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专业好文档土木工程力学(本)$ U! ?1 5 g6 : c9 x形成性考核册& Jp* z, |- m! n+ Z/ OP1 Q: a作业一! w4 E0 C9 X5 w k; o; 4 |6 i说明:本次作业对应于平面体系的几何组成分析和静定结构的受力分析,应按相应教学进度完成。/ N8 E5 V; J n一、选择题(每小题2分,共20分), f3 b7 / |, l# 0 H5 a1三刚片组成几何不变体系的规则是( B )0 z; s& q% R9 j+ SA三链杆相联,不平行也不相交于一点 B三铰两两相联,三铰不在一直线上% - s. yq1 Y6 , A& . j! bC三铰三链杆相联,杆不通过铰 D一铰一链杆相联,杆不过铰1 8 e! x W6 C/ z6 b2在无多余约束的几何不变体系上增加二元体后构成( C )3 H7 0 : Q2 q7 . sYA可变体系 B瞬变体系7 5 Y$ x X8 oC无多余约束的几何不变体系 D有多余约束的几何不变体系1 H. q1 v& u; h+ b5 N0 b5 z: d0 G4 V3瞬变体系在一般荷载作用下,( C )* g, R4 E7 I% U0 JA产生很小的内力 B不产生内力# : S. |1 d6 V6 ; D+ p9 C产生很大的内力 D不存在静力解答8 z, C F/ ; W) J3 O0 7 hU& Q4已知某体系的计算自由度W=-3,则体系的( C)7 m8 H) ?! |! F- F+ LA自由度为3 B自由度等于08 s0 W& L9 n; |C 多余约束数等于3 D 多余约束数大于等于34 P+ % n: V7 l! U8 S2 j( j5不能作为建筑结构使用的是( D ), h0 Lt5 T; y9 V( B9 E0 a4 j l tA无多余约束的几何不变体系B有多余约束的几何不变体系; A; h1 e1 d+ T+ h3 M, mC 几何不变体系 D几何可变体系0 v& , j v5 o0 w* s: l7 V6图示桁架有几根零杆(D )7 S a# n( D; L6 aA0 B21 9 c- k* s X9 u$ o& CC4 D68 B6 p, w k; p& U: G7下图所示结构的弯矩图形状应为(A )* p8 r3 z6 g8 Y8图示多跨静定梁的基本部分是( B)8 z! l& a- c9 o2 UWe. 5 uAAB部分 BBC部分* x; O1 J; T. i. E1 lCCD部分 DDE部分: B% Z8 p6 a5 I$ B# M3 k8 U9荷载作用下产生桁架位移的主要原因是(A )$ n# L3 K: k9 h! 0 z/ G8 MA轴向变形 B弯曲变形3 ?9 q, Q8 - # n5 f y$ g0 ?7 gC剪切变形 D扭转变形1 _) Q1 + S& V0 O$ t+ E10三铰拱在集中力作用下其合理拱轴线形状是( A)6 t6 ! k4 G# t* D- g PA折线 B圆弧% r t/ r6 u! 2 _C双曲线 D抛物线 u+ f; f3 d1 U9 w x二、判断题(每小题2分,共20分), C, _& Yx* J) e1多余约束是体系中不需要的约束。()# A9 K& q+ U Z9 R. S R2如果体系的计算自由度大于零,那么体系一定是几何可变体系。()2 9 ?3 C: Y3 m- h, 3两根链杆的约束作用相当于一个单铰。()+ ?, R/ r; O6 Ho$ * G4 / G4一个体系是有n个自由度的几何可变体系,那么加入n个约束后就成为无多余约束的几何不变体系。()/ T! _2 ! J X5两刚片用三链杆相联,且三链杆平行不等长,则构成瞬变体系。()! u, i$ Cp$ e Z( s5 6图示两个单跨梁,同跨度同荷载。但横截面形状不同,故其内力也不相同。()) x2 K) v _9 D7三铰拱的矢高f越大,水平推力也越大。()1 D/ y, u( X; J6 c8求桁架内力时截面法所截取的隔离体包含两个或两个以上的结点。() R4 ) k: G* G! 9某荷载作用下桁架可能存在零杆,它不受内力,因此在实际结构中可以将其去掉。(). B% ?7 u6 V H10试判断下列弯矩图是否正确。() C! M4 Q j4 三、试对图示平面体系进行几何组成分析。(每小题5分,共20分)/ x6 5 L4 G; v( q C& t; t(1)无多余约束的几何不变体系(2)无多余约束的几何不变体系+ l* X- e( B+ q1 j( T0 _, Q; N. c0 O4 u+ q2 a9 _H$ $ X7 s(3)有2个多余约束的几何不变体系(4)无多余约束的几何不变体系3 z+ a3 W6 & s9 p7 v3 j+ , o( C) 0 v5 l0 L四、绘制下图所示各结构的弯矩图。