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Stresses in a Bent Bar

Stresses in a Bent Bar

Stresses in a Bent Bar

(OP)
A quick design question....

Assume a steel bar is bent into a u-shape, and the two legs of the bar are embedded into concrete such that the yield strength of the bar can be developed (by means of cross bars welded to the legs of the bent bars).

Now if I pick up the concrete section with these u-bars, and if each u-bar has a certain load "x" - say 2000 lbs total load per location, then what is the maximum stress in the bar?

I have always assumed that the maximum stress occurs in the bend of the bar and is about 2X the total stess in each straight leg of the bar assembly.  (if the lifted weight is 2000 lbs per anchor, then the load in each leg of the bar is 1000 lbs and the load at the bend is equal to 2000 lbs, due to the geometry.)

Does anyone have any ways of checking the maximum stress in the bar?

RE: Stresses in a Bent Bar

These contraptions except failing through slippage I read in Fritz Leonhardt's fail where the curve joins the straight part. Hence axial stress plus bendiing locked effects plus friction must be causing the relevant and final crack starting there.

For a non failed state and whilst the bending keeps the bar in the elastic realm (which won't be for the smaller radiuses common in reinforced concrete reinforcement). Leonhardt gives the stress from bending to be

sigma=
diameter of the bar x E / Diameter of the coil to axis

Adding this (or the plastified stress properly derived) to the tensile give the initial status.

When you start to pull the buildup of stresses is bigger then at tangency point since inwards some bend friction must be forcing the allocation of axial deformation to a small longitude right there, where rupture happens.

Leonhardt says that under 20 bar diameter radius for the bent part the capacity is lesser than for the straight bar in about 2 or 3%.

RE: Stresses in a Bent Bar

(OP)
Thanks for your reply!

I assume that to bend the bar into the type of radius I need (a 5/8" diameter bar with a 5.5" outer diameter) the bar would need to yield and there would be the added bonus of now having residual stresses in the bar prior to loading it with the weight of the concrete piece.

Anyone have any info on how to account for the residual stresses that exist in the bar after bending?  In the end, I want to establish a safe working load for these bars so I am a bit lost right now!

KarlT

RE: Stresses in a Bent Bar

Will take a whole length of bar equal to diameter multiplied by pi, this D·pi is the initial common length of all fibers.

After bending, the cog fiber will still have such length. Bur not the outer and inner fibers. The outer, for example, will have (with some approximation) (D+d)·pi length, where d stands for the diameter of the bar. Hence, the total elongation at the outer fiber is d·pi, to be distributed along the total length of the cog circumference.

Hence, epsilon=d·pi/D·pi=d/D

for your numbers D=4.88 in, d=5/8 in

epsilon=0.13

since you attain yield at 0.002 strain you see that almost the entire depth of your bent bar is entirely yielded, and assuming so won't introduce but a very small error.

By the way 0.13 strain is one that only true ductile steels will see without rupture. For nonductile no heated whilst bending you may break them.

Then half of your bent bar is yielded in compression. This is true for the entire curved part but there needs be a transition zone with locked stresses of some kind towards the straight, unbent part, which has not been subject to plastic curvature.

Introduction of the load first will diminish the compressive stress whilst remodelling the locked stresses near the tangency point in the outer half to equilibrate part your pull with the tensile stresses already present there. The process continues by annullating the compression in the inner part then going itself to full yielded in tension.

And Leonhardt says that if well anchored say 97% of A·Fy is attained.

RE: Stresses in a Bent Bar

(OP)
Thanks again Ishvaaag!

I am bending a 15M grade 400W bar(400 MPa yield, weldable low carbon steel in Canada)into a u-shape.  The area of rebar steel per 15M bar is assumed to be 200 mm^2. (I know that it generally isn't a great idea to use rebar for lifting due to embrittlement, but we are using weldable steel so the carbon content should be lower)

Based on what you write above, the maximum ultimate load I can get out of one bent bar is equal to:

0.97*A*Fy*(phi factor) = 0.97*200*400*0.85/1000 = 66kN = 294 kips of load.

Does this mean that I can then pick up a load of 2*(294 kips)/(Appropriate Safey Factor), assuming the rebar is developed in the concrete enough to yield?

Out of curiousity, wouldn't the curve in the bar tend to straighten out as the load increased and form three plastic hinges at failure? (One at the middle of the bend where the load is applied, and one at the junction of the straight and curved portions of the bar)  Does the 0.97*A*Fy account for this?

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