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What are the Different Types of Steel Stacks?
Preet Chovatiya
There are four types of steel stack that generally used in steel stack are as below: Freestanding (Self-supported) steel stack (fig.1) The base support and guyed steel stack (fig.2) The base support and braced steel stack Multi flue stack (fig.3) [caption id="attachment_45199" align="alignnone" widtRead more
There are four types of steel stack that generally used in steel stack are as below:
fig.1
fig.2
fig.3
What is the Procedure to Design the Double Angle Tension Member in Steel Structures with formulas?
Amit Bhuriya
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 lessWhich are the different Types of Welding Process?
RaghavArora
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 best method of Solving Inclined Truss?
RaghavArora
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 lessWhy a Continuous RCC beam should be designed as a Tee beam for Span moments and as a Rectangular Beam for Support Moments?
Abbas Khan Civil Engineer
Bcz , there is a negative moment above the supports.Making it useless and only left is rectangular portion at the bottom, so that's why rectangular beams are used for support moments in continous beams. While in the span case, the top section is under compression and below one is in tension so theRead more
Bcz , there is a negative moment above the supports.Making it useless and only left is rectangular portion at the bottom, so that’s why rectangular beams are used for support moments in continous beams.
While in the span case, the top section is under compression and below one is in tension so the flexural strength is enough to contribute to resist the stresses.
See lessWhat are quickest method to find deflection and slope in beams?
RaghavArora
For continuous loading, Macaulay's method is easiest method. For othe questions, Moment area method is good. If u practice these methods, u will automatically consider them easy. Every method is traditional but u have to be friendly with them to master and the best way is to practice. Engineering itRead more
For continuous loading, Macaulay’s method is easiest method. For othe questions, Moment area method is good. If u practice these methods, u will automatically consider them easy. Every method is traditional but u have to be friendly with them to master and the best way is to practice. Engineering itself means Smart Practice
See lessWhich Type of Cement is used for Pile Foundation?
Madeh Izat Hamakareem
Any type of cement can be used as long as the design and durability requirements of the concrete used to construct the pile are met. For instance, if the sulfate attack is expected, it is desirable to select sulfate resistance cement because it produces concrete that withstand sulfate attack. As a rRead more
Any type of cement can be used as long as the design and durability requirements of the concrete used to construct the pile are met. For instance, if the sulfate attack is expected, it is desirable to select sulfate resistance cement because it produces concrete that withstand sulfate attack. As a result, the produced concrete would durable.
You can visit the following link to read more about Applications of Different Types of Cement for Concrete Construction:
https://test.theconstructor.org/building/applications-cement-types-concrete/7686/
The ACI code 318-18 provides the design and durability requirements of concrete.
Design Requirements of Concrete Based on ACI 318-19
https://test.theconstructor.org/concrete/design-requirements-concrete/39004/
Concrete Durability Requirements Based on ACI-318-19
https://test.theconstructor.org/concrete/concrete-durability-requirements/39036/
See lessWhat is the Minimum Grade of Concrete for Precast Construction?
Madeh Izat Hamakareem
It depends on the structural member. For example, The American Concrete Institute (ACI 318-19) specifies minimum concrete strength for precast pile as 28MPa. The code also provides minimum strength for other structural members. You can also visit: https://test.theconstructor.org/concrete/design-requRead more
It depends on the structural member. For example, The American Concrete Institute (ACI 318-19) specifies minimum concrete strength for precast pile as 28MPa. The code also provides minimum strength for other structural members. You can also visit:
https://test.theconstructor.org/concrete/design-requirements-concrete/39004/
See less