Beam strengthening 1

[andyfabian]

When strengthening steel beams, I have typically use cover plates, WT's, etc. Before the actual design is completed, the welded connection between the two members is easily determined based on the shear flow (VQ/I).

The problem is when I utilize two channels bolted to either side of the web of the W-section. When calculating Q (the first moment of inertia) the ybar (distance from neutral axis of new configuration to the centroid of the attached member) equals zero. Thus, I do not know what the required bolted attachment shall be.

I have a feeling my answer lies somewhere in the actual derivation of the shear flow equation utilizing good ole' calculus.

Thanks in advance.

 

[chicopee]

From the description of your problem it appears that the two channel cross sections occupy a space between the inside of either the upper or lower flanges of the W beam and extend along the web and end either above or below the centroid of the W channel. Both channels maintain a symmetrical cross section of the composite beam and the lengths of both channels run parallel to the length of the W beam. Is that right?

Consequently the centroid of the composite beam no longer matches the centroid of the W beam (obviously) but matches the centroid of the two channels. The new Ybar should be that of the composite cross section above or below the new centroid and that value should not be zero. Am I still on track with your description?

Why dont you weld the two new channels instead of bolting?
You'll get better stress distribution based on VQ/Ib.
If you use bolts calculate shearing resistance (Rs) of bolts(double shear), bearing resistance (Rb) against web, use lowest value of above to determine bolt pitch "e" = (Rs or Rb)*I/VQ.

 

[chicopee]

This is an addendum to my above response. I realized what you were stating w/ the Ybar of the channels as they are at the same levels of the the centroid of the composite beam. Therefore I would assume each channel to consist of two uneven angles and determine their respective centroids which will be the placement of the bolts( this is assuming the channels are deep enough to extend past the centroid of the composite beam). So you will have two Ybar's and two Q's from which you will determine two e's.

 

[trainguy]

That's a great question.

1) I have had the same problem in the past, and did not manage to be able to explicitly quantify the shear flow. There may be an approach if you look at the relative stiffness (Ix) of each of the 3 members of the built-up beam, but I cannot seem to take this any further.

2)Also, try to visuallize the probable load path:
i.e. what happens if you do NOT use any fastening?

3) One last grasp at the straws of structural mechanics:
If each member has, say equivalent stiffness (Ix), then don't you simply need enough fastening to transfer 1/3 of the applied vertical load (from the top flange of the W, I assume) to each reinforcing channel? I do not see much horizontal shear being transferred.

I wonder what the real experts think...

tg

 

[andyfabian]

chicopee and tg thanks for both of your input. got time for another?

i cannot find any design examples for splicing a wood joist that has been badly notched. i realize that replacing the joist is an option, but not in this case. can you point me in the right direction?

 

[FFIELD]

Assuming that the two channels run the full length of the existing beam and the centroid of the channels are at the same vertical elevation as the wide flange, then their is no horizontal shear. This is not a composit beam but rather three beams sharing a common load by relative stiffness conciderations. The force in the connecting bolts is a vertical force rather then a horizontal one. First you must find out how much each channel takes of the vertical load, this is the force in the bolts. The way I see it the channels take 2xI channel divided by the (I Beam + 2x I Channel)x the vertical load per foot. This number times the spacing of the bolts is the vertical force on the bolts in double shear. This also assumes the at the end of the channels enough bolts are added to transfer the vertical reaction load to the bearing point. Hopes this helps.

 

[jambel]

A beam is said to be in a state of pure bending when it is subjected to only bending moment and the shear force is zero. Because the shear force, V = 0 and recalling the expression dM/dx = V = 0, this implies that pure bending refers to flexure of a beam under a constant bending moment. In contrast, nonuniform bending refers to flexure in the presence of shear forces, i.e. the bending moment changes along the axis of the beam.
In deriving expressions for the pure bending of beams, the following assumptions are made:
(1) The beam is prismatic and has an axial plane of symmetry, the x-y plane.
(2) All applied loads act in the plane of symmetry, i.e. the x-y plane.
(3) Bending occurs only in the x-y plane, i.e. the deflection curve lies in the x-y plane
(4) Material of the beam is homogeneous.
(5) Material obeys Hooke's Law with Young's modulus of elasticity in in tension and compression being the same.
(6) Plane section before bending remain plane after bending.
Curvature of Beams
The effect of the transverse load on a beam is to cause the beam to deflect. The deflected shape of the longitudinal axis of the beam is known as the deflection curve. If normals are drawn to the tangents to the deflection curve at two different cross-sections of the beam, these normals will intersect at a point known as the centre of curvature for the deflection curve. The distance from the centre of curvature to the deflection curve is called the radius of curvature, ?. The reciprocal of the radius of curvature is defined as the curvature, K, i.e.

