Subfculty of Civil Engineering rk ech ge with your: Structurl echnics STUDENT NUBER : NAE : Em CT309 STRUCTURAL ECHANICS 4 ril 0, 09:00 :00 hours This em consists of 4 roblems. Use for ech roblem serte sheet of er. Do not forget to mention your nme nd number on ech er. Work net nd tidy, the qulity of the resenttion cn be used in the grding. The use of Phone s or comuters, PDA s nd /or Wifi or Blue Tooth equiment is not llowed. Turn off the equiment nd remove it from your tble. A scientific (rogrmmble) clcultor is llowed All required formuls cn be found on the lst ges of this em Kee n eye on the clock nd use the secified times er roblem s guidnce.
Em CT309 Structurl echnics 4 r 0 Problem : Plsticity ( 45 min ) In the following figure frme is shown which is fied in A nd suorted in B with horizontl roller. Concentrted forces ct t D nd E in the directions shown. Segment ACD hs twice the strength of segment BED. C D E 3 A B Questions: ) Determine the ossible collse mechnisms nd show these with smll sketches. b) Comute the collse lod nd rove the uniqueness of your solution. c) Show the moment distribution t the moment of collse. - -
Em CT309 Structurl echnics 4 r 0 Problem : Influence lines ( 35 min ) The stticlly determinte hinged bem structure form the figure below consists of three bems which re connected though hinges t S nd S. A B S S C D -s z-s 5,0 m,0 m 3,0 m 3,0 m 3,0 m igure : Sttic determinte hinged bem Questions: ) Elin how n influence line for force quntity cn be determined. b) Elin how n influence line for dislcement quntity cn be determined. c) Construct the influence line for the moment t B. d) Construct the influence line for the moment t C. e) Construct the influence line for the sher force directly to the left of C. f) Construct the influence line for the sher force t mid sn of sn AB. g) Construct the influence line for the for the suort rection t C. h) Sketch the influence line for the dislcement of hinge S. i) rk the osition of constnt distributed lod on the structure in order to mimize the bending moment t one of the suorts. - 3 -
Em CT309 Structurl echnics 4 r 0 Problem 3 : work nd energy methods ( 45 min ) A hinged bem with rottionl sring ttched t A is loded by concentrted lod nd distributed lod q. S is hinge connecting the two bem rts loded in bending. The bending stiffness for the totl structure is constnt nd denoted with. The sring stiffness cn be eressed in terms of the bending stiffness by using dimensionless sclr: r ρ ρ r The deformtion energy stored in rottionl sring cn be eressed s: E v r veer r S q A B C Questions: ) ind the moment distribution fort his structure nd drw this oment-line. b) ind the verticl dislcement of the hinge S with work or energy method. Write the dislcement in terms of the given rmeters nd use the given sclr ρ for the stiffness of rottionl sring. c) Secify the mount of deformtion energy stored in the structure due to the secified lod nd eress this in terms of the rmeters, q,, ρ en. d) ind for : 0000 knm ; ρ, 4 m; q 4 kn/m nd 0 kn the deflection t S in mm nd the mimum moment in knm. e) ind the mount of deformtion energy for very smll vlues of r. HINT : use sketch of the deformed structure! - 4 -
