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A double angle is used as the top chord of a given truss shown in figure 17.7. It is composed of 2 - angles 225 x 100 x 12 mm. with short legs back to back. Thickness of gusset plate is 9 mm. Bracing is provided in the plane of the top chord every 2.2 m. Total length of the top chord is 8.8 m. but bracing between trusses is such that lateral support occurs only at the ends of the top chord. Neglect the contribution of roofing to lateral support. Use A 36 steel. K = 1.0, E,= 200000 MPa. 0 Compute the reduction factor due to local buckling of compression member. Compute the allowable compressive stress for the top chord. 2 Compute the maximum axial compressive force which the top chord can carry safely. 3 Properties of 2 angles 2225 x 100 x 12 mm with short legs back to back A 36 steel CIVIL A = 8064.5 mm² Tx = 26.67 mm ry = 115.57 mm F₁ = 248 MPa ENGINEERING STEEL DESIGN 8.8 m 2.2 2.2 2.2 2.2

A double angle is used as the top chord of a given truss shown in figure 17.7. It is composed of 2 - angles 225 x 100 x 12 mm. with short legs back to back. Thickness of gusset plate is 9 mm. Bracing is provided in the plane of the top chord every 2.2 m. Total length of the top chord is 8.8 m. but bracing between trusses is such that lateral support occurs only at the ends of the top chord. Neglect the contribution of roofing to lateral support. Use A 36 steel. K = 1.0, E,= 200000 MPa. 0 Compute the reduction factor due to local buckling of compression member. Compute the allowable compressive stress for the top chord. 2 Compute the maximum axial compressive force which the top chord can carry safely. 3 Properties of 2 angles 2225 x 100 x 12 mm with short legs back to back A 36 steel CIVIL A = 8064.5 mm² Tx = 26.67 mm ry = 115.57 mm F₁ = 248 MPa ENGINEERING STEEL DESIGN 8.8 m 2.2 2.2 2.2 2.2

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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A double angle is used as the top chord of a given truss shown in figure 17.7. It is composed of 2 - angles 225 x 100 x 12 mm. with short legs back to back. Thickness of gusset plate is 9 mm. Bracing is provided in the plane of the top chord every 2.2 m. Total length of the top chord is 8.8 m. but bracing between trusses is such that lateral support occurs only at the ends of the top chord. Neglect the contribution of roofing to lateral support. Use A 36 steel. K=1.0, E, 200000 MPa. Compute the reduction factor due to local buckling of compression member. 2 Compute the allowable compressive stress for the top chord. 3) Compute the maximum axial compressive force which the top chord can carry safely. Properties of 2 angles 225 x 100 x 12 mm with short legs back to back A 36 steel CIVIL A = 8064.5 mm² rx = 26.67 mm ry = 115.57 mm F₁ 248 MPa 2.2 ENGINEERING STEEL DESIGN 8.8 m 2.2 2.2 2.2.
Transcribed Image Text:A double angle is used as the top chord of a given truss shown in figure 17.7. It is composed of 2 - angles 225 x 100 x 12 mm. with short legs back to back. Thickness of gusset plate is 9 mm. Bracing is provided in the plane of the top chord every 2.2 m. Total length of the top chord is 8.8 m. but bracing between trusses is such that lateral support occurs only at the ends of the top chord. Neglect the contribution of roofing to lateral support. Use A 36 steel. K=1.0, E, 200000 MPa. Compute the reduction factor due to local buckling of compression member. 2 Compute the allowable compressive stress for the top chord. 3) Compute the maximum axial compressive force which the top chord can carry safely. Properties of 2 angles 225 x 100 x 12 mm with short legs back to back A 36 steel CIVIL A = 8064.5 mm² rx = 26.67 mm ry = 115.57 mm F₁ 248 MPa 2.2 ENGINEERING STEEL DESIGN 8.8 m 2.2 2.2 2.2.
