RH 331 Note Set 12.1 F2014abn Rigid and raced Frames Notation: E = modulus of elasticit or Young s modulus F = force component in the direction F = force component in the direction FD = free bod diagram G = relative stiffness of columns to beams in a rigid connection, as is I = moment of inertia with respect to neutral ais bending k b c L M V = effective length factor for columns = length of beam in rigid joint = length of column in rigid joint = name for length = internal bending moment = name for a moment vector = internal shear force = summation smbol Rigid Frames Rigid frames are identified b the lack of pinned joints within the frame. The joints are rigid and resist rotation. The ma be supported b pins or fied supports. The are tpicall staticall indeterminate. Frames are useful to resist lateral loads. Frame members will see shear bending aial forces and behave like beam-columns. ehavior The relation between the joints has to be maintained, but the whole joint can rotate. The amount of rotation and distribution of moment depends on the stiffness (EI/L) of the members in the joint. End restraints on columns reduce the effective length, allowing columns to be more slender. ecause of the rigid joints, deflections and moments in beams are reduced as well. Frames are sensitive to settlement because it induces strains and changes the stress distribution. 1
RH 331 Note Set 12.1 F2014abn Tpes Gabled has a peak Portal resembles a door. Multi-stor, multiple ba portal frames are commonl used for commercial and industrial construction. The floor behavior is similar to that of continuous beams. Staggered Truss Full stor trusses are staggered through the frame bas, allowing larger clear stories. Staggered Truss onnections Steel Flanges of members are full attached to the flanges of the other member. This can be done with welding, or bolted plates. Reinforced oncrete Joints are monolithic with continuous reinforcement for bending. Shear is resisted with stirrups and ties. raced Frames raced frames have beams and columns that are pin connected with bracing to resist lateral loads. Tpes of racing knee-bracing diagonal (including eccentric) X diagonal X K or chevron shear walls which resist lateral forces in the plane of the wall Rigid Frame nalsis shear walls K (chevron) Structural analsis methods such as the portal method (approimate), the method of virtual work, astigliano s theorem, the force method, the slope-displacement method, the stiffness method, and matri analsis, can be used to solve for internal forces and moments and support reactions. Shear and bending moment diagrams can be drawn for frame members b isolating the member from a joint and drawing a free bod diagram. The internal forces at the end will be equal and opposite, just like for connections in pinned frames. Direction of the beam-like member is usuall drawn b looking from the inside of the frame. 2
M+ RH 331 Note Set 12.1 F2014abn P M+ M+ Frame olumns ecause joints can rotate in frames, the effective length of the column in a frame is harder to determine. The stiffness (EI/L) of each member in a joint determines how rigid or fleible it is. To find k, the relative stiffness, G or, must be found for both ends, plotted on the alignment charts, and connected b a line for braced and unbraced fames. where G EI l EI l c b E = modulus of elasticit for a member I = moment of inertia of for a member l c = length of the column from center to center l b = length of the beam from center to center For pinned connections we tpicall use a value of 10 for. For fied connections we tpicall use a value of 1 for. raced non-swa frame Unbraced swa frame Frame Design The possible load combinations for frames with dead load, live load, wind load, etc. is critical to the design. The maimum moments (positive and negative) ma be found from different combinations and at different locations. Lateral wind loads can significantl affect the maimum moments. 3
RH 331 Note Set 12.1 F2014abn Plates and Slabs If the frame is rigid or non-rigid, the floors can be a plate or slab (which has drop panels around columns). These elements behave differentl depending on their supports and the ratio of the sides. one-wa behavior: like a wide beam, when ratio of sides > 1.5 two-wa behavior: comple, non-determinate, look for handbook solutions Floor Loading Patterns With continuous beams or floors, the worst case loading tpicall occurs when alternate spans are loaded with live load (not ever span). The maimum positive and negative moments ma not be found for the same loading case! If ou are designing with reinforced concrete, ou must provide fleure reinforcement on the top and bottom and take into consideration that the maimum ma move. 4
