Mechanics of Materials Assignment

Solutions
Q1)
a) Taking ∑ M B = 0
RA * 3.75 – [17.5*3.75 *3.75/2 + 13 * 1.25]
RA = 37.15 KN
Total load = 13 + 17.5 * 3.75 = 78.625 KN
RB = 78.625 – 37.15 = 41.48 KN
Maximum bending moment, shear force (SF) = 0
Therefore,
37.15 – (13 + 17.5 * x) = 0
17.5x = 24.15
X = 1.38 m
b)
37.15 * 1.38 – 17.5 * 1.382
2 = 34.6035 KN/m
Modulus
Z = M
Pt = 34.6035∗106
165
= 2.09718 * 105 mm3
= 209.718 cm3
Try IPN = 200
Having;
Z = 214 cm3
I = 2140 cm4
c) Deflection equation
Y = W 0 x
24 EI (L3−2 L x2 +x3)+ Pb
6 LEI ( l
b ( x−a ) 3 + ( l2−b2 ) x + x3 )
Y =
215.62∗10−18 x
24∗2.140∗107∗2∗1011 (3.753−2∗3.75 x2+ x3 )+ 13∗1.25
6∗3.75∗2.140∗107∗2∗1011 ( l
1.25 ( x−2.5 )3 + (3.752 −1.25
Y = 215.62∗10−18 x−30.6667∗10−18 x3+ 4.08878∗10−18 x4 +¿
Graphical presentation

0 0.5 1 1.5 2 2.5 3 3.5 4
-900
-800
-700
-600
-500
-400
-300
-200
-100
0
100
x in m
y * 10-18 in m
Maximum deflection = -800 *10-18 m
Q2) from the provided table
d = 450 mm
b = 170 mm
t = 24.3 mm
Fs = P
length
Total length = 2*170 + 450*2 * ½ = 790
Fs = 340∗103
790
= 430.38 N/mm
FT = P∗e∗d
2 Ixx
Ixx = 2*225*170 +
4503
12 ∗2
2

= 7.67025 * 106 mm4
FT = 340∗38∗450∗103
2∗7.67025∗106
= 378.997 N/mm
FR = [Fs2 + Fs2]1/2
= [430.382 + 378.9972]1/2
= 573.46 N/mm
b) strength of length = 0.7 * 3000 * 115
241,500 N/mm
Length required = 340∗103
241500
= 1.408 mm
Say 2 mm
The total length = 2 + 3000*2 = 6002 mm
Max permissible, Lmax = 6002
2
= 3001 mm
Q3) M = 1000 KN.m
Let the concrete be of class 30 N/mm2
Hence
Moment = 0.156*fcu*b*d2
1000 = 0.156 8 30 * b * 4002 * 10-6
B = 1335.47
Thake b = 1350 mm
b) we know that
yt = 1350∗400∗200+ 400∗800(400+400)
1350∗400+800∗400
= 423.26
I = 1
12∗1350∗4003 +1350∗400 ( 423.26−200 )2+ 1
12∗400∗8003+ 400∗800 ( 800−423.26 )2

= 9.6601 * 1010 mm4
Shear stress at the bottom of the flange
Area = 1350 * 400 = 540,000 mm2
C.G of the area = yt – 200
= 423.26 – 200 223.26 mm
Width of this level = 1350
Shear force = M/L
= 1000∗106
1200
= 833.333 KN
Bottom of flange = 833.333∗103
1350∗9.6601∗1010 ∗(5400∗223.6)
= 0.077 N/mm2
q at the same level in the web where is 400 mm
= 833.333∗103
400∗9.6601∗1010 ∗(5400∗223.6)
= 1.16 N/mm2