(每小题10分,共30分)5 V7 P4 T: _h) U, 4 g4 _! D(1)- u8 v* O9 V8 e参考答案:) d+ S3 e; t V# L3 4 T h+ a: o% b2 E(2)E, r6 W9 z/ P2 M0 Z1 U参考答案:% U3 A4 F& b0 f3 , _$ . i& S9 i* x4 O* x+ W0 N# O(3)4 h, E1 d9 P( c/ |# 40kN9 d4 p3 M5 N1 O+ S40kN, $ q0 P/ n& * r: T- |4 D20kN/m0 v1 s5 m4 o+ M; Z( 0 A% z2m* p: ?1 F6 g/ d) r/ Y- D0 z2m% R Y5 H. y/ v- |, 2m5 t1 ?; r3 b8 8 U: I X2 t2m. c/ K) D7 ( * O/ n# s2m N9 i& E% E% D: & ) q2m( c# g# / a. hA% % J5 k7 t+ I+ # vC* T& r, g+ W/ z; S- L! ?$ QB3 x8 U3 B7 Z( I/ sXD. o8 i2 N& - N& ?% U# f2 cE: - w4 % x6 J参考答案:# : T7 N. |- A单位:4 l/ |. ?- M7 Y6 N6 R- s五、计算图示桁架中指定杆件的内力。(10分)8 r! L( n& f6 k1 y/ U7 6 g6 FP) B! O8 Z% O5 T* D+ DFP2 |l& y$ x% e& B9 I! 5 |15 |) O8 U+ H& L6 R. T35 L8 7 X0 6 L$ x. t1 q2D H4 t* U I+ b8 V o. a% D/ c/ f6 K5 7 v/ y5 Q0 D n/ ga! E2 W& e7 . L5 B7 1 za9 S : s) Fd- F3 ximg,40,34file:/C:/DOCUME1/ADMINI1/LOCALS1/Temp/msohtml1/01/clip_image032.gif/img参考答案:,6 J. E: G1 r# V, o% k! o7 N2 l- F0 i/ % 1 E& B d0 t6 q m5 I5 K2 C作业二. l+ N. |6 / z说明:本次作业对应于静定结构的位移计算和力法,应按相应教学进度完成。) K/ G$ v* _- J, I6 n A一、选择题(每小题2分,共10分)7 Z# 2 E( p8 # r! ?1用力法计算超静定结构时,其基本未知量为( D)# r3 m F$ g9 c$ IA杆端弯矩 B结点角位移5 T& Y* u# m) z& uC结点线位移 D多余未知力$ t8 D. D( w! |7 A L; W9 b/ q 2力法方程中的系数代表基本体系在作用下产生的(C )5 X: M1 7 # . N4 dA B$ R: x# A3 z6 j0 lC方向的位移 D方向的位移2 B% V+ # A7 s/ D3在力法方程的系数和自由项中(B)( n T& T5 ? s7 M. / C% E) NA恒大于零 B恒大于零1 G G) T* # f1 L7 d( + L! H7 PC恒大于零 D恒大于零; p# h; t8 T5 c5 |) f4下列哪一条不是图乘法求位移的适用条件?(D )2 X: V: F9 I3 5 A直杆 B EI为常数4 i3 P/ T3 T( p) v! * SC、至少有一个为直线形 D 、都必须是直线形 Qy; Z. e; _/ v5下图所示同一结构在两种不同荷载作用下,它们之间的关系是(A) M2 C4 6 g $ g$ p! a% S AA点的水平位移相同 B C点的水平位移相同# N J9 E$ i/ WCC点的水平位移相同 D BC杆变形相同0 z H1 d( ; n6 h4 C; i0 + ?9 n4 E6 k& h& b, ?7 二、判断题(每小题2分,共10分); _N- a2 V1 L: X. K! 5 1静定结构由于支座移动引起的位移与刚度无关。()& o% y1 r- n( B: x k2反力互等定理仅对超静定结构才有使用价值。(). t) 6 V* w. 7 J K7 # j3用力法求解超静定刚架在荷载和支座移动作用下的内力,只需知道各杆刚度的相对值。()2 e9 s2 B_& f7 R: z* D( * F. V# p4同一结构的力法基本体系不是唯一的。()+ t) - : b1 8 N7 J5用力法计算超静定结构,选取的基本结构不同,则典型方程中的系数和自由项数值也不同。() # J) p0 K* J. n7 - E三、求图示简支粱C点的竖向位移,EI =常数。(9分)1 f7 x( S: |; ) H* q, A |, * d4 Z9 A$ FrC! L) f3 N* x9 1 j/ l) o9 _2l/38 h. S4 H6 x! |5 Bl/3. L# y. , ? S! X* i+ W/ s参考答案:()% B! k2 x0 t% C W/ g+ # v4 no3 Kq8 9 m/ E% Z& hB( G- K1 OQ qC 6 % Y) Z m8 el6 N2 P+ T$ E A* z3 * img,40,35file:/C:/DOCUME1/ADMINI1/LOCALS1/Temp/msohtml1/01/clip_image068.gif/img四、计算图示刚架结点C的水平位移和转角,EI =常数。(9分)参考答案:(),(顺时针)# s% i* |# r$ i! K, P7 s9 e1 Uv m7 l4 : u五、试求图所示刚架点D的竖向位移。为常数。(9分)+ - # V4 g. p3 8 R9 P( 8 v0 L6 B3 T% g参考答案:虚设与所求位移相应的单位荷载,并画、图,如图所示。6 a: MP( x3 q. 