN = 12 sin 45=8.48 KN
Then FMY = 12 sin 90=12 KN
Thus reaction at Y(RY) = FMY = 12KN and the nature is tensile (D.Ibrahim, 2013)
Considering joint Y
Y
60° FMY
RYX FNY
So FMY = 12KN
FNY = 12 sin 60=10.392 KN
FYX = 12 cos 60=6 KN
Taking joint Y
FYN FMN
60° 75°
FXN
FYN = 10.392KN
FXN = 10.392sin60 = 8.99KN

For joint x
FXY
FXN
Since joint is in equilibrium
So FXY = FXNsin90 = 10.392KN
1.3 Comparison
Some of values obtained both from the graph and analysis are the same while others differ a little
where by graph values are slightly higher than those done through mathematical analysis.
TASK 2
Data
L = 1.5M
D = 120mm = 0.12m
d = 100mm = 0.1m
F = 50KN = 50000N
T = 20°C
E = 200 ×109N/m2
α = 14×10-6°C-1
Where L-Length,D-External diameter,d-internal diameter,F-Compressive load,T-Temperature,E-
Modulus of elasticity, α- coefficient of linear expansion
(a) The initial stress(6i)
6i = F
A where A is area (D.H.Jama, 2014)
So A= π D2
4 − π d2
4 = π 0.122
4 − π 0.12
4 =3.4558× 10−3 m2
Thus 6i = F
A = 50000 N
3.4558× 10−3 m2 =14468631.19 N /m2