Beam Deflection and Stress Analysis with UDL Load Calculation

This assignment solution focuses on analyzing beam deflection and stress under a uniformly distributed load (UDL). It begins by determining the reactions at the supports (RA and RB) using static equilibrium equations. Shear force and bending moment diagrams are then developed to identify maximum bending moment locations. The solution calculates the center of gravity and moment of inertia for the beam's cross-section. Finally, deflection is calculated using the bending moment equation and integrating to find slope and deflection equations, with boundary conditions applied to determine integration constants and maximum deflection (Ymax) at a specific point (E). The final deflection value YE is calculated and presented.

Solution
Find RA and RB
MA = (0.5 x 0.1) x (0.008+0.1/2) + 1.5 x (0.06 +0.1 + 0.02)
= RB x 0.26
= 0.05 x (0.13) + 1.5 x 0.2 = RB x 0.26
RB = 1.17885 KN
RA + RB = 0.5x0.1 + 1.5
RA = 0.37115 KN
SF bet B & F = 1.117885 K
SF just left = 1.17885 – 1.5 = -0.3215 KN
SF will be between E & D = -0.32115KN
A
RA
B D E B
UDL
0.5 KN/m
0.08m 0.1m 0.02m 0.06m
0.26m
RB
A C D E B
-0.37115 -0.37115
1.17885
-0.32115 KN
1.17885
0.02963
0.064308 0.070731
A C D E B

SF just left of D = 1.17885- 1.5 – 0.5 x 0.1
SF just left of C = -0.37115 KN
Will be calculated up to A
Mmax = RB x 0.06 = 1.17885 x 0.06 = 0.070731 KN-m
MD = 0.064308 KNm
MC = 0.029603 KNm
MA = 0
1. Find center of gravity of section
2. As section is symmetrical about y-y axis CG lies as Y axes
3. Let Bottom of the beam is the axis of reference
Rectangle 1
A1 = 20 x 3 = 60 mm2
Y1 = 3/2 = 1.5 mm
Rectangle 2
A2 = 7 x 15 = 105 mm2
Y2 = 15/2 +3 = 10.5 mm
Dist. between CG from reference axes
y = (a1y1 + a2y2) / (a1 +a2) = 7.2273 mm
Moment of inertia about X-X for rectangle 1 Axis though CG net of X-X axes
3mm
CG2
CG
CG1
20mm
Y1
15mm
7mm
Y

= (20 X 33) /12 = 45 mm4
MI of rectangle 1 about
CG =I
= 45 + 60 x (5.7273) ^2
= 2013 .1179 mm4
Similarly MI of rectangle 2 about an axis through CG and parallel to XY Axes
= 1968.75 mm4
H = y2 y1 = 10.5 – 7.2273
= 3.2727 mm
MI about CG 2
MI of rectangle 1 about
CG =I
= 1968.75 + 105 x (3.2727)^2
= 3093.368 mm4
MI of whole section about XX axis
= 2013.1179 + 3093.3593
=5106.47 mm4
MI about YY axes
= bd^3 / 12 = 3x20^3 / 12
= 428.75 mm^4
I = (bd3)/12
X1X1
G1
X1X1
G1 + a1h2
I = (bd3)/12 = (7x15^3) /12
XX2
G2
X1X1
G2 + a1h2
I = 5106.4772 x 10-12 m4X1X1
CG
I = 2000 + 428.75 = 2428.75 mm4YY
CG

Since beam is like this
And
RA = 0.3711 KN , RB = 1.178846 KN
Consider of section at X from A
Mx = RA x – (0.5 x 0.1) (x – 0.13) – 1.5 (x – 0.13 -0.07)
But EI d2y/dx2 = -Mx
= -0.3711x + 0.05 (x-0.13)+1.5 (x-0.2)
Ei dy/dx = - 0.3711x2/2 + 0.05(x-0.13) ^2 /2 + 1.5/2(x-0.2)^2 /2 +c1
EIy = -0.3711 x3/6 + 0.05(x-0.13)3/6 + 1.5(x -0.2)3 /6 + C1x + C2
Calculate I
& Given E of material
= 71 x 103 N/mm2 = 71 x 10^6 KN/m^2
A
RA
B D E B
UDL
0.5 KN/m
0.13m
0.1m 0.02m 0.06m
0.26m
RB
A C D B
1.17885
(X)
0.5 KN 1.5 KN
0.08m
0.07m
0.06m
x
x
x

Now at x = 0 , y= 0 , y = deflection
Put in equation 3 (c2 = 0)
At x = 0.26, y = 0, put in equation 3
0 = -03711(0.26)^3/6 + 0.05(0.26-0.15)^3 /6 + 1.5 (0.26 -0.2) /6 + C1(0.26)
C1 = 0.003903
Now
EIdy/dx = -0.3711/2 x2 + 0.0/2(x -0.13)2 + 1.5/2 (x – 0.2)2 + 0.003903
EIy =( -0.3711/6) x3 +( 0.05(x-0.13)^3)/6 + 1.5(x-0.2)^3 /6 + 0.003903x
Calculate I for given cross section
I = 5106 .4772 x 10-12 m4
Find x by SFD and BMD method
Ymax = Max Deflection and x
X= 0.2m which is E point
EIy = -4.948 x 10-4 + 2.8583 10-6 + 0 + 7.806 x 10-4 = 2.888583 x 10-4
YE = 7.9616724 x 10-4 m = 0.76167 mm