图的面积:0 F: z& JZe# K+ G图中相应的标距:& U3 p3 m P5 l! H) y可得( S4 E8 q7 h# z5 tH: F% ( J7 A- 7 r L l7 q# Z( ?x六、求图示桁架结点B的竖向位移,已知桁架各杆的EA=21104kN。(9分)X5 B+ z; S* u- T4 T# b5 p6 W+ P) L, r7 0 D: G* h+ Y1 M& x参考答案:0 T4 i! l; R4 M2 X, + l# n七、确定下列结构的超静定次数。(4分) $ D) Fs * u% s) O% A! v& f5 g, M+ ) t(a) (b), K: O& ( V$ g7 b; q, (c) (d): $ W+ S1 g T$ L7 k8 C参考答案:(a)5次(b)1次(c)4次(d)7次5 m) L3 l8 i+ ?9 d! i- S; j八、用力法计算图示结构,并作弯矩图。各杆相同且为常数。(10分)( D+ S5 * P3 p! M+ e: t1 Y3 3 参考答案: 3 L) w U! K. h n! j* K图(单位:) 图(单位:)7 y! Z7 W o7 a9 / u7 Y/ b图(单位:) Nk; p3 L r3 Y九、用力法计算下列刚架,并作弯矩图。为常数。(10分)- x, G z m c$ P5 e% b. i参考答案:/ N* & b6 J8 w1 C2 m y;3 , I) r. |1 O. R8 ?# w% N,- w9 w4 a# h. k$ |7 n3 b/ f6 + s3 u; Y4 n7 7 # / B Q7 F/ b十、用力法计算图示结构,作弯矩图。链杆。(10分)- U1 w% C5 1 j参考答案:2 R! v2 B/ Y8 z* R) T;! y1 N9 oo& S1 E; ?! Q1 d8 f, ( N( 3 十一、利用对称性计算图示刚架,并绘制弯矩图。(10分)8 E2 n. K, C; l- S2 H/ ?参考答案:q# D y! $ J利用对称性) I! D8 M% j6 n3 b& g# C$ C H: g0 oR5 p& ?# _9 j* N L3 j( Q2 K! G4 T F& : 5 f% c! i9 * K4 b3 RV1 x; p4 i. l9 O/ t8 O l8 G作业三* t6 |5 I( ?9 E0 X# G说明:本次作业对应于位移法和力矩分配法,应按相应教学进度完成。/ i + Z! + W一、选择题(每小题2分,共10分)O! 7 g) ) l2 Y8 B$ _# k. C1位移法典型方程实质上是( A)6 H0 h0 T8 F8 P1 |. u8 AA平衡方程 B位移条件2 U$ S% o* L5 H% g; 3 bC物理关系 D位移互等定理5 H) h: d; g! c% 0 X2位移法典型方程中的系数代表在基本结构上产生的(C ) G4 + tE9 B) ?A B1 X5 U( I1 G) h, qC第i个附加约束中的约束反力 D第j个附加约束中的约束反力8 s- & k- O* g1 z( & P) m3用位移法计算刚架,常引入轴向刚度条件,即“受弯直杆在变形后两端距离保持不变”。此结论是由下述假定导出的(D)6 k* K1 m, J; L- g% u1 d$ v2 |- S$ M4 uA忽略受弯直杆的轴向变形和剪切变形 B弯曲变形是微小的, 1 c, s7 W; E. i& GC变形后杆件截面仍与变形曲线相垂直 D假定A与B同时成立% m- k. q) Z9 n T- e9 M: v4在力矩分配法中传递系数C与什么有关( D)% 2 J/ T7 Q/ _. + v5 x% u oA荷载 B线刚度6 k1 Y$ v: h5 g+ k/ UC近端支承 D远端支承2 h$ T3 o v0 w+ D* p5汇交于一刚结点的各杆端弯矩分配系数之和等于(A )% G( _4 R) d# / YA1 B0$ H2 K6 j; s/ Z/ UC1/2 D-16 v g0 * f0 a) |! T$ v9 二、判断题(每小题2分,共10分) L4 k0 0 o9 K1位移法可用来计算超静定结构也可用来计算静定结构。() Z5 b8 w( d+ f4 U2图a为一对称结构,用位移法求解时可取半边结构如图b所示。( ) 2 |2 x, X7 ) O) h3 h7 ?5 w- Q( c+ M2 H2 v+ C- , B4 & 3 |4 t V( J b/ y* ?1 fg4 ! f2 N; d% 9 q! N) y4 O# ( l* S p, u0 F2 K: S图a 图b& q: k, 4 1 L( H% ) y3.用位移法计算荷载作用下的超静定结构时,采用各杆的相对刚度进行计算,所得到的节点位移不是结构的真正位移,求出的内力是正确的。()0 ?+ I. j( ! G2 w4在多结点结构的力矩分配法计算中,可以同时放松所有不相邻的结点以加速收敛速度。()6 h/ 1 y& a+ a( 1 u, I5力矩分配法适用于连续梁和有侧移刚架。()& W+ f. C/ I0 A, o( y( T8 n% L三、用位移法计算图示连续梁,并绘出弯矩图。各杆EI相同且为常数。(10分)$ I9 ?# Y+ u& q0 nf; l$ S) h, r6 V , S1 * j9 s; t% b) _2 p/ Y4 p1 j! y! e# O, * T参考答案:. v5 _) p9 U+ x+ Q+ L, 6 V. v) 9 U8 E7 _* + P4 L% t4 P& RI2 i# j; B# a9 i; a; g四、用位移法计算图示刚架,并绘制弯矩图。