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Consistent Deformations Metode Gaya

Consistent Deformations Metode Gaya

Pendahuluan

Metode konsisten deformasi adalah salah satu teknik analisis struktur statis tak tentu. Metode ini sering juga disebut metode gaya atau metode fleksibilitas. Berikut ini disajikan prosedur yang menjelaskan konsep metode ini untuk menganalisis struktur statis tak tentu berderajat ketidaktentuan satu atau dua.

Menentukan derajat ketidaktentuan

Derajat ketidaktentuan struktur ditentukan oleh banyaknya reaksi yang tak diketahui (r), dikurangi banyaknya persamaan keseimbangan statis (e). Sebagai contoh, perhatikan portal pada gambar berikut. Banyaknya reaksi eksternal yang tak diketahui ( r ), adalah 5 yaitu XA, YA, MA, XB dan YB. Banyaknya persamaan keseimbangan statis (e) adalah 3 yaitu (Fx = 0, Fy = 0 and M = 0). Dengan demikian derajat ketidaktentuan struktur ( n ) tersebut dihitung sebagai berikut:

n = r - e = 5 - 3 = 2

Gambar 1 Struktur portal statis tak tentuPada gambar 2, portal didukung oleh perletakan jepit pada kedua perletakannya dan pada batang horizontal dihubungkan oleh sendi. Karena momen pada sendi sama dengan nol, maka dengan menggunakan syarat keseimbangan di kanan dan kiri sendi akan diperoleh tambahan persamaan keseimbangan. Persamaan yang diperoleh dari kondisi ini diberi simbul ek. Pada model struktur ini reaksi yang tidak diketahui (r) sebanyak 6, banyaknya persamaan statik adalah sebanyak 3 persamaan. Dari kondisi sendi penghubung diperloleh satu kondisi lagi yaitu ek = 1, sehingga derajat ketidaktentuan (n) menjadi :n = r - (e + ec) = 6 - (3 + 1) = 2

Figure 2 - Indeterminate frame structure with hinge

Memilih redundan

Penentuan redundan pada prinsipnya mengubah model struktur statis tak tentu menjadi statis tertentu dengan menganggap sebagian reaksi perletakan sebagai redundan. Dengan demikian perlu memilih reaksi perletakan sebanyak derajat ketidaktentuan (n) sebagai redundan. Pilihan redundan dapat ditentukan dengan bebas sesuai dengan keinginan. Walaupun demikian pemilihan redundan sebaiknya mempertimbangkan kemudahan yang diberikan apabila redundan itu dipilih. Perhatikan gambar 1. XB dan YB dapat dipilih sebagai redundan. Alternatif lainnya adalah XB dan MA sebagai redundan. Cobalah tentukan alternatif lainnya! Apabila reaksi yang dianggap redundan telah ditentukan, bagaimanakah model struktur tersebut setelah beberapa reaksi dianggap redundan?

Hapus reaksi perletakan pada redundanHapus reaksi perletakan (restraint) yang terkait dengan redundan terpilih pada struktur statis tak tentu untuk mendapatkan model struktur statis tertentu. Model statis tertentu ini harus stabil dan berupa sistem struktur yang logis dan dapat diterima.

Sket garis elastis model struktur

Sket garis elastis atau garis lendutan model struktur statis tertentu akibat beban luar, dan tandai deformasi yang terjadi pada reaksi perletakan yang dihapus (restraint)(lih. gambar 3).

Figure 3(a) - Primary structureFigure 3(b) - Primary structure deflected shape

Hitung deformasi pada redundan

Hitung deformasi yang berhubungan dengan redundan, misalnya rotasi pada perletakan A (A0) dan translasi/pergeseran pada perletakan B (B0). Semua ini dapat dihitung dengan menggunakan metode yang sudah dipelajari sebelumnya seperti : metode kerja virtual berikut :

(a)Gambar diagram momen (M0), untuk model struktur statis tertentu dengan beban luar (lih. Gambar 4(a)(i)). Gambar diagram momen juga dapat dihitung dengan metode superposisi seperti ditinjukkan oleh gambar 4(a)(ii). Hal ini dapat menyederhanakan perhitung integral untuk menghitung A0 dan B0.

Figure 4(a)(i) - Moment diagram of primary structure

Figure 4(a)(ii) - Moment diagram by superposition

(b)Berikan beban satuan pada lokasi redundan yang terkait. Ingat untuk redundan berupa momen, berikan momen satuan. Kalau redundan di A berupa momen, berikan beban satuan berupa momen di A ( MA= 1 ft-k. Selanjutnya sket garis defleksi atau garis elastisnya. Berikan notasi deformasi pada setiap reaksi yang dihilangkan ( fba untuk translasi di titik B dan faa untuk rotasi di titik A). Gambar diagram momen untuk model statis tertentu akibat beban satuan, liahat gambar 4(b).

Figure 4(b)(i) - Moment diagram with MA = 1 ft-kFigure 4(b)(ii) - Deflected shape with MA = 1 ft-k

(c)Hitung rotasi , A0, pada perletakan A dengan persamaan berikut:

(d)Berikan beban satuan pada redundan berikutnya, misalnya gaya satuan XB = 1 k pada perletakan B. Sket garis elastisnya, kemudian berikan label setiap deformasi yang terjadi pada setiap reaksi yang dihilangkan.Kemudian gambar diagram momen untuk model struktur statis tertentu dengan beban satuan tersebut. Lihat gambar 4(c).