Solutions: Ł = 225 = 18.75 12 Qa=1.0 Q₁ = 1.34-0.76 [4/4] Fy E = 1-34-0.76 (18.75). Q=0.84 a.] k² rx = кг K2 - 1.0x2200 115.57 10x8800 = 329.96 26.67 248 200,00 - Use k2 largest ř < 200 = 19.036 Assume K=1.0 8m X axis 4.71√ √2x105 my 248 x0.84 195.93 2007 145.93 :. Elastic Fe: 11² x 200 ow (200)² 2.2 2.2 2.2 ✓ 4.71√ 2.2 y Fe=49.35 kindly Recheck work thank you CIVIL ENGINEERING STEE DESIGN aki 2x105 FyxQs Per= 0.877 Fe = 0.877 99.35 6. Pcr= 43.28 mPa (n = 43.28 x 8044.5 lov c) → Pn= 349.03 kW Slender Unstiffened Klements, Q, The reduction factor Q, for slender unstiffened elements is defined as follows: 505.7.1 1. 2 For flanges, angles, and plates projecting from rolled columns or other compression members: a. When b. a 3. b. When L When 0.56 Q. = 1.415-0.74 ≤ 0.56 2. = 1.0 When b. 0.69E Fy (2) For flanges, angles, and plates projecting from built-up columns or other compression members: When ≤0.64 When 0.64 where: ke Q8== Q=1.415-0.65 a. When Q. = 1.0 For single angles Q;= ¬OF L.03 When When where b- Ek <1.03 Eke -1.17 ER 0.90Ek Fy (2² 2x = 1.0 0.45 0.45 Q,= 1.34-0.76 另 ≤1.17 Qs= The and shall not be taken loss than 0.35 nor greater than h/t 0.76 for calculation purposes 34-0.76 (2) (505.7-4) > 0.91 (505.7-5) 0.53E Fy()* (505.7-6) (505.7-7) Eke 7 (505.7-8) (505.7-9) 91 (505.7-10) (505.7-11) (505.7-12) full width of longest angle leg. mm
Transcribed Image Text:Solutions: Ł = 225 = 18.75 12 Qa=1.0 Q₁ = 1.34-0.76 [4/4] Fy E = 1-34-0.76 (18.75). Q=0.84 a.] k² rx = кг K2 - 1.0x2200 115.57 10x8800 = 329.96 26.67 248 200,00 - Use k2 largest ř < 200 = 19.036 Assume K=1.0 8m X axis 4.71√ √2x105 my 248 x0.84 195.93 2007 145.93 :. Elastic Fe: 11² x 200 ow (200)² 2.2 2.2 2.2 ✓ 4.71√ 2.2 y Fe=49.35 kindly Recheck work thank you CIVIL ENGINEERING STEE DESIGN aki 2x105 FyxQs Per= 0.877 Fe = 0.877 99.35 6. Pcr= 43.28 mPa (n = 43.28 x 8044.5 lov c) → Pn= 349.03 kW Slender Unstiffened Klements, Q, The reduction factor Q, for slender unstiffened elements is defined as follows: 505.7.1 1. 2 For flanges, angles, and plates projecting from rolled columns or other compression members: a. When b. a 3. b. When L When 0.56 Q. = 1.415-0.74 ≤ 0.56 2. = 1.0 When b. 0.69E Fy (2) For flanges, angles, and plates projecting from built-up columns or other compression members: When ≤0.64 When 0.64 where: ke Q8== Q=1.415-0.65 a. When Q. = 1.0 For single angles Q;= ¬OF L.03 When When where b- Ek <1.03 Eke -1.17 ER 0.90Ek Fy (2² 2x = 1.0 0.45 0.45 Q,= 1.34-0.76 另 ≤1.17 Qs= The and shall not be taken loss than 0.35 nor greater than h/t 0.76 for calculation purposes 34-0.76 (2) (505.7-4) > 0.91 (505.7-5) 0.53E Fy()* (505.7-6) (505.7-7) Eke 7 (505.7-8) (505.7-9) 91 (505.7-10) (505.7-11) (505.7-12) full width of longest angle leg. mm
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