## Solved problem-5-1-part 1-2.

Solved problem-5-1-part 1-2 for deflection is the title of this post, which is a design example for a simply supported beam.

In this post, termed Deflection of Steel Beams Part 2, we will select the W section for the beam based on the Zx value.

### A solved problem-5-1-part 1-2 for the design of a beam.

The solved problem 5-1 is quoted from the Book of Prof. Fredrick Roland’s book.

Zone -1 bending, we know that zone 1 has M plastic moment for which, Mp=Fy*Zx, you can check the plastic theory in the previous posts.

We have a simply supported beam with a span L=30 ft with ASTM A992 where Fy=50 Ksi and Fult=65 ksi.

This solved problem is a design example with given deflection criteria.

This problem is selected after reviewing the previous information for The LRFD and the ASD for design under bending.

The beam is laterally supported for its entire length. Therefore, the section will be considered compact and in zone 1.

If we start with the LRFD design, we need to estimate the beam’s moment as M=W*L^2/8, cb=1, which is the moment coefficient. The section is fully braced.

#### Factored Load for LRFD design and the ultimate moment.

First, the Ultimate load =*1.20D+1.60 L* for D=0.30 kips/ft and L=0.70 Kips/ft, then Wult=1.20*0.30+1.60*0.70=1.48 Kips/ft, The value of Mult will be equal to 1.48*30^2/8=166.50 ft. kips. To convert to Inch kips, multiply by 12, 66.50*12= 1998 inch. Kips in the LRFD design.

#### How do we get Zx, which is based on LRFD?

The reduction factor for strength based on the lRFd Φb=0.90 while Ωb=1.67, to check Φb* Ωb=1.50. for step-1. We will estimate Zx from the known formula Mn=Fy*Zx, Fy=50 ksi, then Φb*Mn =Φb*Fy*Zx=Mult, The yield street Fy=50 ksi. The plastic section modulus Zx= Mult/0.90*50* Zx=1998/45=44.50 inch3.

Once we have estimated the required Zx, we will get the required W section based on Table 3-2, which shows sections arranged based on Zx. The first selection is W14x30, but we must check the inertia based on serviceability requirements.

### Determine the moments based on Live load and Moment for total load.

We will estimate the Total load for ASD design as the sum of D plus L, which is equal to 1.0 kip/Ft. We will estimate the live load moment together with the moment due to the total load. We will need both to find the deflection values. We will estimate the required Zx.

#### Inertia for the beam to give a safe deflection case of L.

Deflection based on Live load criteria, where the deflection delta is equal to L/360. The length is 30 feet. We convert to inches by multiplying by 12.

The deflection Δ-L=L/360=360/360=1, and the WLive load is 0.70 ft/kips. The moment used is the moment value due to LL load equal to 78.75 Ft.kips.

Use the AISC relation for deflection as equal to ML^2/C1*Ix, where c1=161, equate M*L^2/C1*Ix to 1 inch. M is in Ft.KIps, L in Feet. We will get the Ix value as equal to 440.22 inch4.

#### Inertia for the beam to give a safe deflection case of D+L.

Checking the deflection of the steel beam on the floor based on L/240 for D+L is required.

We are in the serviceability criteria; we use the D+L, not the ultimate loads. D+L=1 kips/ft. The deflection due to total load equals(30*12/240)=1.50 inch.

The moment used is the moment value due to the total load equal to 112.50 Ft.kips.

Use the AISC relation for deflection as equal to ML^2/C1*Ix, where c1=161, equate M*L^2/C1*Ix to 1 inch. M is in Ft.KIps, L in Feet. We will get the Ix value as equal to 419.25 inch4. Please refer to the next slide image for more details.

#### The final selected Inertia for the beam.

We have two values: the first one is 440.22 inch4, and the second estimated inertia value is 419.25 inch4; we will select the bigger value, 440.22 inch4.

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In the next post, we will continue our estimation for selecting the final economic section.

As an external resource for the deflection. **Chapter 8 – Bending Members**.

This is the next post, solved problem 5-1 part 2-2, how to find LL for a given slender section?