Przykłady realistyczne of Flexural Silniejsze obliczenia Steel Kod Beams Per Aisc

Flexural measurants are essential in determinaing thee load- carrying capacity of steel beams according to AISC codes. These calculations ensure safety andd compleance in structural design. Real- exterd examples illustrate how accorders applicy these principles in practice.

Egzamin 1: Simple Beem Under Uniform Load

A steel beom with a span of 6 meters is subieted to a uniform load of 10 kN / m. The beom 's cross- section is a W- shaped section with a momento of inertia (I) of 1500 cm present 1; British 1; FLT: 0 presentable 3; 4 presentation 1; FLT: 1 presentation 3; 3. Thee goal is to verify if thee beam can with stand the bending momento.

Te maximum bendim moment (M) for a simple supported beem under uniform load is calculated as:

M = (w * L = 1; F = 1; F = 1; F = 3; F = 3; F = 3; F = 3; F = 3; F = 3; F = 3; F = 3; F = 3; F = 3; F = 3; F = 3; F = 3; F = 3; F = 3; F = 3; F = 3; F = 3; F = 3; F = 3; F = 3; F = 3; F = 3; F = 3; F = 3; F = 3; D = 3; B = 3; B = 3; B = 3; B = 3; B = 3; B; B = 3; B = 1; 2; 2; B = 1; F = 1; F = 1; F = 1; F = 1; F = 1; F = 1; F = 1; F = 1; F = 1; F = 1; F = 1; F = 1; F = 1; F = 3; F = 0; F = 0; F = 0; F = 0; F = 0; F = 0; F = 3; F = 0; F = 0; F

Where w = 10 kN / m and L = 6 m, so:

M = (10 * 6 XI1; XI1; FLT: 0 XI3; XI3; 2 XI1; XI1; FLT: 1 XI3; XI3;) / 8 = 45 kNm

Using AISC formulas, the required d section modulus (S) is calculated as:

S = M / Fy

Założenie Fy = 250 MPa, then:

S = 45,000 / 250 = 180 cm precidi1; precidil; precidil: 0 precidil; precidial; precidial; precidial; precidial; precidial; precidial; precidial; precidial; precidial; precidial; precidial; precidial; precidial; precidial; precidial; precidial; precidial; precidition; precidition; precidition; precidition; preciditionary; preciditionary; preciditionary; preciditionary; preciditionary; preciditionary; precidireline; precidirecidiured; procidirecision; procision; procision; procion; precision; precision; precision; procision; procision; procision; procision) (1) (1)

To section 's S przekracza wartość, indicating confidentacy.

Badanie 2: Bending Stress Check

A steel beam wigh a prostotular cross- section (width 200 mm, height 300 mm) is supported over a 5- meter span. It carries a contributed load of 20 kN at mid- span. The engineer needs to o verify the bending stress.

Te maximum bending moment (M) at mid- span is:

M = (P * L) / 4 = (20 * 5) / 4 = 25 kNm

Te moduły section (S) for a prostotular section is:

S = (b * h = 1; x = 1; x = 1; x = 3; x = 3; x = 3; x = 3; x = 3; x = 3; x = 3; x = 3; x = 3; x = 3; x = 3; x = 3; x = 3; x = 3; x = 3; x = 1; x = 1; x = 1; x = 1; x = 1; x = 1; x = 1; x = 3; x = 3; x = 3; x = 3; x = 3; x = 3; x = 3; x = 3; x = 1 + 2; x = 1 + 2; x = 1; x = 1 + 2; x + 2 = 1 + 2; x + 2 + 2 + 2; x + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 3 + 3 + 3 + 3 + 3 + 3 + + + + + + + + + +

Kalkulating S:

S = (0.2 * 0,3 XI1; XI1; FLT: 0 XI3; XI3; 2 XI1; XI1; FLT: 1 XI3; XI3;) / 6 = 0,003 m XI1; XI1; FLT: 2 XI3; XI3; FLT: 3 XI3; XI3; FL3; Or 3000 cm XI1; XI1; FLT: 4 XI3; X3; XI1; FLT: 5 XI3; XI3; FLS: 3; XIXIX3;

Te bending stress (∞) is:

-------------------------------------------------- = M / S = (25,000 * 10 Xi1; Xi1; FLT: 0 Xi3; Xi3; 3 Xi1; Xi1; FLT: 1 Xi3; Xi3;) / 3000 = 8.33 MPa

Od tej pory te stresy i te dopuszczalne limit (np. 250 MPa), te beem i s approphable for thee load.

Egzamin 3: Shear Force and d Shear Stres

A steel beom spins 8 meters andd supports a point load of 50 kN at mid- span. The cross- section is a I- beum with a web squatness of 8 mm. The engineer checks shear capacity.

Te maximum shear force (V) at mid- span is equal to the load:

V = 50 kN

Thee shear stress (τ) in thee web is calculated as:

τ = V / A (1); (1); (1); (1): (1); (1): (1); (1) - (1); (1) - (1); (1) - (1); (1) - (1); (1) - (1) - (1); (1) - (1); (1) - (1) - (1); (1) - (2) - (2) (3); (1) - (3) - (3); (2) - (3) - (3) (3) (3); (3) (3) (4) (4) (4) (4) (4) (4) (4) (4) (4) (4) (4) (4) (4) (4) (4) (4) (4) (4) (4) (4) (5) (5) (5) (5) (5) (5) (5) (5) (5) (5) (5) (5) (5) (5) (5) (5)

Were A is 1; Xi1; FLT: 0 Xi3; Xi3; web Xi1; Xi1; FLT: 1 Xi3; Xi3; = web area = web xixness * web hight.

A = 1; A = 1; FLT: 0 = 3; FLT: 0 = 3; B = 1; FLT: 1; FLT: 1 = 3; = 0, 008 m * 0, 3 m = 0, 0024 m = 1; FLT: 2 = 3; FLT: 2 = 3; FLT: 2 = 3; Flight: 3; FLT: 3; Flight: 3; Flight: 3; Flight: 3; Flight: 3; Flight: 3; Flight: 3; Flight: 3; FS: 3; FS: 3; FS: 3; FS: 3; FS: 3; FS: 3; FS: 1; FS: 1; FS: 1; FS: 1; FS: 1; FS: 1; FS: 1; FS: 1; FS: 1: 1: 1: 1: 1: 1: 1: 1: 1: 1: 1: 1: 2

τ = 50,000 N / 0,0024 m

Od tej pory te wszystkie rodzaje mocy (np. 250 MPa), te same rodzaje mocy (b) i ich właściwości.