### Transcription of 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.). 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.

2 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. 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 ! YS = 50 ksi . The maximum **beam** deflection max is not specified.

3 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 ! 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 =.

4 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. 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.

5 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. 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 !

6 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. 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. = kips . Since the loading is symmetrical, ft.

7 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. Notes for Strength of Materials, ET 200 **beam** **Design** Step 6 Use Vapplied ! " YS dt w to find the shear strength. A. W18 40 **beam** can support a shear load of kips ! 50 2 ! in. ! in. = 113 kips . Since the actual in. shear load of kips is less than 113 kips, the **beam** will not fail in shear. Examples #1 and #2 are the easiest types to solve because the weight of the **beam** is a uniform distributed load, therefore the load, shear, and moment diagrams have the same shape after the **beam** weight is included.

8 Example #3. Select the lightest W- **beam** that will support a point load of 40 kips at the midspan of a simply-supported 30 foot span. Step 1 P = 40 kips , L = 30 ft. , ! YS = 50 ksi , max is not speci- fied. Step 2 The total load on the **beam** is 40 kips. Since the loading is 40 kips symmetrical, R A = R B = = 20 kips . 2. Step 3 The shear diagram for a point load at the midspan is two rectangles. The moment diagram is a triangle, where Mmax is the area of the shear diagram up to the midspan, or the area of the left- hand rectangle: M max = 20 kips ! 15 ft. = 300 kip ft. Step 4 Calculate the needed plastic section modulus: " 300 kip ft. 12 in. **beam** Zx ( ). Required Z x = =. ! YS 50 kips ft. W21 62 144. = W14 74 126. Select W18 60, with a weight of 60 , or kips/ft. W18 60 123. W21 50 110.

9 2011 Barry Dupen 4 of 8 Revised 4 May 2011. Notes for Strength of Materials, ET 200 **beam** **Design** Step 5 Redraw the load, shear, and moment diagrams to include the weight of the **beam** . Add the **beam** weight to the point load for kip 20 ft. a total load of 40 kips + = kips . Since the ft. loading is symmetrical, R A = R B = kips . V1 = R A = R B = kips kip 15 ft. V2 = V1 ! = 20 kips ft. V3 = V2 ! 40 kips = !20 kips V4 = V3 + kips = 0 kips The maximum moment is the area of the left-hand trapezoid, which is the average height times the base: kips + 20 kips M max = 15 ft. = kip ft. 2. ! kip ft. 12 in. Required Z x = = , 50 kips ft. which is slightly less than Zx of the selected **beam** , so the **beam** is strong enough in bending. Step 6 Use Vapplied ! " YS dt w to find the shear strength. A.

10 W18 60 **beam** can support a shear load of kips ! 50 2 ! in. ! in. = 151 kips . Since the actual in. shear load of kips is less than 151 kips, the **beam** will not fail in shear. Another way to solve this problem is to use the Formula Method and Superposition. For Step 2 and Step 3 , Case #5 in Appendix H. gives us P 40 kips R A = R B = Vmax = = = 20 kips 2 2. PL 40 kips ! 30 ft. M max = = = 300 kip ft. 4 4. 2011 Barry Dupen 5 of 8 Revised 4 May 2011. Notes for Strength of Materials, ET 200 **beam** **Design** Solve Step 4 as before. For Step 5, we can add the reaction forces for the two cases: P wL 40 kips kips 30 ft. RA = RB = + = + = kips 2 2 2 ft. 2. The maximum shear load occurs at the same place in both shear diagrams (the ends of the beams) and is equal to the reactions, so Vmax = R A = R B = kips.