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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-Yo

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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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