Unformatted text preview: SECTION 10.4 Method of Superposition 643 Problem 10.4-2 The propped cantilever beam shown in the figure supports a uniform load of intensity q on the left-hand half of the beam. Find the reactions RA, RB, and MA, and then draw the shear-force and bending-moment diagrams, labeling all critical ordinates. q MA RA A B L — 2 L — 2 RB Solution 10.4-2 Propped cantilever beam SHEAR-FORCE AND BENDING-MOMENT DIAGRAMS RB MA qL 8
2 Select RB as redundant. EQUILIBRIUM RA qL 2 RBL V RA RELEASED STRUCTURE AND FORCE-DISPLACEMENT
RELATIONS O x1 RB q A L 2 L 2 B ( B)1 7qL4 384EI x1 57L 128
Mmax A L B RB ( B)2 RBL3 3EI M O MA COMPATIBILITY B ( B)
1 1 ( B)
2 2 0 Mmax Substitute for ( B) and ( B) and solve for RB: RB 7qL 128 945qL2 32,768 OTHER REACTIONS (FROM EQUILIBRIUM) RA 57qL 128 MA 9qL2 128 Problem 10.4-3 The figure shows a propped cantilever beam ABC having span length L and an overhang of length a. A concentrated load P acts at the end of the overhang. Determine the reactions RA, RB, and MA for this beam. Also, draw the shear-force and bending-moment diagrams, labeling all critical ordinates. MA A RA L B P C RB a 644 CHAPTER 10 Statically Indeterminate Beams Solution 10.4-3 Beam with an overhang Select MA as redundant. EQUILIBRIUM RELEASED STRUCTURE AND FORCE- SECTION 10.4 Method of Superposition 651 Solution 10.4-11 Beam supported by a beam COMPATIBILITY ( B)
1 Let RB interaction force between beams Select RB as redundant. RELEASED STRUCTURE AND FORCE-DISPL. EQS.
( B) 1 L 2 ( B)2 A B P C ( B)1 5PL3 48EI ( B) 2 ( B) 3 Substitute and solve: RB 40P 17 SYMMETRY AND EQUILIBRIUM RD RA RE P RB 2 RD 20P 17 RE A L 2 B C ( B)2 RB RB ( B)3 E 23P 17 (minus means downward) MA RB ¢ L ≤ 2 PL 3PL 17 MB MB PL 2 5PL 17 RBL3 24EI RBL3 384EI BEAM ABC: Mmax BEAM DE: Mmax Mmax PL 2 D L 4 L 4 Problem 10.4-12 A three-span continuous beam ABCD with three equal spans supports a uniform load of intensity q (see figure). Determine all reactions of this beam and draw the shear-force and bending-moment diagrams, labeling all critical ordinates. q A D B C RA L RB L RC L RD Solution 10.4-12 Three-span continuous beam FORCE-DISPLACEMENT RELATIONS 11qL4 5 RB L3 ( B)1 ( B)2 12 EI 6 EI COMPATIBILITY
B SELECT RB AND RC AS REDUNDANTS. SYMMETRY AND EQUILIBRIUM RC RB RA RD 3qL 2 RB ( B)1 ( B)2 0 ∴ RB RELEASED STRUCTURE
q FORCE-DISPLACEMENT RELATIONS 11qL 10 OTHER REACTIONS
D ( B)1 A 11qL 5 RB L (2 B ( B )B)1 12 EI 6 EC I
L L () (B B)2 B 2 0 RB 4 3 From symmetry and equilibrium: RC RB RD 11qL 10 2 qL 5 L COMPATIBILITY
B ( B)1 ∴C R B 11qL 10
D RC (Continued) 652 CHAPTER 10 Statically Indeterminate Beams LOADING, SHEAR-FORCE, AND BENDING-MOMENT DIAGRAMS
A D B C RA L RB L RC L RD 2 5 V qL O 2L 5 1 2 3 5 2L 5 2 5 2 25 3 5 M qL2 O 2 25 1 40 1 2 MB Mmax
