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Question Number 157139 by chuxx last updated on 20/Oct/21
The figure given below is the section  of the reinforced concrete short column.  Calculate the stress in the concrete  and the stress in the steel,if the  axial load of 780KN is applied to  the column.    Young′s modulus of steel=210KN/mm^2   Young′s modulus of concrete=14KN/mm^2   diameter of the steel bars=20mm
$${The}\:{figure}\:{given}\:{below}\:{is}\:{the}\:{section} \\ $$$${of}\:{the}\:{reinforced}\:{concrete}\:{short}\:{column}. \\ $$$${Calculate}\:{the}\:{stress}\:{in}\:{the}\:{concrete} \\ $$$${and}\:{the}\:{stress}\:{in}\:{the}\:{steel},{if}\:{the} \\ $$$${axial}\:{load}\:{of}\:\mathrm{780}{KN}\:{is}\:{applied}\:{to} \\ $$$${the}\:{column}. \\ $$$$ \\ $$$${Young}'{s}\:{modulus}\:{of}\:{steel}=\mathrm{210}{KN}/{mm}^{\mathrm{2}} \\ $$$${Young}'{s}\:{modulus}\:{of}\:{concrete}=\mathrm{14}{KN}/{mm}^{\mathrm{2}} \\ $$$${diameter}\:{of}\:{the}\:{steel}\:{bars}=\mathrm{20}{mm} \\ $$
Commented by chuxx last updated on 20/Oct/21
Answered by mr W last updated on 20/Oct/21
F=780 KN  α=(Y_(steel) /Y_(concrete) )=((210)/(14))=15  cross−section a×b  n steel bars with diameter d  A_(total) =ab+(α−1)((nπd^2 )/4)  stress in concrete σ_(concrete) =(F/A_(total) )=(F/(ab+(α−1)((nπd^2 )/4)))  stress in steel σ_(steel) =ασ_(concrete) =((αF)/(ab+(α−1)((nπd^2 )/4)))
$${F}=\mathrm{780}\:{KN} \\ $$$$\alpha=\frac{{Y}_{{steel}} }{{Y}_{{concrete}} }=\frac{\mathrm{210}}{\mathrm{14}}=\mathrm{15} \\ $$$${cross}−{section}\:{a}×{b} \\ $$$${n}\:{steel}\:{bars}\:{with}\:{diameter}\:{d} \\ $$$${A}_{{total}} ={ab}+\left(\alpha−\mathrm{1}\right)\frac{{n}\pi{d}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${stress}\:{in}\:{concrete}\:\sigma_{{concrete}} =\frac{{F}}{{A}_{{total}} }=\frac{{F}}{{ab}+\left(\alpha−\mathrm{1}\right)\frac{{n}\pi{d}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$${stress}\:{in}\:{steel}\:\sigma_{{steel}} =\alpha\sigma_{{concrete}} =\frac{\alpha{F}}{{ab}+\left(\alpha−\mathrm{1}\right)\frac{{n}\pi{d}^{\mathrm{2}} }{\mathrm{4}}} \\ $$
Commented by chuxx last updated on 20/Oct/21
please recommend any textbook for  me. The ones I′ve seen arent   comprehensive=
$${please}\:{recommend}\:{any}\:{textbook}\:{for} \\ $$$${me}.\:{The}\:{ones}\:{I}'{ve}\:{seen}\:{arent}\: \\ $$$${comprehensive}= \\ $$
Commented by mr W last updated on 20/Oct/21
A_(total)  is the “cross−section” as if  all materials were concrete. since  in fact the steel is α times strong  as the concrete, in A_(total)  we replace  the steel area with a concrete area  which is equal to α times the steel  area.
$${A}_{{total}} \:{is}\:{the}\:“{cross}−{section}''\:{as}\:{if} \\ $$$${all}\:{materials}\:{were}\:{concrete}.\:{since} \\ $$$${in}\:{fact}\:{the}\:{steel}\:{is}\:\alpha\:{times}\:{strong} \\ $$$${as}\:{the}\:{concrete},\:{in}\:{A}_{{total}} \:{we}\:{replace} \\ $$$${the}\:{steel}\:{area}\:{with}\:{a}\:{concrete}\:{area} \\ $$$${which}\:{is}\:{equal}\:{to}\:\alpha\:{times}\:{the}\:{steel} \\ $$$${area}. \\ $$
Commented by chuxx last updated on 20/Oct/21
Thank you so much.  It′s been a long  time.
$${Thank}\:{you}\:{so}\:{much}.\:\:{It}'{s}\:{been}\:{a}\:{long} \\ $$$${time}. \\ $$
Commented by chuxx last updated on 20/Oct/21
Sir, while I was personally trying  to solve it before I got confused  and sent it here I got the   A_(concrete) = A_(rectangle)  − 6A_(circle)   where A_(circle) =region bounded by the  steel rods    how then did you come about    A_(total) =ab + (α−1)((nπd^2 )/4)?  Thanks in advance.
