Civil Engineering Assignment: Structural Analysis and Design

This document contains solutions to a civil engineering assignment, focusing on structural analysis and design principles. The solutions cover a range of problems, starting with the calculation of reactions, bending moments, and shear forces in a beam under various loads. The assignment then delves into the analysis of structural elements, including the determination of shear stress and the strength of materials. The solutions also include calculations for moment of inertia, deflection, and concrete design, with considerations for dimensions, material properties, and applied loads. The document provides step-by-step calculations and formulas to arrive at the final answers, making it a valuable resource for students studying civil engineering and structural analysis.

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