Steel Beam Design - IPFW

1 Notes for Strength of Materials, ET 200 beam Design Steel beam Design Six Easy Steps Steel beam Design is about selecting the lightest Steel beam that will support the load without exceeding the bending strength or shear strength of the material, and without exceeding the maxi- mum allowable deflection for the beam . We want the lightest beam because it is generally the cheapest. We can solve these problems with a 6-step process. Step 1 Identify all loads and Design constraints (yield strength, maximum allowable deflection max, beam length L, etc.)

2 Step 2 Draw the load diagram and calculate all reactions. Step 3 Draw the shear and moment diagrams , and calculate Vmax and Mmax. If the loading conditions are right, use the Formula Method to find these values. Step 4 Calculate the plastic section modulus Zx required to sup- port the applied moment . Select the lightest Steel beam from the Appendix that supports Mmax and has enough stiffness to limit max (if deflection is a constraint). Step 5 Include the beam weight in new drawings of the load, shear, and moment diagrams .

3 Check that the beam can support the applied loads and its own weight, and that it still meets the maxi- mum deflection constraint. Step 6 Calculate the shear strength of the selected beam , and check that the beam will support more shear load than is applied. Example #1. Select the lightest W- beam that will support a uniformly distrib- uted load of 3 kip/ft. on a simply-supported span of 20 ft. The beam is rolled high-strength, low-alloy Steel (HSLA). Step 1 We know the loading and length; the Steel has a yield strength !

4 YS = 50 ksi . The maximum beam deflection max is not specified. 3 kip 20 ft. Step 2 The total load on the beam is = 60 kips . Since ft. the loading is symmetrical, R A = R B = 30 kips . 2011 Barry Dupen 1 of 8 Revised 4 May 2011. Notes for Strength of Materials, ET 200 beam Design Step 3 The shear diagram for a uniform distributed load is two triangles. The moment diagram is a parabola, where Mmax is the area of the shear diagram up to the midspan, or the area of the left- hand triangle. Since the area of a triangle is the base times the 30 kips !

5 10 ft. height divided by two, M max = = 150 kip ft. 2. Step 4 The moment strength of a Steel beam is M R = ! YS Z x . We can rewrite the equation to find the value of Zx required to beam Zx ( ). support the applied moment . W18 40 M Required Z x = = W12 50 ! YS ! YS. " 150 kip ft. 12 in. W10 54 =. 50 kips ft. W16 36 = in. 3. W12 40 Appendix I lists W-beams in decreasing order of plastic section modulus Zx. Look for a beam with a slightly larger Zx than the required value. In this case, the lightest beam is W16 36, with a weight of 36 , or kips/ft.

6 Step 5 We can add the beam weight to the applied uniform dis- tributed load, for a total of kips/ft. The total load on the kip 20 ft. beam is = kips . Since the loading is sym- ft. metrical, R A = R B = kips . The maximum moment is kips ! 10 ft. M max = = kip ft. 2. ! kip ft. 12 in. Required Z x = = , which 50 kips ft. is less than Zx of the selected beam . As long as we have more than we need, the beam will survive. If the new required Zx had been 66 , then we would have to select a different beam . Step 6 We know the beam will support the load without exceeding its bending strength; now we need to check shear strength.

7 For wide-flange Steel W-beams, Vapplied ! " YS dt w where d is the beam depth and tw is the thickness of the web. Find these dimen- sions in Appendix A. A W16 36 beam can support a shear load of kips ! 50 2 ! in. ! in. = kips . Since the actual in. shear load of kips is less than kips, the beam will not fail in shear. 2011 Barry Dupen 2 of 8 Revised 4 May 2011. Notes for Strength of Materials, ET 200 beam Design Example #2. Select the lightest W- beam that will support a uniformly distrib- uted load of 3 kip/ft.

8 On a simply-supported span of 20 ft. and de- flect no more than inches. The beam is rolled high-strength, low-alloy Steel (HSLA). Steps 1-4 The first few steps are identical to Example #1 because the beam loading and length are the same. However, we have an additional constraint of ! max = in. From Appendix H, Case #1, the maximum deflection for a simply-supported beam with a uni- 5wL4. form distributed load is ! max = . We can rewrite the equa- 384EI. tion to find the moment of inertia required to limit the maximum deflection.

9 5wL4. Required I =. 384E! max beam Zx ( ) Ix ( ). 5 3 kip ( 20 ft.) in. (12 in.) = 600 4 2 3. W18 40 612. =. 384 ft. 30 " 10 3 kip in. W12 50 394. 4. The moment of inertia of a W16 36 is 448 in. , which is not W10 54 303. enough. Instead, we need to select a beam with a moment of iner- tia greater than 600 , such as W18 40, which has a weight of W16 36 448. 40 or kip/ft. W12 40 310. Step 5 Add the beam weight to the applied uniform distributed load, for a total of kips/ft. The total load on the beam is kip 20 ft.

10 = kips . Since the loading is symmetrical, ft. R A = R B = kips . The maximum moment is kips ! 10 ft. M max = = 152 kip ft. 2. ! 152 kip ft. 12 in. Required Z x = = , which 50 kips ft. is less than Zx of the selected beam , so the beam is strong enough in bending. Checking for deflection, 5 kip ( 20 ft.) in. (12 in.) = 608 4 2 3. Required I =. 384 ft. 30 ! 10 kip in. 3. Since the beam 's moment of inertia is more than the required value, the beam meets the deflection criterion. 2011 Barry Dupen 3 of 8 Revised 4 May 2011.