- Solved problem-5-1-part 1-2.
- A solved problem-5-1-part 1-2 for the design of a beam.
- Estimation of the factored loads for the solved probem 5.1.
- Estimation of the required Inertia for the beam to give a safe deflection case of D+L.
- Estimation of the required Inertia for the beam to give a safe deflection case of L.
- The required Zx value for the steel beam in the solved probem 5.1.
- What is the deflection value for the selected Zx section for the steel beam in the solved probem 5.1?

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

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

Solved problem-5-1-part 1-2 for deflection is the title of this post in which we start solving problem 5.1 which is a design example for a simply supported beam. In this post which is termed deflection of steel beams part 2, we will start to 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.

We will check the solved problem 5-1 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 previous lectures.

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 as a revision of the previous information for The LRFD and the ASD for design under bending.

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

#### Estimation of the factored loads for the solved probem 5.1.

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

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 , Mult=1.48*30^2/8=166.50 ft.kips.

While for the ASD, the w-total=D+L=0.30+0.70 =1.0 kips/ft, Mt=1*30^2/8 = 112.50 ft.kips, we convertconvert the ft.kips into inch.kips.

Mt=112.50*12=1350 inch.kips in the case of the ASD, while Mult=166.50*12= 1998 inch.kips in the *LRFD design*.niform loading.

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, fy=50 ksi. Zx= Mult/0.90*50* Zx=1998/45=44.50 inch3.

Mn/(Ωb which is= 1.67) should be>=Mt. Mt from the ASD we have estimated as= 1350 inches. kips.

Zx value=(1350*1.67/50)=45.09 inch3.

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

Since in the solved problem. It is required to check the deflection of the steel beam on the floor based on L/360 for live loads and L/240 for D+L, the deflection of the steel beam should be estimated with care due to conversion between units as we will see together. For the floor beams E=29000 Ksi.

We are in the serviceability criteria, we use the D+L, not the ultimate loads. D+L=1 kips/ft, delta=(5/384)*W* L^4/EI.

We need I value for the beam and we will compare the estimated deflection of a steel beam with L/240 criteria for the floor beams for total loads.

We will convert the Length of the beam which is L in feet=30′ to inches, we have L=30′ =30*12=360 inches.

So L/240=360/240=1.50″, permitted deflection due to (D+LL). We will equate the deflection due to Total load(D+L)wt=1.00 kips/ft acting on a span of 360 inches, convert the load to kips/ inch, by dividing by 12.

Our load is=1.0 kips/ft, we convert to Wt= 1/12=kips/inches. We kept the kips to match with the E units and lengths in inches, then permitted deflection due tp (D+LL)= 1.50= (5/384)* (W=1/12) *(360)^4 In inches /29000kips/inch2.

The inertia for the section, I=5*360^4/ (29000*384*12) *1/(1.50), I =418.96 inch4, this value will give permitted deflection based on total load.

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

The second criteria is the Live load criteria where the deflection delta is=L/360. Δ-L=L/360=360/360=1, WLive load=0.70 ft/kips or =0.70/12 kips/inch. Δ-L=1=(1/I)*(5/384)*(0.70)* (360)^4 /29000/12. Then inertia value I due to Live load L=439.91 inch4.

We have two values the first one is 418.96 inch4 and the second estimated value of inertia is 439.9 inch4, we will select the bigger value

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

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 , Mult=1.48*30^2/8=166.50 ft.kips.

While for the ASD, the w-total=D+L=0.30+0.70 =1.0 kips/ft, Mt=1*30^2/8 = 112.50 ft.kips, we convertconvert the ft.kips into inch.kips. Mt=112.50*12=1350 inch.kips in the case of the ASD, while Mult=166.50*12= 1998 inch.kips in the LRFD design due to uniform loading.

#### The required Zx value for the steel beam in the solved probem 5.1.

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, fy=50 ksi. Zx= Mult/0.90*50* Zx=1998/45=44.50 inch3.

Mn/(Ωb which is= 1.67) should be>=Mt. Mt from the ASD we have estimated as= 1350 inches. kips.

Zx value=(1350*1.67/50)=45.09 inch3.

#### What is the deflection value for the selected Zx section for the steel beam in the solved probem 5.1?

Since in the solved problem. It is required to check the deflection of the steel beam on the floor based on L/360 for live loads and L/240 for D+L, the deflection of the steel beam should be estimated with care due to conversion between units as we will see together. For the floor beams E=29000 Ksi.

We are in the serviceability criteria, we use the D+L, not the ultimate loads. D+L=1 kips/ft, delta=(5/384)*W* L^4/EI.

We need I value for the beam and we will compare the estimated deflection of a steel beam with L/240 criteria for the floor beams for total loads.

We will convert the Length of the beam which is L in feet=30′ to inches, we have L=30′ =30*12=360 inches.

So L/240=360/240=1.50″, permitted deflection due to (D+LL). We will equate the deflection due to Total load(D+L)wt=1.00 kips/ft acting on a span of 360 inches, convert the load to kips/ inch, by dividing by 12.

Our load is=1.0 kips/ft, we convert to Wt= 1/12=kips/inches. We kept the kips to match with the E units and lengths in inches, then permitted deflection due tp (D+LL)= 1.50= (5/384)* (W=1/12) *(360)^4 In inches /29000kips/inch2.

The inertia for the section, I=5*360^4/ (29000*384*12) *1/(1.50), I =418.96 inch4, this value will give permitted deflection based on total load. The second criteria are the Live load criteria where the deflection delta is=L/360.

Δ-L=L/360=360/360=1, WLive load=0.70 ft/kips or =0.70/12 kips/inch. Δ-L=(5/384)*(0.70)* (360)^4 /29000/12.

Here is the detailed calculation for the deflection of the steel beam given in the solved problem for the deflection of a steel beam.

Then inertia value I due to Live load L=439.91 inch4, we have two values the first one is 418.96 inch4 and the second estimated value of inertia is 439.9 inch4, we will select the bigger value

This is the pdf file used in the illustration of this post.

This is an external resource for 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?