(10分)( xs$ X0 Y. C7 H I( q5 h7 ( 8 O* q1 4 N % L6 D参考答案:;; N6 H7 O4 ; M1 f+ d# C2 Q4 l; m/ m0 F. P五、用位移法计算图示刚架,并绘出弯矩图。(10分)+ X0 ! x1 y5 j8 x参考答案:利用对称性取半边结构如下图示/ 7 g8 f$ R* I# b# J* ?% H;; E1 Q K4 Z* F/ b3 : / S1 Y6 u: _: k# W5 g, D. Y ; M- H( # t$ ? , A) u1 T六、用位移法计算图示刚架(利用对称性),并绘出弯矩图。各杆EI相同且为常数。(10分)9 x) YY( M5 V W7 3 x P U; w* W参考答案:利用对称性取半边结构如下图示0 C% A+ S( w+ n/ w+ c u/ c_8 U: M3 r: C# r5 E七、用力矩分配法计算图示结构,并绘出弯矩图。各杆相同且为常数。(10分)6 ) a# y- X( f$ i B6 z0 r/ ?8 h9 s0 u# S% v3 N q9 b& C1 e, p ga( |& v5 L% |1 1 b( H ?7 八、用力矩分配法计算图示结构,并绘出弯矩图。各杆相同且为常数。(10分)0 i7 5 b& M3 I! ! J d- % d, c9 K. ?# K1 x7 ?$ i& p! E2 k4 a5 D4 & V! f3 P5 A& C) T5 # L九、用力矩分配法计算图示结构,并绘出弯矩图。各杆相同且为常数。(10分) R$ E3 A, v# o5 F: h0 o6 v$ B: g c十、用力矩分配法计算图示结构,并绘出弯矩图。各杆相同且为常数。(10分), |+ D# ?K( Z! X: B& Q8 C) _, q0 G S1 q 0 O m6 , C( fu1 J8 A- T. q) a作业四# n. Q( E4 O. Z$ X说明:本次作业对应于影响线及其应用和结构动力计算,应按相应教学进度完成。- v6 d( W# U! W7 i2 z一、选择题(每小题2分,共10分)& r, bS z+ s: K7 8 u- C2 m2 g d1静定结构的影响线的形状特征是(A )7 Y% 5 m0 r+ b+ M $ IA直线段组成 B曲线段组成M) l7 J2 D& P6 oC直线曲线混合 D变形体虚位移图1 l& U8 V9 T( v& y3 l8 m2图示结构某截面的影响线已做出如下图所示,其中竖标yc是表示(C)$ F1 s. ut0 I/ U/ s9 Z$ UA P=1在E时,C截面的弯矩值 B P=1在C时,A截面的弯矩值) V C; U( q! Z0 k: s6 QC P=1在C时,E截面的弯矩值 D P=1在C时,D截面的弯矩值. q0 |) . A/ 6 Z& d1 8 , S B j7 0 k8 N W, R8 ? N7 s B6 S$ 8 a0 a% n5 o, c; ( H+ : 7 H; h, Q. B I3 l1 E; O# Q( p+ y: g$ C( K( P, I/ + f3绘制任一量值的影响线时,假定荷载是(A)4 m9 FP8 x+ T$ W$ pA一个方向不变的单位移动荷载 B移动荷载+ g$ ; h% O) W) C动力荷载 D可动荷载5 Z$ A& dS* w, s4.如下图所示,若要增大其自振频率w值,可以采取的措施是(B)# e( # x- b. D( HA 增大L B 增大EI K- v$ n; C 增大m D增大P : A0 b, K7 A- a9 j% R# 4 V8 l1 n, z! X4 o5无阻尼单自由度体系在简谐荷载作用下,共振时与动荷载相平衡的是(D)/ B6 y! S* x3 T* e- x9 ) j& E) sA 弹性恢复力 B惯性力 8 H( u( H3 a) r- c. gbC 惯性力与弹性力的合力 D没有力) j) , r+ A) p) f8 U& Y二、判断题(每小题2分,共10分)G. p) k2 W5 g1静定结构和超静定结构的内力影响线均为折线组成。()6 , z j+ G8 O2图示结构 C 截面弯矩影响线在 C 处的竖标为。()3 H/ Y) ! n: m6 q! I, i) P* N. ; i# 3简支梁跨中C 截面弯矩影响线的物理意义是荷载作用在截面C 的弯矩图形。() g g |, t/ s( ) O, o d( J4图示体系有5个质点,其动力自由度为5。(设忽略直杆轴向变形的影响)( )1 P& v5 5 T+ # g0 U& v! |$ g0 l2 ! q7 K, u- E7 p- Y5 D7 * q 5 U3 s0 R* L4 c) b: T, j& F5结构的自振频率与结构的刚度及动荷载有关。(): X/ O; S& t) C% a3 m+ - C/ o) Zs% t三、画图示伸臂梁,FRA的影响线。(10分)* c. I3 q( E E4 T- u# n; j6 8 M参考答案:/ K& x3 v; ! 3 + s: V四、绘制伸臂梁C截面、D截面的弯矩影响线和剪力影响线。(10分)2 T+ t) H8 o* q1 qE: v, 参考答案:7 V/ l6 h4 I4 v. N r- 6 G) : i$ Z2 l& U b* H t% J- B/ f, I- I7 x五、作图示梁FyA、MC的影响线,并利用影响线计算图示荷载作用下的FyA、MC值。