$${Sir},\:{while}\:{I}\:{was}\:{personally}\:{trying} \\ $$$${to}\:{solve}\:{it}\:{before}\:{I}\:{got}\:{confused} \\ $$$${and}\:{sent}\:{it}\:{here}\:{I}\:{got}\:{the}\: \\ $$$${A}_{{concrete}} =\:{A}_{{rectangle}} \:−\:\mathrm{6}{A}_{{circle}} \\ $$$${where}\:{A}_{{circle}} ={region}\:{bounded}\:{by}\:{the} \\ $$$${steel}\:{rods} \\ $$$$ \\ $$$${how}\:{then}\:{did}\:{you}\:{come}\:{about} \\ $$$$ \\ $$$${A}_{{total}} ={ab}\:+\:\left(\alpha−\mathrm{1}\right)\frac{{n}\pi{d}^{\mathrm{2}} }{\mathrm{4}}? \\ $$$${Thanks}\:{in}\:{advance}. \\ $$
Commented by mr W last updated on 20/Oct/21
A_(concrete) = A_(rectangle)  − 6A_(circle)   this is the section area of concrete.  but the steel bars also carry the load.  and the steel is α times as good as the  concrete with α=Young′s of steel /  Young′s of concrete. so the total  section area is  A_(total) =A_(concrete) +αA_(steel)   A_(total) =A_(concrete) +αnA_(circle)   A_(total) =A_(rectangle) −nA_(circle) +αnA_(circle)   A_(total) =A_(rectangle) +(α−1)nA_(circle)   A_(total) =ab+(α−1)((nπd^2 )/4)  this is very basic knowledge, better  try to read a text book.
$${A}_{{concrete}} =\:{A}_{{rectangle}} \:−\:\mathrm{6}{A}_{{circle}} \\ $$$${this}\:{is}\:{the}\:{section}\:{area}\:{of}\:{concrete}. \\ $$$${but}\:{the}\:{steel}\:{bars}\:{also}\:{carry}\:{the}\:{load}. \\ $$$${and}\:{the}\:{steel}\:{is}\:\alpha\:{times}\:{as}\:{good}\:{as}\:{the} \\ $$$${concrete}\:{with}\:\alpha={Young}'{s}\:{of}\:{steel}\:/ \\ $$$${Young}'{s}\:{of}\:{concrete}.\:{so}\:{the}\:{total} \\ $$$${section}\:{area}\:{is} \\ $$$${A}_{{total}} ={A}_{{concrete}} +\alpha{A}_{{steel}} \\ $$$${A}_{{total}} ={A}_{{concrete}} +\alpha{nA}_{{circle}} \\ $$$${A}_{{total}} ={A}_{{rectangle}} −{nA}_{{circle}} +\alpha{nA}_{{circle}} \\ $$$${A}_{{total}} ={A}_{{rectangle}} +\left(\alpha−\mathrm{1}\right){nA}_{{circle}} \\ $$$${A}_{{total}} ={ab}+\left(\alpha−\mathrm{1}\right)\frac{{n}\pi{d}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${this}\:{is}\:{very}\:{basic}\:{knowledge},\:{better} \\ $$$${try}\:{to}\:{read}\:{a}\:{text}\:{book}. \\ $$
Commented by chuxx last updated on 20/Oct/21
I′ll also like to ask sir, what is the  difference between the cross-section  and A_(total) ; I thought they were  supposed to be the same.
$${I}'{ll}\:{also}\:{like}\:{to}\:{ask}\:{sir},\:{what}\:{is}\:{the} \\ $$$${difference}\:{between}\:{the}\:{cross}-{section} \\ $$$${and}\:{A}_{{total}} ;\:{I}\:{thought}\:{they}\:{were} \\ $$$${supposed}\:{to}\:{be}\:{the}\:{same}. \\ $$
Commented by mr W last updated on 20/Oct/21
i can′t recomment you a text book,  since in fact i don′t read book either.  what i told you above is directly from  my own understanding of the physics.
$${i}\:{can}'{t}\:{recomment}\:{you}\:{a}\:{text}\:{book}, \\ $$$${since}\:{in}\:{fact}\:{i}\:{don}'{t}\:{read}\:{book}\:{either}. \\ $$$${what}\:{i}\:{told}\:{you}\:{above}\:{is}\:{directly}\:{from} \\ $$$${my}\:{own}\:{understanding}\:{of}\:{the}\:{physics}.\: \\ $$
Commented by chuxx last updated on 20/Oct/21
I am most grateful. Thank you for  always helping.
$${I}\:{am}\:{most}\:{grateful}.\:{Thank}\:{you}\:{for} \\ $$$${always}\:{helping}. \\ $$

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