{"id":6019,"date":"2018-11-21T09:52:20","date_gmt":"2018-11-21T04:22:20","guid":{"rendered":"https:\/\/www.onlinecivilforum.com\/site\/?p=6019"},"modified":"2022-12-13T11:32:48","modified_gmt":"2022-12-13T06:02:48","slug":"bar-bending-schedule-of-column","status":"publish","type":"post","link":"https:\/\/onlinecivilforum.com\/site\/bar-bending-schedule-of-column\/","title":{"rendered":"Bar Bending Schedule of Column"},"content":{"rendered":"

Bar Bending Schedule of Column<\/strong><\/h2>\n

The calculation of bar bending schedule of column given below:<\/strong><\/p>\n

The height of the column is 4 m and having a cross-sectional area is 300 x 400 mm and having a 40 mm of clear cover.Six bars are going to use has a dia of 16 mm.The 8 mm dia of stirrups is going to use @ 150mm and @200 mm at L\/3.So calculate the cutting length bars and stirrups?<\/p>\n

\"Column<\/p>\n

\"buy<\/a><\/p>\n

Given Data:<\/strong><\/h3>\n

Height = 4 m<\/p>\n

Cross section = 300 x 400 mm<\/p>\n

Clear Cover = 40 mm<\/p>\n

No of vertical bar = 6 no\u2019s<\/p>\n

Diameter of vertical bar = 16 mm<\/p>\n

Diameter of Stirrup = 8 mm<\/p>\n

Stirrups center to center spacing = @150 or @ 200 mm<\/p>\n

BBS of Column = ?<\/p>\n

<\/p>\n

Solution:<\/strong><\/h3>\n

The Calculation was to proceed into two steps.
\n1.Vertical bar calculation<\/p>\n

2.Cutting length of stirrups<\/p>\n

Step 1: Vertical Bar Calculation<\/strong><\/h3>\n

Length of 1 bar = H + Ld<\/p>\n

Where<\/strong><\/p>\n

Ld = development length<\/p>\n

H = Height of column<\/p>\n

Length of 1 bar<\/strong><\/h3>\n

= 4000 mm + 40d (d is dia of the bar)<\/p>\n

= 4000 + 40 x 16<\/p>\n

= 4000 + 640<\/p>\n

= 4640 mm or 4.640 m<\/p>\n

So the length of 1 vertical bar is 4.640 m and we have a total 6 number of bars,<\/p>\n

Total length <\/strong><\/h3>\n

= 6 x 4.640<\/p>\n

= 27.84 m long vertical bar is required.<\/p>\n

Step 2: Cutting the length of stirrups<\/strong><\/h3>\n

Cross-sectional area column is 300 mm* 400mm<\/p>\n

A is vertical cross-section area of the stirrup<\/p>\n

B is a horizontal cross-section area of the stirrup<\/p>\n

<\/p>\n

Calculation of A<\/strong><\/h3>\n

= 300 \u2013 2 x clear cover<\/p>\n

= 300 \u2013 2 x 40<\/p>\n

= 300 \u2013 80<\/p>\n

= 220 mm<\/p>\n

Calculation of B<\/strong><\/h3>\n

= 400 \u2013 2 x clear cover<\/p>\n

= 400 \u2013 2x 40<\/p>\n

= 400 \u2013 80<\/p>\n

= 320 mm<\/p>\n

No of stirrups<\/p>\n

= 4000\/3<\/p>\n

= 1333.3mm or 1.33m<\/p>\n

<\/p>\n

Formula = L\/3 \/ spacing + 1<\/p>\n

(No of stirrups in end zone)<\/p>\n

= 1333.3 \/ 150<\/p>\n

= 8.8 nos say 9 nos<\/p>\n

There are total two zones of 150mm spacing and one zone of 200mm spacing<\/p>\n

= 2 x 9<\/p>\n

= 18 nos (at end zones)<\/p>\n

At Mid Zones<\/p>\n

= 1333.3 \/ 200<\/p>\n

= 6.6 nos say 7 nos<\/p>\n

Total no of stirrups<\/p>\n

= 18 + 7<\/p>\n

= 25 nos<\/p>\n

Cutting Length of one stirrup<\/p>\n

Formula:<\/strong><\/h3>\n

= (2 x A) + (2 x B) + hook \u2013 bend<\/p>\n

Cutting length<\/p>\n

= (2 x A) + (2 x B) + 2 x 10d \u2013 5 x 2d<\/p>\n

Where<\/strong><\/p>\n

Hook = 10d<\/p>\n

Bend = 5x 2d (because we have 5 bends in stirrup)<\/p>\n

d = dia of bar<\/p>\n

= (2 x 220) + (2 x 320) + 2 x 10 x 8 \u2013 2 x 5 x 8<\/p>\n

= 440 + 640 +160 – 80<\/p>\n

= 1160 mm or 1.16 m<\/p>\n

So we have a total 25 nos of stirrups are going to use,<\/p>\n

Total length<\/p>\n

= 25 x 1.16<\/p>\n

= 29m long 8 mm bar<\/p>\n","protected":false},"excerpt":{"rendered":"

Bar Bending Schedule of Column The calculation of bar bending schedule of column given below: The height of the column is 4 m and having a cross-sectional area is 300 x 400 mm and having a 40 mm of clear cover.Six bars are going to use has a dia of 16 mm.The 8 mm dia …<\/p>\n","protected":false},"author":2264,"featured_media":6023,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"two_page_speed":[],"footnotes":""},"categories":[58],"tags":[136,247],"class_list":["post-6019","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-structural","tag-bar-bending-schedule","tag-column"],"yoast_head":"\nBar Bending Schedule of Column<\/title>\n<meta name=\"description\" content=\"Bar Bending Schedule of Column The calculation of bar bending schedule of column given below\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link 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