Consider the cantilever beam subjected to a point load at the end. Consider two x-sections initially at a distance dx prior to the application of the load. After the application of the load, the distance ?? the curve between the normals to the tangents to the deflection curve at the two section is ds. The angle dt is the small angle between the normals to the deflection curve at the two sections. From geometry, we have that rdt = ds which is approximately equal to dx.
Also observe that if ? is the angle between the normals at the ends of the beam,
Sign Convention for Curvature
If the coordinate system is chosen such that the x-axis is positive to the right and the y-axis is positive upwards, then the curvature is positive if the deflection curve is concave upwards. The curvature is negative if the deflection curve is concave downwards. Note that positive curvature is produced by positive bending moments and negative curvature by negative bending moments. However, observe that positive bending moments produce negative deflections while negative bending moments produce positive deflections.
Longitudinal Strains in Beams
Because the bending moment is constant (uniform) in a beam subjected to pure bending, the bending deformation will also be uniform and the deflection curve will be in the form of a circular arc. Consider the beam element shown below which is bent into a circular arc by the couples M0.

As a result of the bending deformation, fibres on the convex side are elongated (i.e. in tension) while those on the concave side are shortened slightly (i.e. in compression). Somewhere between the top and bottom of the beam, there is a layer of fibres which remain unchanged in length. This surface is called the neutral surface of the beam. The intersection of the neutral surface with the longitudinal/axial plane of symmetry (i.e. the x-y plane) is called the neutral axis of the beam. Its intersection with the plane of any x-section is called the neutral axis of that cross-section.

To obtain the normal strain, consider a longitudinal fiber ef that is located at a distance y from the neytral surface. Prior to the bending of the beam, ef = gh. After bending, however, ef shortens to ef'. Because gh lies on the neutral surface, it's length is unchanged. Hence,
Original length of ef = gh= rdq
Final length of ef = ef'= (r-y)dq
Change in length = final length - original length = ef' -ef
= (r-y)dq - rdq = -ydq

Normal Strain

This shows that the normal strains in the beam are proportional to the curvature k and vary linearly with the distance y from the neutral surface. Note that the expression above applies to beams with a positive curvature. For such beams, all fibres below the neutral surface have a positive y value and hence will have compressive strains. However, fibres above the neutral surface will have a negative y value and hence it will have normal tensile strains.

 

[trainguy]

We understand beam theory, but thanks for the refresher jambel. Do you have any comments on the original question?

tg

 

[AndyZ]

Expanding on FField, the there is no horizontal shear if the bolts are coincident with the cg in a section that is doubly symmetric (both horizontal and vertical axes). If the bolts are above or below the cg, then there will be horizontal shear. And you would use the method you always have.

Other than this clarification, I agree with FField's response.

 

[etompos]

FFIELD is correct. It is NOT a 'composite' beam if all members' ybars coincide. It is a 'load sharing' situation. The connection only needs to be designed to transfer sufficient vertical load between the members. Deflection compatibility must be maintained between members by proportioning the load based on the relative stiffnesses of the members. I design for this situation all the time. It is important to consider where the load is applied and how it will transfer to ALL members.

 

[ChipB]

I too, agree with FFIELD & ANDYZ.

The way I would find my load distribution is:
W=design load
V=Max Shear from design load
tw=Web thickness of W-section
tc=Web thickness of c-section
1) Calculate my deflection on the built up section.
2) Keeping the deflection calculated above constant, calculate what uniform load (W1) placed strictly on the W-beam, would produce the same deflection.
3) Required Vertical load transfer to channels would be W-W1
4) Calculate your shear flow AT YOUR FASTENERS. Q=Area of Steel above your bolt line * the distance from the centroid of the composite area TO the centroid of the Area of Steel above your bolt line. We'll call this V(sf)
5) Calculate horizontal shear transfer required. We'll call this V(hs). This is proportionate to the thickness of the members. Therefore, V(hs)=(2*tc)/(tc+tc+tw) * V(sf). This would be coming from the shear flow equation VQ/Ib where b is the thickness of the material.
6) Space bolts to transfer the vertical and horizontal forces. Remember V(res)=(V(horiz.)^2 +V(vert.)^2)^0.5. And, ughh, as you don't want any slippage between members, design and specify Slip Critical connection.

