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radius of gyration for unreinforced masonry pier
how do you calculate the radius of gyration for an unreinforced masonry pier 16"x16" plan dimensions?

(i/a)^0.5
you could also just pull it off the the ncma section properties for a 16" wall. i=((15.625)^4)/12; a=15.625^2
(i/a)^0.5 = 4.5105 in^4.
ncma says 16" fully grouted (or solid) units have an ravg of 4.51 in^4.
in plan, i placed two blocks on top of each other and calculated the ix for the section buy using the parallel axis therom and the given ix for an 8" hollow wall. came up with an rx of 25.67. area of 1 8x16 block is 39.9 in^2 and ix is 444.22 in^4
ix of pier is 2048.4 in^4, area is 79.8 in^2.
it is unreinforced and ungrouted? that doesn't sound very good. you should at least grout it solid. either way, if it is ungrouted, your r is still coming up too high. you forgot to take the square root of i/a. are you using 2-8" blocks to make the pier or are you using 1-16" block?
bldgdesign: you got too many mistakes in your calcs.
firstly, if it is fully grouted, two block area is an=244in^2, secondly, the ix of two block is ix=4967 in^4,
thirdly after ix/an, you need square root it to get r=4.52.
did anybody ever wonder why you would want to calculate radius of gyration for a masonry pier?
to determine axial load capacity?
is this a trick question?

i am checking someone elses drawings. they used two 8x16x8 blocks placed side by side which makes a square 16x16. these are trailer support piers, gravity load only. hollow, dry stacked and face bonded.
i did forget to hit the square root key on the calculator. didn't change much in my check, though. allowable stress dropped 1/2%. new r is 5.07.
yes this is to determine axial load capacity. the r is used in the slenderness reduction portion of the equation.

2009-09-15, 04:08 PM	#1
huangyhg 超级版主注册日期: 04-03 帖子: 18592 精华: 36 现金: 249466 标准币资产: 1080358888 标准币	radius of gyration for unreinforced masonry pier radius of gyration for unreinforced masonry pier how do you calculate the radius of gyration for an unreinforced masonry pier 16"x16" plan dimensions? (i/a)^0.5 you could also just pull it off the the ncma section properties for a 16" wall. i=((15.625)^4)/12; a=15.625^2 (i/a)^0.5 = 4.5105 in^4. ncma says 16" fully grouted (or solid) units have an ravg of 4.51 in^4. in plan, i placed two blocks on top of each other and calculated the ix for the section buy using the parallel axis therom and the given ix for an 8" hollow wall. came up with an rx of 25.67. area of 1 8x16 block is 39.9 in^2 and ix is 444.22 in^4 ix of pier is 2048.4 in^4, area is 79.8 in^2. it is unreinforced and ungrouted? that doesn't sound very good. you should at least grout it solid. either way, if it is ungrouted, your r is still coming up too high. you forgot to take the square root of i/a. are you using 2-8" blocks to make the pier or are you using 1-16" block? bldgdesign: you got too many mistakes in your calcs. firstly, if it is fully grouted, two block area is an=244in^2, secondly, the ix of two block is ix=4967 in^4, thirdly after ix/an, you need square root it to get r=4.52. did anybody ever wonder why you would want to calculate radius of gyration for a masonry pier? to determine axial load capacity? is this a trick question? i am checking someone elses drawings. they used two 8x16x8 blocks placed side by side which makes a square 16x16. these are trailer support piers, gravity load only. hollow, dry stacked and face bonded. i did forget to hit the square root key on the calculator. didn't change much in my check, though. allowable stress dropped 1/2%. new r is 5.07. yes this is to determine axial load capacity. the r is used in the slenderness reduction portion of the equation. __________________ 借用达朗贝尔的名言：前进吧，你会得到信心! [url="http://www.dimcax.com"]几何尺寸与公差标准[/url]