18 ft RH 331 Note Set 12.1 F2014abn Eample 1 40 k 15 ft The rigid frame shown has been analzed using an advanced structural analsis technique. The reactions at support are: = -28.6 k, = -15.3 k, M = 208 k-ft. The reactions at support D are: D = -11.4 k, D = 15.3 k, M D = 110 ft-k. Draw the shear and bending moment diagrams, and identif V ma & M ma.. 12 ft 18 ft Solution: NOTE: The joints are not shown, and the load at joint is put on onl one bod. D 40 k 15 ft 12 ft 208 k-ft 15.3 k 28.6 k 110 k-ft D 11.4 k 15.3 k deflected shape (based on +/- M) 5
RH 331 Note Set 12.1 F2014abn Eample 2 The rigid frame shown has been analzed using an advanced structural analsis technique. The reactions at support are: = 2.37 kn, = 21.59 kn, M = -4.74 knm. The reactions at support are: = -2.37 kn, = 28.4 kn, M = -26.52 knm. Draw the shear and bending moment diagrams, and identif V ma & M ma. Solution: M M M M 10 kn/m 26.52 knm 2.37 kn 28.41 kn 6 m 10 kn/m 6 m 4.74 knm 2.37 kn 21.59 kn Reactions These values must be given or found from non-static analsis techniques. The values are given with respect to the global coordinate sstem we defined for positive and negative forces and moments for equilibrium. Member End Forces The free-bod diagrams of all the members and joints of the frame are shown above. The unknowns on the members are drawn positive on one bod for a joint, and the opposite directions are drawn on the other bod of for the joint. We can begin the computation of internal forces with either member or, both of which have onl three unknowns. Member With the magnitudes of reaction forces at known, the unknowns are at end of,, and M, which can get determined b appling F 0, F 0, and M 0. Thus, F 2. 37kN 0 = -2.37 kn, F 21. 59kN 0 = -21.59 kn M 2. 37kN( 6m) 4. 74kN m M 0 M = -9.48 knm Joint The joint is shown to illustrate that the forces and moments on the cuts will be equal and opposite, resulting in the forces and moment at joint be on bod to be opposite from those on bod : = 2.37 kn, = -21.59 kn and M = -9.48 knm Member ll forces are known, so equilibrium can be checked: F 2. 37kN 2. 37kN 0 21. 59kN 28. 49kN ( 10kN / m) 5m 0 M F 28. 41kN( 10kN / m( ( 2. 26. 52kN m9. 48kN m 0 21.59 kn 23.3 knm (2.16 m) (2.84 m) -28.41 kn -9.48 knm -9.48 knm -26.52 knm -2.37 kn 4.74 knm 6 deflected shape (based on +/- M)
30.95 knm -7.42 kn -6.12 knm RH 331 Note Set 12.1 F2014abn Eample 3 The rigid frame shown has been analzed using an advanced structural analsis technique. The reactions at support are: = -3.42 kn, = -5.44 kn, M = 12.63 knm. The reactions at support D are: D = 7.42 kn, D = 25.44 kn, M D = -6.12 knm. Draw the shear and bending moment diagrams, and identif V ma & M ma. 800 N/m 8 m 20 kn 60 kn-m Solution: M 0.8 kn/m 12.63 knm 3.42 kn M 8 m 20 kn M M D 6.12 knm 60 knm D 7.42 kn D 5.44 kn 25.44 kn Reactions The values given were found from non-static analsis techniques and put on the FD s. Member End Forces The free-bod diagrams of all the members and joints of the frame are shown above assuming joint force directions on one bod and opposite directions on the joining bod. ecause there is a force and moment at, these must be put on onl one bod of the joint, and it doesn t matter which bod. Member F 3. 42kN 0. 8kN / m5m 0 = 7.42 kn Joint F 5. 44kN 0 = 5.44 kn M 3. 42kN( 0. 8kN / m( ( 2. 12. 63kN m M 0 M = 14.47 knm M = M = 14.47 knm Member F 7. 42kN 0 = 7.42 kn Joint F 5. 44kN 0 = 5.44 kn M 14. 47kN m5. 44kN( 8m) M 0 M = -29.05 knm M = M D = -29.05 knm Member D 7. 42kN 7. 42kN 0 F (checking) 5. 44kN 20kN 25. 44kN 0 F ( 29. 05kN m) 60kN m6. 12kN m7. 42kN( 0 M 7.42 knm -5.44 kn 14.47 knm 14.47 knm -29.05 knm 3.42 kn -12.63 knm 7 deflected shape (based on +/- M)
RH 331 Note Set 12.1 F2014abn Eample 4 Find the column effective lengths for a steel frame with 12 ft columns, a 15 ft beam when the support connections are pins for a) when it is braced and b) when it is allowed to swa. The relative stiffness of the beam is twice that of the columns (2I). 12 ft I 2I I 15 ft 8