Continue to Site

Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

  • Congratulations GregLocock on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!

Weld Capcity Basics on Tension Member force 2

Status
Not open for further replies.

palk7 EIT

Structural
May 12, 2020
150
Hi,

In the bottom chord of a truss due to overstressed, it needs a plate reinforcement on the top side of the WT section for the force below in the image.

The new reinf. plate will be welded to the WT section for increased capacity. The tension force is 62 Kips from model. so while doing the weld length calculation.

1). Do we have to consider 62 kips and divide it bu the weld capacity of a particular size to get our weld length?

2). Do we have to consider 62+62 Kips as the tension force as its pulling apart in both ends, so sum up that force for both ends 124 Kips and divide by weld capacity to get our weld length

3). Since the weld direction is joined parallel to the direction of loading force can we ignore the weld design as per AISC Table J2.5 for fillet welds.

What is the realistic approach for the situation.

Weld_Capcity_cj1gdf.png


Thank you!
 
Replies continue below

Recommended for you

Or that 62 Kips will be divided by 2 = 31 Kips at each end to get a overall tension force of 62 Kips? If this is the case then it will be considering like a shear force of 31 Kips that needs to be resisted on one side and similarly 31 kips on other side and do the calc. as per the formula for weld design considering them as shear forces
 
First question: how are you getting that load into your plate? If it's through the existing WT, you're in trouble. Generally speaking, we consder the tension in a member such as a truss chord as being uniform along its length. So if it fails in the middle due to pure tension, it will also fail at the connections. If you're reinforcing it starting at x=1", you'll get a failure somewhere between X=0 and X=0.9999999"
 
PhamENG, its being considered as a built up section with the plate addition. The forces are through the typical truss format, I didn't show the entire truss with diagonals, verticals, top chord etc. As it was mostly concerned to the weld lengths. The one I showed is just in the middle part of the long truss section and those pinned supports are not actually the end columns they are the panel points where all the diagonals, verticals converge at one location so just considered them as pinned there. As the tension force is exceeding the allowable capacity at the middle section of the T section we are reinforcing only in the middle section and not throught the entire bottom chord length.
 
1. To transfer 62 kips using only longitudinal welds to the plate, the weld length will need to be 62 kips / weld strength in kips per inch. You likely have longitudinal welds on both sides of the plate so you can cut your weld length in half (unless you already accounted for this by doubling the weld strength). If you're also using transverse welds at the end of the plate along with the longitudinal welds, you'll need to use AISC 360-16, Section J2.4b(2), which accounts for the different strengths and stiffnesses of the welds in the two different directions.

2. A member that has 62 kips of tension in it will need to have a connection AT EACH END that can support 62 kips.

3. Tension or compression parallel to the weld axis per Table J2.5 can be neglected in weld design, but your fillet welds will transfer the tension force through shear, so you still must design them for this. See this thread if you want to understand this more, but the important thing is just that you cannot use this to skip out on designing your welds.

Is 62 kips the force in the plate or the force in the combined section? The weld only needs to be designed for the force that is going to be in the plate. I would recommend, though, that you design your welds to transfer the total tensile capacity of the plate regardless of what the actual force in the plate is expected to be in order to ensure that the plate will yield before the welds fail.

As phamENG mentioned, the plate has to be fully developed before any tension in the plate is required. This means the plate will have to extend beyond the panel points where the reinforcing is required, and the welded connection will have to be able to fully develop the plate before reaching the panel points. Between the panel points you can probably switch to intermittent welds.

Is this an existing truss that you are strengthening? If that's the case, any loads that are currently in the WT section (DL + construction LL) will remain exclusively in the WT section and any loads applied after strengthening will go into both the WT section and reinforcing plate.

Structural Engineering Software: Structural Engineering Videos:
 
The reinforcing plate must have a length equal to the length of the middle section plus weld length at each end. It cannot be on top of the bottom chord, because it would run into web members along its length.

palk7 said:
1). Do we have to consider 62 kips and divide it by the weld capacity of a particular size to get our weld length?

2). Do we have to consider 62+62 Kips as the tension force as its pulling apart in both ends, so sum up that force for both ends 124 Kips and divide by weld capacity to get our weld length

3). Since the weld direction is joined parallel to the direction of loading force can we ignore the weld design as per AISC Table J2.5 for fillet welds.

What is the realistic approach for the situation.

1. No, the plate will only carry a portion of the 62k.

2. Are you serious? The tension is 62k, part of which has to be carried by the reinforcing plate. You can't have tension in a member unless it's pulling apart at both ends; if only one end, it's not in equilibrium.

3. I'm not familiar with that table, so don't know what you want to ignore.

The realistic approach is to provide a plate on the bottom of the bottom chord, to determine what portion of the 62k tension must be carried by the plate and to provide the correct length of weld to each side of the flange of the tee.




BA
 
Agree, there will be end welds near the panel points continuous and the plate extends 12" beyond the panel point both sides. And in between intermittent welds spaced. Actually its combined, but I agree the plate will take up only a certain portion of the capacity. The max. capcity of the plate is 59 Kips. The force is 62 Kips, so just for ease considering the max. force as the weld capcity (conservative). So it would be 62 Kips divided by weld capcity per inch to get the overall weld length and then divide them accordingly at the ends as continous and the intermittent stitching?
 
Its just an example of 62 Kips like a rough number throwed in for example, it was more intended for the weld length calculation.
 
The weld length that you calculated should be for the continuous welds at each end of the plate, and it all has to be outside of the panel points (so if you're only extending the plate 12" past the panel points then your calculated weld length needs to be less than that). The intermittent welds between the panel points should not count towards that required weld length.

Structural Engineering Software: Structural Engineering Videos:
 
palk7 said:
Its just an example of 62 Kips like a rough number throwed in for example, it was more intended for the weld length calculation.

It is not quite as simple as that. Suppose the truss has 12 panels of length 'h', so the truss span is 12h. The bottom chord is a tee shape with factored tensile capacity of 100k.

Panels 1,2,3,10,11 and 12 are each assumed to have a factored bottom chord tension value less than 100k, so no remedial action is required for those panels.

The middle section, consisting of panels 4,5,6,7,8 and 9, have a factored bottom chord tension of 130,145,150,150,145 and 130k respectively. The deficit is 30, 45 and 50k for panels 4,5,6 and 9,8,7 respectively. A bottom reinforcing plate with factored tensile capacity of 50k is selected.

The weld must resist 30, 15 and 5k for panels 4, 5 and 6 respectively (same for panels 9, 8 and 7). If the weld size selected has a resistance of 2.5 k/", the weld lengths required are 12", 6" and 2", placed just outside the panel named, half on each side of the reinforcing plate. The length of plate would be 2(3h + 12/2 + c) = 6h + 12" + 2c, where c is clearance required.

BA
 
Palk7 EIT:
Note also, that you can cause some serious problems when welding on a fully loaded member or on an already overstressed or yielding member. The bottom chord should really be unloaded in some way to prevent these problems. Furthermore, unless this unloading happens, the new combined built-up member does not gain stresses, or element forces, or distribute stresses in the normal fashion for the whole built-up member. The original WT bottom chord member will continue to grow in tensile stress from its original high stress level, while the new reinforcing pl. force/stress level starts growing basically from zero. So, the original WT may start yielding or continue yielding while the new pl. starts picking up its share of the load, at a pretty low stress level.
 
I agree with dhengr. The existing truss must be shored until all welding is completed.

BA
 
Status
Not open for further replies.

Part and Inventory Search

Sponsor