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How to treat the under design poured concrete on shearwall? example: the designed strength of concrete (f’c) must be 7000 psi however it was poured by only f’c=5000psi?
Please design a G+1 building with an example.
Please provide a sufficient information about your query and mention specific data related to building design. Thankyou
Please provide a sufficient information about your query and mention specific data related to building design.
Thankyou
See lessWhich are the different types of welding processes in steel structures and which one is the most common and affordable?
The most commonly used and Affordable welding is Stud Welding because it is cost-efficient. The different types of Welding processes in steel structures are as follows:- 1. Stud Welding:- It involves the electric process of welding a fastener on a metal base by heating both the components with an arRead more
The most commonly used and Affordable welding is Stud Welding because it is cost-efficient.
The different types of Welding processes in steel structures are as follows:-
1. Stud Welding:- It involves the electric process of welding a fastener on a metal base by heating both the components with an arc.
2. Flux Core Welding:- this involves a consumable tubular electrode containing a flux and a constant voltage welding power supply. It’s A portable and fast welding process, and less-skilled workers can easily undergo this.
3. Shield metal arc welding (SMAW):- it involves a consumable fixed-length electrode covered with a flux, and an electric power source is used to weld two metals together. After the completion of this process, mineral coating flux Of electrode disintegrates and releases a gas Commonly known as Shielding gas, which saves joint from unfavorable atmospheric conditions.
See lessWhat is double R.C. beam?
The beam, which is reinforced from both zones i.e., compression and tension, that beam is said as Doubly Reinforced beam. It is usually given when the depth of the beam is constricted. It is obviously not safe to provide a singly r/f beam with min depth resulting in failure due to insufficient resisRead more
The beam, which is reinforced from both zones i.e., compression and tension, that beam is said as Doubly Reinforced beam. It is usually given when the depth of the beam is constricted. It is obviously not safe to provide a singly r/f beam with min depth resulting in failure due to insufficient resistance to oppose the bending moment.
Moment of resistance could not be increased by increasing the steel in the tension zone. Rather it can be increased by increasing the reinforcement, but it should not exceed 25% of the strained side. That’s why a doubly reinforced beam is provided to increase the moment of resistance of a beam having limited dimensions.
See lessbeam failing , how to fix or remedial solution
The beam should be analyzed to find out the cause of deflection, then decide on the repair technique. one solution could be increasing the size of the beam and adding extra reinforcement in order load carrying capacity of the beam, and increase flexural resistance to reduce deflection. Before applicRead more
The beam should be analyzed to find out the cause of deflection, then decide on the repair technique. one solution could be increasing the size of the beam and adding extra reinforcement in order load carrying capacity of the beam, and increase flexural resistance to reduce deflection. Before application of this repair technique, the beam should be released from certain amount of loads through the provision of jacks for the secondary beams. The new cross section with the added reinforcement ratio should be analyzed for compute its load carrying capacity.
See lessWhat is the procedure to design the double angle tension member in steel structures with formulas? Kindly elaborate on it?
As per Section 6 (Design of Tension Members) of IS 800:2007 – Code of Practice for Construction in Steel, The design strength of the tension member is the minimum of following, Design strength due to yielding of the gross section (Tdg) (Clause 6.2 of IS 800:2007). Design Strength Due to Rupture of CRead more
As per Section 6 (Design of Tension Members) of IS 800:2007 – Code of Practice for Construction in Steel,
The design strength of the tension member is the minimum of following,
Let’s take an example to understand the designing of double angle tension member:
Data Known:
Service Load, T = 200 N/mm2
STEP 1:
Factored Load, Tu = Tdg = 1.5 x 200 = 300 N/mm2
Considering the tension member fails due to yielding of gross section, determine the gross area of angles required.
Tdg = AgFy/ꙋm0 → Ag = Tdgꙋm0/Fy
Ag = (300 x 103 x 1.1)/250 = 1320 mm2
The total gross area of tension member (Ag) required is 1320 mm2. Remember this is the area of two angle sections. Therefore,
Gross area of single angle section (Ag1) = (Ag)/2 = 1320/2 = 660 mm2
From steel table (SP 6-1), choose an angle having gross area of single angle about 25% to 40% more than computed above.