Em CT309 Structurl echnics 4 r 0 Problem 4 ( 45 min ) An element subject to bending nd sher is comosition of two mterils. The geometry of the cross section is symmetricl nd shown in the figure below. Both mterils re erfectly bonded nd the cross section cn be regrded s n inhomogeneous cross section. The dimensions nd Young s moduli of the mterils cn be found below. The bem is coincides with the -is nd the origin of the coordinte system is tken t the norml force center of the cross section. The cross section is loded in the -z-vlk with sher force of 9 kn nd bending moment of 90 knm. y NC? mteril mteril R S z U T Given : 3 E 00 0 N/mm ; 3 E 800 0 N/mm ; 300 mm Questions: ) Why is the origin of the coordinte system tken t the norml force center? b) ind the loction of the norml force center. c) The constitutive mtri of the cross section is given below. Show the correctness of this mtri. 0, 6 0 0 0 080 430 0 0 430 080 units N, mm d) ind the osition of the neutrl is, the lne of loding m nd the lne of curvture k nd drw these lines in one grh. e) Show in second grh the stress distribution of the cross section for both mterils. f) ind the totl longitudinl force in the interfce RST between mteril nd er unit of length. g) ind the most outer left oint of the kernel of this cross section nd mrk this oint in one of the grhs of the cross section. h) In which lne will this cross section hve the mimum bending stiffness nd wht is the mgnitude (vlue) of this stiffness? - 5 -
Em CT309 Structurl echnics 4 r 0 ORULAS Inhomogeous nd/or unsymmetricl cross sections : ε ( y, z) ε + κ y + κ z nd: σ ( y, z) E( y, z) ε ( y, z) y z y yy yz κ y ey yy yz / y nd κ e EA / z z yz zz z z yz zz V ES s s V ( ) ( ) ( ) ( ) ( ) y y VzESz ( ) R s ; or: ; σ t ( ) yy zz b yz tn α ; + ± + ( ) ( yy zz ) ( ) ( ) ( ), yy zz yy zz yz Deformtion energy: Kinemtic reltions: du Ev EAε d (etension) ε d Ev κ d (bending) d w κ d Comlementiry energy: Constitutive reltions: N Ec d (etensie) EA N EA. ε. κ Ec d (buiging) Cstiglino s theorem s: Work method with unit lod: Ev Ec ( ) m( ) d i ui u u i i - 6 -
Em CT309 Structurl echnics 4 r 0-7 -
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Em CT309 Structurl echnics 4 r 0 Problem : Plsticity ANSWERS ) The given structure is stticlly indeterminte with degree of one. To obtin mechnism, two dditionl hinges re required. Hinges cn occur t four ositions. Thus, si mechnisms hve to be investigted. However, moments cn not occur t E due to the loding nd suort conditions t B thus only three mechnisms re left to be considered. Possible mechnisms re shown below. Ech mechnism is shown in detil on the following ges. 3 4 5 4 9-9 -
Em CT309 Structurl echnics 4 r 0 θ 5θ δ A 0 4 δθ δθ 0(!) + 5δθ 0 5 θ θ δ A 0 4 δθ δθ δθ + δθ + 7δθ 0 9-0 -
Em CT309 Structurl echnics 4 r 0 θ θ θ 3 δ A 0 δθ δθ δθ + δθ + δθ 0 ) The ultimte lod is the lowest lod found: 4 9 0,444 b) To find the correct moment distribution strt with the suort rections. The moment distribution shows stticlly dmissible distribution in which t no cross section the bending moments eceeds the utltimte lstic ccity of the cross section. (uniqueness theorem) 9 4 9 7 8 8 - -
Em CT309 Structurl echnics 4 r 0 Problem : Influence lines A B S S C D -s z-s 5,0 m,0 m 3,0 m 3,0 m 3,0 m,0 q q q B m, q 5 5,0 i-lijn B q (etreme loding) 3,0 q q 3 6 9 C m, q,0 i-lijn C,0 i-lijn V C-L 0,5 0,4 i-lijn V AB 0,5 i-lijn C V,0,0,0 i-lijn A V 8/ - -
Em CT309 Structurl echnics 4 r 0 Problem 3 : Work nd Energy ethods ) The moment distribution cn be found bse don equilibrium. The distributed lod q is in fct blnced thus only the concentrted lod is contributing to the moment in the sring nd the member AS. ( first yers knowledge!!) r ρ ρ r r S q A B C Q 7 knm -line due to nd q -line due to Q 40 r Q S q b) To find the dislcement t S dummy lod Q is required. In red the moment distribution due to the dummy lod is resented. Since the moment distribution over SB nd BC re ehibiting mirror symmetry, only rt SB hs to be emined. Using locl coordinte for rt AS nd SB denoted with nd, the deformtion energy cn esily be eressed in terms of, q nd Q : (( )( )) + q Q v 0 0 Ev + d + d r ( Q) v met: Using Cstiglino s theorem solves the deflection t S: (note: nd integrl is irrelevnt) w S A ( ) ( ) E Q Q v Q ρ 3 3 3 The dummy lod is zero which results in: Hinge S will move uwrds due tot the lod. w S 3 3 3 (3 + ρ) 3ρ ρ 3 c) Using the given vlues results in dislcement of 0,0533 m uwrds. This dislcement is indeendent of the distributed lod q! B - 3 -