(20分)9 lT, s( A+ j, b2 P% y, Y- L! I% O9 l, Z0 4 w s C- 4 I8 : R7 _参考答案:3 Y P2 # c& R$ ?( n6 zFyA=23kN,MC=30 kNm1 I$ E# C; K$ # e* pP1 E& A6 C7 3 m O) t0 b8 E+ S i# v! s# G: m( l: i# G六、试求图示体系的自振频率。EI=常数,杆长均为。(10分) A3 c9 E% J8 Y) # M5 d f/ N% ?& Ed) s7 g1 _ x/ 参考答案:0 Q( a5 p9 x5 L/ e. 1 J9 T5 c9 i1 s8 S% V7 A O- j1 H3 m/ M七、求图示体系的自振频率。忽略杆件自身的质量。(15分)9 . n5 , O7 f6 u参考答案:4 a* V) D- P7 l% C9 c+ e! k八、求图示体系的自振频率。质量m集中在横梁上。各杆EI=常数。(15分)# t- w# 8 A+ h0 H( d# w* P3 Q3 / W7 E, d参考答案:Winger Tuivasa-Sheck, who scored two tries in the Kiwis 20-18 semi-final win over England, has been passed fit after a lower-leg injury, while Slater has been named at full-back but is still recovering from a knee injury aggravated against USA.Both sides boast 100% records heading into the encounter but Australia have not conceded a try since Josh Charnleys effort in their first pool match against England on the opening day.Aussie winger Jarryd Hayne is the competitions top try scorer with nine, closely followed by Tuivasa-Sheck with eight.But it is recently named Rugby League International Federation player of the year Sonny Bill Williams who has attracted the most interest in the tournament so far.The Kiwi - with a tournament high 17 offloads - has the chance of becoming the first player to win the World Cup in both rugby league and rugby union after triumphing with the All Blacks in 2011.Id give every award back in a heartbeat just to get across the line this weekend, said Williams.The (lack of) air up there Watch mCayman Islands-based Webb, the head of Fifas anti-racism taskforce, is in London for the Football Associations 150th anniversary celebrations and will attend Citys Premier League match at Chelsea on Sunday.I am going to be at the match tomorrow and I have asked to meet Yaya Toure, he told BBC Sport.For me its about how he felt and I would like to speak to him first to find out what his experience was.Uefa hasopened disciplinary proceedings against CSKAfor the racist behaviour of their fans duringCitys 2-1 win.Michel Platini, president of European footballs governing body, has also ordered an immediate investigation into the referees actions.CSKA said they were surprised and disappointed by Toures complaint. In a statement the Russian side added: We found no racist