Taking Rolled Steel Equal Angle ISA 60x60x8 having following properties,
Sectional Area, A= 896mm2
Total Gross Area, Ag0 = 2×896 = 1792 mm2 > 1320 mm2 (O.K)
STEP 2:
Designing Connections: – We can provide either bolted or welded connections, so let us provide bolted connections.
Total thickness of angles having outstanding legs placed back to back,
ta = 8+8 =16mm
Let us provide 20mm diameter bolts of grade 4.6 and Steel of grade Fe415,
Diameter of bolt, d = 20mm
Diameter of bolt hole, dh = 20 + 2 = 22mm (Table 19 of IS 800:2007)
Fu = 410 N/mm2
Fub = 400 N/mm2
Fy = 250 N/mm2
Kb = 0.606
= 2 x (Fub/√3) x (Anb/ꙋmb) where, Anb = 0.78 x (πd2/4)
= 2 x (400/√3) x ((0.78 x (π(20)2/4))/1.25)
= 90.545 KN
= (2.5 Kb d t Fub)/ ꙋmb
= (2.5 x 0.606 x 20 x 16 x 400)/1.25
= 155.136 KN
Therefore, Bolt Value = Least of (90.545, 155.136) = 90.545 KN
No. of bolts required, N = (Tu/Bolt Value) = (300/90.545) = 3.31 ≈ 4 nos
STEP 3:
Check of Strength due to rupture of critical section,
The design strength,
Tdn = 0.9Ancfu/ꙋm1 + βAgofy/ ꙋm0
Where,
β = 1.4 – 0.076(w/t)( fy/fu)( bs/Lc) ≤ (fu ꙋm0/fy ꙋm1)
≥ 0.7
w = outstand leg width = 60mm
t = total thickness of angles = 16mm
w1 = end distance = 40mm
bs = shear lag distance = w + w1 – t = 60 + 40 – 16 = 84mm
Lc = length of end connection = 3 x 60 = 180mm
β = 1.4 – 0.076(60/16)(250/410)(84/180)
= 1.4 – 0.081
= 1.319 ≤ (410×1.1/250×1.25) = 1.4432 (OK)
≥ (0.7) (OK)
Anc = (60+60-2 x 22) x 8 = 608mm2
Ago = (60 x 16) = 960mm2
Tdn = 0.9Ancfu/ꙋm1 + βAgofy/ ꙋm0
= ((0.9 x 608 x 410)/1.25) + ((1.319 x 960 x 250)/1.1)
= 179481.6 + 287781.81
= 467343.6 N
= 467.34 KN > 300 KN (O.K)
STEP 4:
Check for Strength Due to Block Shear (Tdb),
Tdb = [(Avgfy)/(√3ꙋm0) + (0.9Atnfu)/( ꙋm1)] or [(0.9Avnfu)/(√3ꙋm1) + (Atgfy)/(ꙋm0)]
Avg = 220 x 16 = 3520 mm2
Avn = (220-3×22-(22/2)) x 16 = 2288 mm2
Atg = 40 x 16 = 640 mm2
Atn = (40-(22/2)) x 16 = 464 mm2
Tdb = [((3520×250)/ (√3x1.1)) + ((0.9x464x410/1.25)]
= 461880.22 + 136972.8
= 598853.02 N
= 598.85 KN
Tdb = [((0.9x2288x410)/ (√3x1.25)) + ((640×250/1.1)]
= 389952.53 + 145454.54
= 535407.07 N
= 535.41 KN
Tdb = min (535.41, 598.85) = 535.41 KN > 300 KN (O.K)
Therefore, the selected section is safe.
So, Provide 2 ISA 60x60x8 angles placed in such a way that outstanding legs are placed back to back and attached with 20mm diameter bolts of grade 4.6. Edge distance and End distance is 40mm and pitch is 60mm.
See lessI want to brace my steel structure. what is the standard?
For better discussion, assume a Wide Flange column section. To carry the moment more efficiently, you need to apply a moment about the Major axis. Now consider a portal frame, say all the joints are pin joint. Now use a lateral load. It will collapse. Because as all the joints are pinned, they willRead more
For better discussion, assume a Wide Flange column section. To carry the moment more efficiently, you need to apply a moment about the Major axis. Now consider a portal frame, say all the joints are pin joint. Now use a lateral load. It will collapse. Because as all the joints are pinned, they will rotate infinitely, causing the structure to collapse. To Resist this collapse against lateral load, you need to provide – Either fixed support instead of pinned support or make the beam-column junction by moment connection. Or you can provide both. Now, your structure is stable under lateral load. How fixed support looks like when I section or complete flange section is used as a column section?
