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What is the introductory rate analysis of SS Masonry?
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 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 lessWhat is the best method of solving inclined truss? And how do can I get the magnitude of the forces in the members?
Hey, To solve the inclined trusses, Methods of joints is the appropriate method. With the help of this, one can easily identify the force in respective member of truss. But before solving the truss, some points should be kept in mind and these are listed below :- 1. If two members are meeting at joiRead more
Hey, To solve the inclined trusses, Methods of joints is the appropriate method. With the help of this, one can easily identify the force in respective member of truss.
But before solving the truss, some points should be kept in mind and these are listed below :-
1. If two members are meeting at joint, with no external force acting on them, then forces in both the members will be zero.
2. If the magnitude of force will come positive then force will be considered as Tensile And if magnitude of force will come negative then force will be considered Compressive.
3. If two members are collinear then force in the third member ,connected but not in straight line with two members, will be zero: F3 = zero
See lessWhat is bundled bar in reinforced concrete?
A group of reinforced bars, parallelly set to each other to act as a unit should be limited four in a bundle used as reinforcement in reinforced concrete, enclosed by stirrups or ties. Bars greater than 32mm should not be bundled in beams. The development length of every bar of bundled bars should bRead more
A group of reinforced bars, parallelly set to each other to act as a unit should be limited four in a bundle used as reinforcement in reinforced concrete, enclosed by stirrups or ties. Bars greater than 32mm should not be bundled in beams.
The development length of every bar of bundled bars should be that for the individual bar, increased by 10 % for two bars in contact, 20 % for three bars in contact, and 33 % for four bars in contact.
Structural Drawing is included
Sir, I will share some information related with different drawings: Architectural Drawing can be termed as the mother drawing for all the other drawing. Structural drawing is the backbone drawing of the building. Electrical drawing represent the details of electric fixtures, location of switches. PRead more
Sir,
I will share some information related with different drawings:
See less
Komal Bhandakkar
The rate analysis varies from place to place & it depends on the cost of construction material as well as labor cost. Roughly, we can say Ru. 134 per Cft. Required when PCC of 1:4:8 is used.
The rate analysis varies from place to place & it depends on the cost of construction material as well as labor cost.
Roughly, we can say Ru. 134 per Cft. Required when PCC of 1:4:8 is used.
See less