A palindrome is a number or a text phrase that reads the same backwards as forwards. For example, each of the following five-digit integers is a palindrome: 12321, 55555, 45554 and 11611. Write a C++ program that reads in a five-digit integer and determines whether it is a palindrome .

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 Step 1

Assumption

• Horizontal shear acting on the inner column is equal to the double of horizontal shear on outer columns.
• Shear is acting at midpoint of each member.

Shear forces assumed on each member

IJ=R

JK=S

KL=T

IE=P

JF=2P

KG=2P

LH=P

EF=U

FG=V

GH=W

EA=Q

BF=2Q

CG=2Q

DH=Q

Step 2

Horizontal shear

For storey first

$$\begin{gathered} \mathrm{P }+ \mathrm{2 }\mathrm{P }+ \mathrm{2 }\mathrm{P }+ \mathrm{P }= \mathrm{20 } \\ \mathrm{6 }\mathrm{P }= \mathrm{20 } \\ \mathrm{P }= \mathrm{3 }. \mathrm{33 }\mathrm{k }N \end{gathered}$$

For bottom storey

$$\begin{gathered} \mathrm{Q }+ \mathrm{2 }\mathrm{Q }+ \mathrm{2 }\mathrm{Q }+ \mathrm{Q }= \mathrm{40 }+ \mathrm{20 } \\ \mathrm{6 }\mathrm{Q }= \mathrm{60 } \\ \mathrm{Q }= \mathrm{10 }\mathrm{k }N \end{gathered}$$

Taking joint I

$$\begin{gathered} \underset{}{\overset{}{\sum }}{\mathrm{M }}_{\mathrm{I }}= \mathrm{0 } \\ \mathrm{R }\times \mathrm{4 }- \mathrm{P }\times \mathrm{2 }+ \mathrm{50 }\times \mathrm{0 }= \mathrm{0 } \\ \mathrm{R }= \frac{\mathrm{3 }. \mathrm{33 }\times \mathrm{2 }}{\mathrm{4 }} \\ \mathrm{R }= \mathrm{1 }. \mathrm{665 }\mathrm{k }N \end{gathered}$$

Taking joint J

$$\begin{gathered} \underset{}{\overset{}{\sum }}{\mathrm{M }}_{\mathrm{I }}= \mathrm{0 } \\ - \mathrm{S }\times \mathrm{3 }+ \mathrm{2 }\mathrm{P }\times \mathrm{2 }- \mathrm{R }\times \mathrm{4 }= \mathrm{0 } \\ - \mathrm{S }\times \mathrm{4 }+ \mathrm{2 }\times \mathrm{3 }. \mathrm{33 }\times \mathrm{2 }- \mathrm{1 }. \mathrm{665 }\times \mathrm{4 }= \mathrm{0 } \\ \mathrm{S }= \mathrm{2 }. \mathrm{22 }\mathrm{k }N \end{gathered}$$

Taking joint K

$$\begin{gathered} \underset{}{\overset{}{\sum }}{\mathrm{M }}_{\mathrm{I }}= \mathrm{0 } \\ - \mathrm{S }\times \mathrm{3 }+ \mathrm{2 }\mathrm{P }\times \mathrm{2 }- \mathrm{T }\times \mathrm{4 }= \mathrm{0 } \\ - \mathrm{2 }. \mathrm{22 }\times \mathrm{3 }+ \mathrm{2 }\times \mathrm{3 }. \mathrm{33 }\times \mathrm{2 }- \mathrm{T }\times \mathrm{4 }= \mathrm{0 } \\ \mathrm{T }= \mathrm{1 }. \mathrm{665 }\mathrm{k }N \end{gathered}$$

For bottom storey

Taking joint E

$$\begin{gathered} \underset{}{\overset{}{\sum }}\mathrm{M }= \mathrm{0 } \\ \mathrm{3 }\times \mathrm{Q }+ \mathrm{2 }\times \mathrm{P }- \mathrm{4 }\times \mathrm{U }= \mathrm{0 } \\ \mathrm{3 }\times \mathrm{10 }+ \mathrm{2 }\times \mathrm{3 }. \mathrm{33 }- \mathrm{4 }\times \mathrm{U }= \mathrm{0 } \\ \mathrm{U }= \mathrm{9 }. \mathrm{165 }\mathrm{k }N \end{gathered}$$

Taking joint F

$$\begin{gathered} \underset{}{\overset{}{\sum }}{\mathrm{M }}_{\mathrm{F }}= \mathrm{0 } \\ \mathrm{3 }\times \mathrm{2 }\mathrm{Q }+ \mathrm{2 }\times \mathrm{2 }\mathrm{P }- \mathrm{4 }\times \mathrm{U }- \mathrm{3 }\times \mathrm{V }= \mathrm{0 } \\ \mathrm{3 }\times \mathrm{2 }\times \mathrm{10 }+ \mathrm{2 }\times \mathrm{2 }\times \mathrm{3 }. \mathrm{33 }- \mathrm{4 }\times \mathrm{9 }. \mathrm{165 }- \mathrm{3 }\times \mathrm{V }= \mathrm{0 } \\ \mathrm{V }= \mathrm{12 }. \mathrm{22 }\mathrm{k }N \end{gathered}$$

Taking joint G

$$\begin{gathered} \underset{}{\overset{}{\sum }}{\mathrm{M }}_{\mathrm{F }}= \mathrm{0 } \\ \mathrm{3 }\times \mathrm{2 }\mathrm{Q }+ \mathrm{2 }\times \mathrm{2 }\mathrm{P }- \mathrm{4 }\times \mathrm{W }- \mathrm{3 }\times \mathrm{V }= \mathrm{0 } \\ \mathrm{3 }\times \mathrm{2 }\times \mathrm{10 }+ \mathrm{2 }\times \mathrm{2 }\times \mathrm{3 }. \mathrm{33 }- \mathrm{4 }\times \mathrm{W }- \mathrm{3 }\times \mathrm{12 }. \mathrm{22 }= \mathrm{0 } \\ \mathrm{W }= \mathrm{9 }. \mathrm{165 }\mathrm{k }N \end{gathered}$$

Shear forces

P= 3.33 kN

Q= 10 kN

R= 1.665 kN

S= 2.22 kN

T= 1.665 kN

U= 9.165 kN

V= 12.22 kN

W= 9.165 kN

Step 3

Axial force calculations

In column AEI

Axial force at AEI=10.33 kN

In column BFJ

Axial force at BFJ=-2.5 kN (Upward direction)

In column CGK

Axial force at CGK= 2.5 kN

In column DHL

Axial force at DHL=8 kN