A steel beam is simply supported across a span of length L

A steel beam is simply supported across a span of length L = 3 m. A concentrated lateral load W acts on the beam at a distance of 1 m from the left hand

support.The cross sectional area of the beam is defined as a hollow square bow beam with side b = 40 mm and constant thickness t = 5 mm and is constructed of steel

with yield stress Y = 250 MPa.

a. Determine the magnitude WY of the load for which yield first occurs.

b. Determine the magnitude WC of the load for which collapse occurs.

c. Determine the maximum working load Wmax and maximum working stress W based on a safety factor

of 2 against yield.

d. Determine the maximum working load Wmax and maximum working stress W based on a load factor

of 2 against collapse.

[Ans. a: 2748 N;

b: 3469 N;

c: 1374 N, 126 MN/m2 ;

d: 1734 N, 159 MN/m2 ]

Solution

The general bending stress equation is =My/I

where M=Bending Moment exerted at the cross section under study

y=distance of the neutral axis from the end surface

I=Area Moment of Inertia of the cross section under study

The bending moment can be calculated as=2000W/3(W/3Nx2000mm)

Moment of Inertia can be calculated as :(Outer dimension⁴-Inner dimension⁴)/12=145833.33 mm⁴

y=40/2=20mm

Bending stress with yield=250 Mpa or 250 N/mm²

Answer a)Putting the above values in the bending stress equation

250=(2000W*20)/(3*145833.33)

W=2734N

Answer b)Ultimate strength(collapsing strength) as 1.2 xYield strength and solving for W we get the value as 3281 N (_collapsing=300N/mm²)

Answer c)Stress factor 2 means max allowable stress is =250/2=125N/mm2

Again solving for W with limiting stress as 125 N/mm2

we get W as 1367 N

Answer d) Stress factor 2 for collapse means max allowable stress is =300/2=150N/mm2

As _collapsing=300N/mm²

Solving for W we get the values as 1640.5 N