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What is the best method of solving inclined truss? And how do can I get the magnitude of the forces in the members?
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 lessUnder what situation shall engineers use jacking at one end only and from both ends in prestressing work?
Prestressing is now a must to have phenomenon in every mega project whether it includes flyovers or dam or anything. During prestressing operation at one end, frictional loses will occur and the prestressing force decreases along the length of tendon until reaching the other end. These frictional loRead more
Prestressing is now a must to have phenomenon in every mega project whether it includes flyovers or dam or anything.
During prestressing operation at one end, frictional loses will occur and the prestressing force decreases along the length of tendon until reaching the other end. These frictional losses include the friction induced due to a change of curvature of tendon duct and also the wobble effect due to the deviation of duct alignment from the centerline.
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 lessWhat is the compressive strength of Fe500 rebars?
What 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.
How will Poison’s Ratio of a material Affect its Strength?
Poisson's ratio is defined as the ratio of the change in the width per unit width of a material, to the change in its length per unit length as a result of strain. Poisson ratio measures the deformation in the material in a direction perpendicular to the direction of the applied force. MathematicallRead more
Poisson’s ratio is defined as the ratio of the change in the width per unit width of a material, to the change in its length per unit length as a result of strain. Poisson ratio measures the deformation in the material in a direction perpendicular to the direction of the applied force.
Mathematically, poissons ratio is equal to the negative of the ratio of lateral strain and longitudinal strain. Therefore, if the poisson’s ratio is greater than the strength is greater.
What is fatigue in any structural member?
Concrete fatigue refers to the phenomenon of rupture under repeated loadings, each of which is smaller than a single static load that exceeds the strength of material.
Concrete fatigue refers to the phenomenon of rupture under repeated loadings, each of which is smaller than a single static load that exceeds the strength of material.
See less
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 less