This is a fixed column base. Here you can see that the bolts are outside of the flange, just like your moment connection. So when the moment is applied about the major axis, it’s converted into push and pull, and these bolts carry this push-pull. Now, what will happen if a moment is applied about the minor axis? When the moment is being applied about the minor axis, there is no flange about this minor axis. So, The carrying capacity is very small. So, the moment can not be converted into push and pull efficiently. And these bolts will have almost nothing to carry for this moment. So, always this moment connected base will behave as a pinned joint about the minor axis. You have already learned that if a lateral load is applied on a frame with pinned support and pinned joint, it will collapse. But In a framed structure, there always remain lateral load, which may come from wind load or friction load or any other type of load. It is also clear that even a moment connected base can not prevent that force is applied across the minor axis of the Wide Flange column. So the solution is, add any diagonal member. Now, after applying a lateral load, will it collapse? No. Why?.. 5thstd math For given three lengths, only a single Triangle can be formed. And that’s why this structure doesn’t collapse under lateral load. If it collapses, this triangle has to be deformed, which is not possible. So, I think now it is clear to you why bracing is used. But how the lateral force is carried? The force travel from this point to this support through his bracing directly.
Why? Because… Force is also clever like us. If I ask you to travel from this point to this point, you will not follow this path. Instead, You will take this shortcut route to reach from this point to this point. Just like that, the force also follows the shortest path. If you want more scientific reason, well, it is because of the “theory of least work.” Now come to this Steel frame. Here along this Transverse direction, The Moment Connection has been used, and it is stable under the lateral load. But Along this longitudinal direction, there is no moment connection. All this joint is pinned connected. Even the support condition is also pin connected. Why this is pin connected, you have already learned that. Let say Lateral Load is applied here at this point. And you have a point to provide bracing. So, all this lateral travel through this member up to this point. Then this goes to this point to the support through this bracing, and as this load travel through this member, each of this member are loaded axially. And it would help if you designed all this member for axial load only. So, it is clear to you why and when bracing needs to be used.
What is the reason? i. When all the joints are pinned joint and ii. The lateral load needs to be carried. In those cases, we need to use bracing.
See lessWe have noticed some distress in the concrete columns in our apartment, and we have chosen to strengthen it using steel. We are using C section SAIL steel to be fixed on the two sides of the column till it ...
There could be broadly two types of corrective treatment: a. Support the concrete column with adjoining steel column ( an ISMB section will be better.), for ordinary house 12 feet height column, suggested steel ISMB is 250X 150 mm section. b. if column is found reparable, it is renovated with ‘ JackRead more
There could be broadly two types of corrective treatment:
a. Support the concrete column with adjoining steel column ( an ISMB section will be better.), for ordinary house 12 feet height column, suggested steel ISMB is 250X 150 mm section.
b. if column is found reparable, it is renovated with ‘ Jacketing ‘ around it. That is, scarp old peeling plaster, reach to reinforcing material, apply corrosive preventive liquids available from a good hardware store. Then, tie new reinforcement over the old reinforcement. Over it place 4-inch fresh cement concrete with aggregate of 12mm. Of course, steal or wooden scaffolding will be required.
It will increase the size of the column but will give strength to it.
BUT Note: if the column has already buckled down.. it is advisable to immediately get the help of an engineer and show him the site because if the column collapses, building or roof over it is in danger.
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Komal Bhandakkar
Design of poured concrete on the shear wall: When the final f'c extrapolated from a break of 3 days then an additional break of 5 days (it is usually 3, 7 & 28 days), it is directly indicated that the first cylinder itself was defective. When the f'c of 5000psi was one of the 28th day (i.e. finaRead more
Design of poured concrete on the shear wall:
But after all, the testing fell short of the design, the general contractor, and batch plant manager approach to the officer in charge (Airforce colonel which is assigned to oversee the project) to propose one of the serious field testings to demonstrate that the cantilever would work.
Finally, he decided to redo it.
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