If the beam is long, I typically design & change the spacing in the middle third of it. Shear is less therefore spacing can be greater. I do the same for WT reinforcements.

If the beam to be modified is in place, I would shore it at mid-span to remove the existing deflection prior to making this modification.

Your question on a notched wood beam:

Always check the reduced capacity and compare to the loads. You may not have to do anything.

Check notch location in relation to nuetral axis.

Notch on the bottom:
I have used Simpson Ties across the notch if it is in the middle third of the joist.
If the notch falls near or at the end, I still do as above, but also build a "seat" at the top of the joist. I check this seat for its shear and bending capacity.

Notch in the top:
Typically not pretty. Try to turn a 2x member flat-wise between the notch and notch each end of it so it overlaps the joist. Size overlap and space nails to transfer compressive force in top of joist.
If this can't be done, try to frame it to the two adjacent joists. If near the end, usually not a problem to do this, but be careful at mid-span.

As I haven't really come across anyone else that does this around here, any thoughts on its validity?

Hope I've helped you
Chip

 

[jambel]

OK Trainguy this is very plain! but for people that are not a smart as you I have to state with:

Shear Stress and Bearing Stress
A force acting parallel or tangential to a section taken through a material (i.e. in the plane of the material) is called a shear force. The shear force intensity, i.e. shear force divided by the area over which it acts, is called the average shear stress t,
i.e. t =
where V = shear force;
A = cross-sectional area

Because the distributions of shear stresses, unlike normal stresses, is not uniform the expressions above more precisely defines the average shear stress over an area. When shear stress arises as a result of the direct action of forces trying to cut through a material, it is known as direct shear force. Shear stresses can also arise indirectly as a result of tension, torsion or bending of a member.

Depending on the type of connection, a connecting element (bolt, rivet, pin) may be subjected to single shear or double shear.

Sign Convention for Shear Stress, t acting on sections perpendicular to a coordinate axes
A shear stress acting on a positive face of an element is positive if it acts in the positive direction of one of the co-ordinate axes and negative if it acts in the negative direction of the axis.
A shear stress acting on a negative face of an element is positive if it acts in the negative direction of an axis and negative if it acts in the positive direction of the axis.


Observe that bolts, pins and rivets will create bearing stresses sb in the members they connect, as the connecting element bears against the members. The distance from the edge of the member to the connecting element must the enough to withstand the bearing stresses, otherwise the connection will fail in bearing. The average bearing stress sb is defined as the ratio of the bearing force F to the product of the bolt diameter d and the plate thickness t. Note that the bearing resistance of a connection is determined using the edge distance, e because that represents the likely failure path.


Consider the bolt in single shear
sb =
where sb = bearing stress; F = force on rivet = applied load, P; d = bolt diameter; and t = plate thickness

 

[boo1]

Andy
Guidelines For the Analysis of Drilled Holes or Notches in Structural Glued Laminated Timber Beams can be found at:

https://www.aitc-glulam.org/shopcart/details.asp?category=technotes&item=TN-19

 

[andyfabian]

guys....thanks for all of your input. not only have i received a refresher course in beam theory, i think i have the answer i needed for the original question.

and thanks boo1 for the website, but that website is for glued laminated beams. i am looking for a reference that will help me determine what is required to strengthen a badly notched member back to its original strength, not determine the strength of the member after being notched.

can anyone help? i realize replacement of the member is an option, but not in my case. sistering the members is the only option, but how far do i develop the sister joist past the notches? if this is not clear, i can elaborate.

 

[boo1]

Andy, the design criteria is material based, wood. Check the references at the end of the article.

also see:

http://www.fpl.fs.fed.us/documnts/FPLGTR/fplgtr113/Ch08.pdf

 

[FFIELD]

Further to my response on the shear in bolts. No matter where the bolts are, either above or below the common center line of the channels, the horizontal shear will be zero. Check the horizontal strain at any bolt location and you will see that it is the same for the channel and the beam. No change in strain, no horizontal shear in the bolts.

 

[chicopee]

To splice badly notched beams run two parallel beams of similar depth along the sides of the notched beams. Allow at least 3' to 6' of overlap over each notched ends of the beams. For nailing or bolting of the slices best best to talk to an experienced carpenter as to the pitches and staggers. This method of repair is often seen on heavy timber beams damaged by fire.