1 10 1 10 MC 2 qL2 25 qL2 10 Problem 10.4-13 A beam AC rests on simple supports at points A and C (see figure). A small gap 0.4 in. exists between the unloaded beam and a support at point B, which is midway between the ends of the beam. The beam has total length 2L 80 in. and flexural rigidity EI 0.4 109 lb-in.2 Plot a graph of the bending moment MB at the midpoint of the beam as a function of the intensity q of the uniform load. Hints: Begin by determining the intensity q0 of the load that will just close the gap. Then determine the corresponding bending moment (MB)0. Next, determine the bending moment MB (in terms of q) for the case where q q0. Finally, make a statically indeterminate analysis and determine the moment MB (in terms of q) for the case where q q0. Plot MB (units of lb-in.) versus q (units of lb/in.) with q varying from 0 to 2500 lb/in. q A B RA RB L = 40 in. L = 40 in. = 0.4 in. C RC SECTION 10.4 Method of Superposition 653 Solution 10.4-13 q0 (MB) Beam on a support with a gap ( B) COMPATIBILITY B 1 4 3 5qL RB L or ¢ RB 24 EI 6 EI EQUILIBRIUM
q load required to close the gap magnitude of gap
0 bending moment when q q q0 q0 ( B) 2 5qL 4 6 EI ¢ L3 CASE 1 RA
C RC RC RAL 2RA 3qL 8 qL2 2 2qL 3EI ¢ L3 3EI ¢ L2 RB 0 A B RA MB RA RB L L qL2 8 RC NUMERICAL VALUES 0.4 in. L 40 in. EI 0.4 109 lb-in.2 Units: lb, in. From Eqs. (1) and (2): q0 300 lb in. (MB)0 240,000 lb-in. For q q0: MB 800 q (3) For q q0: MB 300,000 200 q (4) B MB RA 5 qL4 24 EI qL2 2 RC qL q
4 CASE 2 q0 (1) (2) 5q0 L 24 EI ¢ ¢ q0 B 24 EI 5 L4 2 q0 L 12EI ¢ ( MB ) 0 2 5 L2 CASE 3 q q0 (statically indeterminate) Select RB as redundant. RELEASED STRUCTURE
q A GRAPH OF BENDING MOMENT MB (EQS. 3 AND 4)
MB (lb-in.) 300,000 200,000 100,000 O 100,000 200,000 q0 (MB)0 240,000 EQ. (3) EQ. (4) 1000 MB q 0 at 1500 2000 q (lb/in.) 300 ( B)1 C A ( B)2 C RB ( B)1 ( B)2 5qL4 24EI RBL3 6EI 654 CHAPTER 10 Statically Indeterminate Beams Problem 10.4-14 A fixed-end beam AB of length L is subjected to a moment M0 acting at the position shown in the figure. (a) Determine all reactions for this beam. (b) Draw shear-force and bending-moment diagrams for the special case in which a b L /2. M0 A MA RA a L b RB B MB Solution 10.4-14 Fixed-end beam (M0 applied load) COMPATIBILITY
B Select RB and MB as redundants.
M0 A MA RA a b RB B MB ( B) 2RBL3 ( B) RB L2
1 1 or
B 2 3MBL2 ( B) ( B) ( B) 3 0 2b) (1) 3M0a(a
3 ( B) 2 0 (2) or 2MB L 2M0 a L a b SOLVE EQS. (1) AND (2): RB MA MB RBL M0 FROM EQUILIBRIUM: 6M0 ab L3 M0 b ( 3a L2 b MA
3M0 2L O RB ( B)2 RBL3 3EI B ( B)3 MB ( B)3 MBL2 2EI M O M0 4 L 6 M0 2 M0 4 EQUILIBRIUM RA RB 6M0 ab L3 MB M0 a ( 3b L2 L) RELEASED STRUCTURE AND FORCE-DISPL. EQS.
( B)1 ( B)1 A ( B)1 M0a EI M0 ( B)1 B M0a (a 2EI 2b) RA MA L) SPECIAL CASE a RA RB 3M0 2L L2 MB M0 4 ( B)2 A ( B)2 RBL2 2EI B ( B)2 V A M0 2 ( B)3 MBL EI ( B)3 ( uB ) 1 ( uB ) 2 ( uB ) 3 M0 a ( B)1 EI RB L2 ( B)2 2 EI MB L ( B)3 EI M0 a (a 2 EI RB L3 3 EI MB L2 2 EI 2b ) ...
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