Using Douglas Fir-Larch No. 2 dimension lumber, design joists at 16 inches o.c., and a girder (span = 6 feet). Assume that the girder load is distributed uniformly. Use L = 40 psf and D = 15 psf. Use CH = 1.0.
Assignment #4r
contact | homepage | spring 2008 index for ARCH 264/564 | current index for ARCH 264/564 | homework index | textIssued April 24, 2008.
Due: April 29, 2008 (not collected or graded)
Design the following beams/joists/girders for bending; then check for shear and deflection. Do not re-design if shear or deflection are not OK. Use L/360 criteria for deflection. Do not use live load reduction. Assume simple spans (i.e., hinge and roller constraints).
1. Wood joist and girder.
Using Douglas Fir-Larch No. 2 dimension lumber, design joists at 16 inches o.c., and a girder (span = 6 feet). Assume that the girder load is distributed uniformly. Use L = 40 psf and D = 15 psf. Use CH = 1.0.
Joist design: First, draw load, shear, and moment diagrams, as shown below.
1. Design for bending. Required CFSx = M / Fb* = 15,840 / (875 x 1.15) = 15.74 in3.
In the equation above, Fb* is the allowable bending stress with all adjustments except for the size factor, CF. The adjustment included is for repetitive use (Cr = 1.15). From Table A-8.3, select a 2x10.
2. Check for shear. Required A = 1.5 (V / F'v) = 1.5 (384 / 95) = 6.1 in2. The value of 384# is based on the shear force at a distance "d" (the width of the 2x10 joist) from the support, found with similar triangles as shown below:
Since this required area is less than or equal to the actual area (Table A-4.1) of 13.875 in2, the joist is OK for shear.
3. Check for deflection. The allowable deflection (Table A-8.1) is L / 360 = 12 x 12 / 360 = 0.4 inches. The actual deflection (Table A-8.2) is based on the live load only, as shown below:
Δ = CPL3 / (EI) = (22.46)(640)(123) / (1,600,000 x 98.93) = 0.16 inches. Since the actual deflection is less than or equal to the allowable deflection, the joist is OK for deflection.
Girder design: First, draw load, shear, and moment diagrams, as shown below.
1. Design for bending. Required CFSx = M / Fb* = 35,640 / (875) = 40.73 in3.
In the equation above, Fb* is the allowable bending stress with all adjustments except for the size factor, CF. There are no adjustments included. From Table A-8.3, select 2 - 2x10s.
2. Check for shear. Required A = 1.5 (V / F'v) = 1.5 (1471 / 95) = 23.23 in2. The value of 1471# is based on the shear force at a distance "d" (the width of the double 2x10 girder) from the support, found with similar triangles as shown below:
Since this required area is less than or equal to the actual area (Table A-4.1) of 2 x 13.875 = 27.75 in2, the girder is OK for shear.
3. Check for deflection. The allowable deflection (Table A-8.1) is L / 360 = 6 x 12 / 360 = 0.2 inches. The actual deflection (Table A-8.2) is based on the live load only, as shown below:
Δ = CPL3 / (EI) = (22.46)(2880)(63) / (1,600,000 x 2 x 98.93) = 0.04 inches. Since the actual deflection is less than or equal to the allowable deflection, the girder is OK for deflection. Note that the moment of inertia for the double 2x10 is found by multiplying Ix for a single 2x10 by two.
2. Shear stress adjustment factor.
What shear stress adjustment factor, CH, should be specified for a 6x12 girder subjected to a shear force of 6400#, if F'v = 95 psi.
The required area (for shear) = 1.5 (V / F'v) = 1.5 V / (Fv x CH) = 1.5 (6400) / (95 x CH) = 101 / CH in2. The actual area of a 6x12 is 63.25 in2. Equating the two areas, we can find how big CH needs to be:
101 / CH = 63.25; so CH = 1.60. Examining the choices in Table A-3.8, we would specify CH = 1.67.
3. Steel beam and girder.
Using A-992 steel, design the typical beam and cantilever girder shown in the sketches above, by selecting the lightest W-shape. Use L = 60 psf; and D = 50 psf. The girder is modeled, per load diagram, as a determinate cantilever beam with concentrated loads corresponding to the points where the beams frame in. Note that the load at the end of the cantilever is half that of the other loads (because its tributary area is half). Neglect the weight of the curtain wall; only include live and dead loads.