insults from fans of CSKA. Baumgartner the disappointing news: Mission aborted.The supersonic descent could happen as early as Sunda.The weather plays an important role in this mission. Starting at the ground, conditions have to be very calm - winds less than 2 mph, with no precipitation or humidity and limited cloud cover. The balloon, with capsule attached, will move through the lower level of the atmosphere (the troposphere) where our day-to-day weather lives. It will climb higher than the tip of Mount Everest (5.5 miles/8.85 kilometers), drifting even higher than the cruising altitude of commercial airliners (5.6 miles/9.17 kilometers) and into the stratosphere. As he crosses the boundary layer (called the tropopause),e can expect a lot of turbulence.The balloon will slowly drift to the edge of space at 120,000 feet ( Then, I would assume, he will slowly step out onto something resembling an Olympic diving platform.They blew it in 2008 when they got caught cold in the final and they will not make the same mistake against the Kiwis in Manchester.Five years ago they cruised through to the final and so far history has repeated itself here - the last try they conceded was scored by Englands Josh Charnley in the opening game of the tournament.That could be classed as a weakness, a team under-cooked - but I have been impressed by the Kangaroos focus in their games since then.They have been concentrating on the sort of stuff that wins you tough, even contests - strong defence, especially on their own goal-line, completing sets and a good kick-chase. Theyve been great at all the unglamorous stuff that often goes unnoticed in the stands but not by your team-mates.It is as though their entire tournament has been preparation for the final.In Johnathan Thurston, Cooper Cronk, Cameron Smith and either Billy Slater or Greg Inglis at full-back they have a spine that is unmatched in rugby league. They have played in so many high-pressure games - a priceless asset going into Saturday.The Kiwis are a lot less experienced but winning a dramatic match like their semi-final against England will do wonders for their confidence.They defeated Australia in the Four Nations final in 2010 and the last World Cup, and know they can rise to the big occasion.
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