Engineering Mathematics 1

This document provides solutions for various problems in Engineering Mathematics 1. It covers topics such as horizontal reaction, maximum moment, uniform strain component, curvature, vertical displacement, relative rotation, and more.

Engineering Mathematics 1
Engineering Mathematics
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Engineering Mathematics 2
1. Horizontal reaction at A
First, we draw the free body diagram as follows.
Take the moment about C from the left side to be zero.
a RA −h H A −w × h× h
2 =0
5.5 RA −4.1 H A−18.1× 4.1× 4.1
2 =0
5.5 RA −4.1 H A=152.1305… equation 1
Take the moment about point D to be zero.

Engineering Mathematics 3
10 RA −h H A−w ×h × h
2 =0
10 RA −4.1 H A −18.1× 4.1× 4.1
2 =0
10 RA −4.1 H A =152.1305… equation2
Solving equations 1 and 2 we get:
5.5 RA −4.1 H A=152.1305
10 RA −4.1 H A =152.1305
RA =0 , H A =152.1305
−4.1 =−37.105 kN
Therefore, the horizontal reaction at point A is −37.105 kN
2. Maximum moment in AB
The free body diagram given the horizontal reaction becomes:

Engineering Mathematics 4
M xx =37.105 x − w x2
2 =37.105 x−9.05 x2
Differentiating the equation with respect to x yields:
d M xx
dx =37.105−18.1 x
For maximum moment, d M xx
dx =0
d M xx
dx =37.105−18.1 x=0
18.1 x=37.105
x= 37.105
18.1 =2.05 m
Maximum moment∈ AB, Mxx =37.105 ( 2.05 )−9.05 ( 2.05 )2=38.03262 5 kNm
3. Uniform strain component ε T in member BCD
Uniform strain component , δT =α T ∆ T =α T ( ∆ T 2−∆T 1 )
¿ 0.00001 ( 12−−4 )=1.6× 10−4
Hence , theuniform strain component , δ T=1.6× 10−4
4. Curvature ϕT in member BCD
ϕT = 1
lBCD α T ∆ T = 1
( a+b ) δT
¿ 1
( 5.5+ 4 ) 1.6 × 10−4

Engineering Mathematics 5
¿ 657.8947
5. Vertical displacement ∆E at point E
The vertical displacement is zero
6. Relative rotation rel φC at point E
Since point C is pin connected, rel φC= δ
c = 0
2.2 =0
Section 2
1. Case 1: Distributed load acting on member BC only
The free body diagram of member BC is shown below.

Engineering Mathematics 6
RB=RC = wb
2 = 11.5 ×5.7
2 =32.775 kN
The maximum moment in member BC is m= w b2
8 = 11.5 ×5.72
8 =46.704375 kNm
Strain energy stored in member BC is U = m2 b
2 EI = 46.7043752 ×5.7
2× 20625 =0.301416 kNm
From Castigliano’s theorem, δc= U
RC
=0.301416
32.775 =9.19652× 10−3 m
The horizontal displacement, δ c=9.19652 ×10−3 m
2. Case 2: Initial curvature in member AB only
¿ the given figure , tan θ= h
a , θ=tan−1
( h
a )=ta n−1
( 4.5
3 )=56.3099 °
The curvature ϕ AB=ϕ0 × I AB =ϕ0 × a
cos θ
¿ 4 ×10−4 × 3
cos 56.3099° =2.1633 ×10−3 rad m−1
The horizontal displacement due to curvature in member AB, δc H =ϕAB × sinθ
¿ 2.1633 ×10−3 ×sin 56.3099 °

Engineering Mathematics 7
δ c H =1.8 ×10−3 m
Case 3: Support settlement at point D
Support settlement , ∆ CδD=δD × b
h =0.019× 5.7
4.5 =24 . 067 ×10−3 m
Case 4
The free body diagram to calculate the support reactions is shown below.
The horizontal reactions at the supports will cancel out each other.
Now, considering the vertical forces,
A y+ D y=wb=11.5 × 5.7=65.55 kN
Considering the moment equilibrium at point D.
A y ( a+ b )− w b2
2 =0

Engineering Mathematics 8
A y ( 3+5.7 ) −11.5 ×5.72
2 =0
A y=11.5 ×5.72
2 × 1
(3+5.7) =21.4733 kN
But A y+D y=65.55 kN
so that , D y=65.55 kN− A y=65.55 kN −21.4733 kN =44.0767 kN
From the free body diagram above, F AB= A y
sinθ = 21.4733 kN
sin 56.3099 =25.8077 kN
Therefore, the force in member AB, F AB=25.8077 kN
Horizontal displacement at point C=δc−δc H +∆ CδD
¿ ( 9.19652−1.8+24.067 ) ×10−3 =31.46352×10−3
Therefore , theh orizontaldisplacement at point C=0.03146352 m