When checking girder deflection, model the cantilever girder as two separate beams, as shown below, in order to be able to use the available deflection tables. The "revised model" has moments and deflections close to that of the original model, as can be seen from the approximate deflected shapes sketched in the diagrams. Check deflection for both parts ("a" and "b"). Remember to use only the live load portion of the concentrated loads when checking deflection. Also, note that the span used in the calculation of allowable deflection is taken as 2 x L for the cantilever -- see note in Table A-8.1.
Beam design: First, draw load, shear, and moment diagrams, as shown below.
1. Design for bending. Required Sx = M / Fb = 1237.5 / (0.66 x 50) = 37.5 in3.
In the equation above, Fb is the allowable bending stress, equal to 0.66Fy (see Table A-3.11 and A-3.12). From Table A-8.4, select a W16x26.
2. Check for shear. Required Aw = V / Fv = 16.5 / (0.4 x 50) = 0.825 in2. The value of 16.5 k is the maximum shear force (see diagram above). It is not necessary to find the shear force at a distance "d" from the support for steel beams and girders.
Since this required area is less than or equal to the actual "web" area (Table A-4.3) = 15.69 x 0.25 = 3.92 in2, the beam is OK for shear.
3. Check for deflection. The allowable deflection (Table A-8.1) is L / 360 = 25 x 12 / 360 = 0.83 inches. The actual deflection (Table A-8.2) is based on the live load only, as shown below:
Δ = CPL3 / (EI) = (22.46)(18)(253) / (29,000 x 301) = 0.72 inches. Since the actual deflection is less than or equal to the allowable deflection, the beam is OK for deflection. Note that the modulus of elasticity for steel (29,000 ksi) can be found in Table A-3.11; the moment of inertia is found in Table A-4.3. The units for load (kips) and modulus of eleasticity (ksi) are consistent.
Girder design: First, draw load, shear, and moment diagrams, as shown below. The concentrated load, P = (60 + 50)(25 x 12) = 33,000# = 33 k. The reactions are found by taking moments about either the left or right reaction point. For example, taking moments about the right reaction, we get: VL(48) - 33(36) - 33(24) - 33(12) + 16.5(12) = 0; from which: VL = 45.375 k.
1. Design for bending. Required Sx = M / Fb = 8316 / (0.66 x 50) = 252 in3.
In the equation above, Fb is the allowable bending stress, equal to 0.66Fy (see Table A-3.11 and A-3.12). From Table A-8.4, select a W30x99.
2. Check for shear. Required Aw = V / Fv = 53.625 / (0.4 x 50) = 2.68 in2. The value of 53.625 k is the maximum shear force (see diagram above).
Since this required area is less than or equal to the actual "web" area (Table A-4.3) = 29.65 x 0.52 = 15.42 in2, the beam is OK for shear.
3. Check for deflection. As per instructions, we approximate the deflection analysis by breaking the cantilever beam into two parts.
Part A:
The allowable deflection (Table A-8.1) is L / 360 = 48 x 12 / 360 = 1.6 inches. The actual deflection (Table A-8.2) is based on the live load only (see diagram above): P = 60(25 x 12) = 18,000# = 18 k.
Δ = CPL3 / (EI) = (36.12)(18)(483) / (29,000 x 3990) = 0.62 inches. Since the actual deflection is less than or equal to the allowable deflection, this part of the beam is OK for deflection. Note that the modulus of elasticity for steel (29,000 ksi) can be found in Table A-3.11; the moment of inertia is found in Table A-4.3. The units for load (kips) and modulus of eleasticity (ksi) are consistent.
Part B:
The allowable deflection (Table A-8.1) is L / 360 = 2 x 12 x 12 / 360 = 0.8 inches (see note about using 2 x L for cantilever spans). The actual deflection (Table A-8.2) is based on the live load only (see diagram above): The typical concentrated load was previously found to be = 60(25 x 12) = 18,000# = 18 k, so we use half that value for the end condition: P = 9 k.
Δ = CPL3 / (EI) = (576)(9)(123) / (29,000 x 3990) = 0.08 inches. Since the actual deflection is less than or equal to the allowable deflection, this part of the beam is OK for deflection. Note that the modulus of elasticity for steel (29,000 ksi) can be found in Table A-3.11; the moment of inertia is found in Table A-4.3. The units for load (kips) and modulus of eleasticity (ksi) are consistent.
Since both parts of the girder are OK for deflection, the entire girder must be OK.
© 2008 Jonathan Ochshorn. First posted 25 April 2008. Last updated: 13 May 2008