Em CT309 Structurl echnics 4 r 0 d) The totl deformtion energy stored due to the loding is: (note : nd integrl is essentil) + q ( ( )) v 0 0 Ev + d + d r E v ( 0 ( + ρ) + ρ(5 + ρ ) 3 q q 0ρ e) If the sring stiffness becomes very smll, oint A turns into hinge. A mechnism will occur. The lod cn move infinitely nd thus roduce infinite work. The bending deformtion will be smll comred to the deformtion energy stored in the sring. So rcticlly ll deformtion energy hs to be tken by the sring(!) which results in hrdly ny deformtion energy (nd curvture) in the elements loded in bending. The brs will remin stright: E ( ) w r v lim ρ 0-4 -
Em CT309 Structurl echnics 4 r 0 Problem 4 : Non-symmetricl cross sections ) See the lecture notes. b) The il stiffness of the cross section cn be found with: EA E E + E 6 0 The origin of the coordinte system used is locted t the NC. The verticl osition of the NC with resect to the uer side of the cross section is: z NC 400 mm The horizontl osition with resect to the left side of the cross section is: y NC 00 mm c) The cross sectionl constitutive reltion reltes the sectionl forces to the deformtions if the cross section. The bending stiffnesses cn be found using the strtegy outlined in the lecure notes. This emle is very bsic so only nswers re resented here: N EA 0 0 ε y 0 yy yz κ y cross sectionl constitutive reltion [N, mm] z 0 yz zz κ z 0, 6 0 0 0 K 0 0 430 080 f 0 6 0 080 430 90 0 d) Since this structure is loded in bending only, the strin ε t the NC must be zero. The curvtures cn be found with the constitutive reltion: N ε 0 EA κ 0,79 0 ( ) y zz y yz z yy zz yz κ + 0, 476 0 ( ) z yz y yy z yy zz yz The direction of the lne of loding nd the lne of curvture cn be obtined with: κ α α α α z o z o tn m m 90 ; tn k k 76 y κ y The stresses for ech oint of the cross section cn be comuted with: σ ( y, z) E ( ε + κ y + κ z) N/mm y z The neutrl is n.. cn lso be found with this ltter eression by: ε ( y, z) ε + κ y + κ z 0 κ y + κ z 0 y 4z 0 y z y z e) The stress distribution cn be visulized with few oints. Only the four vlues mrked in bold in the tble on the net ge were essentil for the grhs. - 5-9 6 6
Em CT309 Structurl echnics 4 r 0 Tbel : Stress in the secified oints teril oint y [mm] z [mm] E [N/mm ] Stress [N/mm ] R 00-00 00000-4, S -00-00 00000-7, T -00 00 00000, V -400 00 00000 8,3 W -400-400 00000-8,3 X 00-400 00000-4,4 R 00-00 800000-7,3 S -00-00 800000-63,7 T -00 00 800000 9,0 U 00 00 800000 7,3 The neutrl is goes through the NC since the norml force N is zero. The red stress distribution reresents the stresses in mteril nd the blue one reresents mteril. The moment nd thus the lod cts in the -m lne. The curvture κ cts in the -k lne. f) The longitudinl force er unit length of bem in the interfce between mteril nd cn be obtined with: 7 n 4 - + 8 mteril 9 y X R U k NC m mteril S T z W n V teril (RSTU) is tken s the sliding element with cross sectionl re (): ( ) ( σ σ σ σ ) ( ) R 4 R S T U s V Vz z s + + + 4 7,3 300 300 3 6 9 0 86 N/mm 90 0 g) The outer left kern oint cn be found by tking neutrl is long VW. The loction of the kernel oint cn be found with: ey yy yz /( 400) 50 mm e z EA yz zz 0,5 h) The rincile direction of this cross section is t n ngle of 45 degrees. or the rincile coordinte system this cross section hs one is of symmetry. The mimum stiffness of this section then becomes: ( ) ( ) ( ) + ± +, yy zz yy zz yz 5400 0 Nmm ; 340 0 Nmm imum bending stiffness is therefore 5400 0 Nmm. - 6 - k