Figure 4(c)(i) - Moment diagram with XB = 1 k

Figure 4(c)(ii) - Deflected shape with XB = 1 k

(e)Hitung translasi, B0, pada perletakan B menggunakan persamaan berikut:

(f)Hitung deformasi model struktur statis tertentu akibat redundan MA (Gambar 4(b)) atau redundan XB (Gambar 4(c)). Ini semua dapat diselesaikan dengan menggunakan persamaan berikut:

Persamaan diatas akan menghasilkan koefisien-koefisien faa, fab, fba, dan fbb. Koefisien fij menyatakan koefisien yang berkaitan dengan redundan i, yang ditimbulkan oleh satu satuan beban redundan j.

Tulis persamaan konsisten deformasi Tulis persamaan deformasi konsisten yang berkaitan dengan redundan. Dalam hal ini :

(a)Rotasi di perletakan A = 0 karena perletakan A adalah jepit. (1)

(b)Translasi pada perletakan B = 0 karena perletakan di B adalah sendi sehingga tidak boleh terjadi translasi horizontal. (2)

Menyelesaikan persamaan deformasi

Selesaikan persamaan (1) dan (2) pada tahap sebelumnya untuk mengetahui redundan yang tidak diketahui MA dan XB. Ingat bahwa jika nilai MA dan XB positif, ini berarti bahwa arah gaya yang diasumsikan pada awal perhitungan ( gambar 4(b) dan 4(c) adalah benar. Jika negatif, ini berarti arah yang sebenarnya berlawanan dengan yang diasumsikan pada awal perhitungan.

Menentukan Reaksi Perletakan

Reaksi perletakan XA, YA dan YB dari model struktur statis tak tentu dapat ditentukan dengan memberikan nilai MA dan XB dengan arah sesuai dengan arah sebenarnya, kemudian gunakan syarat kesetimbangan untuk membuat persamaan kesetimbangan. (Fx = 0, Fy = 0 and M = 0).

Gambar diagram M, D, N untuk struktur statis tak tentu

Begitu semua reaksi perletakan dapat ditentukan, diagram gaya normal, lintang dan momen dapat digambarkan. Berdasarkan data ini, sket garis elastis dapat diperkirakan.

Contoh 1: Gunakan metoda deformasi konsisten (MDK) untuk menghitung reaksi perletakan, diagram M, D, dan N model struktur dengan pembeban seperti gambar berikut. Modulus elastisitas balok adalah E dan momen inersia penampang balok(I) konstan sepanjang balok.

Gambar 1 Balok diatas dua perletakan

Menentukan derajat ketidaktentuan

n= r-e = 4-3 = 1 ((r = 4, e = 3).

Pilih redundan dan hilangkan reaksi perletakan ybs(restraint)Pada contoh ini reaksi yang dihilangkan adalah RB, dengan demikian struktur dasarnya menjadi :

Figure 2 - Primary structure

Hitung reaksi perletakan dan diagram momennyaPada sistem struktur statis tertentu ini akan terjadi deformasi pada B sebesar B0 sebagai akibat beban luar yang bekerja pada struktur.

Figure 3 - Support reactionsDiagram momen akibat beban luar pada struktur ini adalah sbb:

Figure 4 - M0 - Moment diagram due to applied loads

Hitung deformasi pada redundan Berikan beban satuan pada B, kemudian hitung deformasi vertikal pada titik B (fbb).

Figure 5 - Primary structure with unit load applied and resulting deflected shape

Figure 6 - Moment diagram mb with YB = 1 k

Hitung translasi vertikal yang terkait dengan redundan YB pada perletakan B dengan persamaan berikut.

Hitung deformasi pada redundan, B0.

No.A/EI (A) (k-ft2)/EI (h) pada diagram mb(ft-k)Ai*hi (k2-ft3)/EI

A011/3 x 20 x -400/EI = -2666.67/EI15-40000/EI

A021/2 x 20 x -120/EI = -1200/EI13.33-16000/EI

A0320 x -36/EI = -720/EI10-7200/EI

A041/2 x 6 x -36/EI = -108/EI00

Total = Q(B0) = -63200/EI

Dengan demikian, dengan Q = 1 k ;Q(B0) = -63200 (k2-ft3)/EIB0 = -63200 (k-ft3)/EI

Menghitung koefisien fleksibilitas (fbb)dengan menentukan deformasi model struktur statis tertentu akibat beban satuan pada redundan YB = 1 k No.A/EI (A) (k-ft2)/EI (h) pd mb diagram mb (k-ft)Ai*hi (k2-ft3)/EI

A111/2 x 20 x 20/EI = 200/EI2/3 x 20 = 13.332666.67/EI

Total = Q(fbb) = 2666.67/EI

Dengan Q = 1 k ;diperoleh Q(fbb) = 2666.67 (k2-ft3)/EIfbb = 2666.67 (k-ft3)/EI

Tulis persamaan deformasi konsisten

Persamaan deformasi konsisten yang terkait dengan redundan YB pada perletakan B adalah :

B0 + fbb * YB = 0 (1)

Persamaan ini sama dengan 0 karena sendi tidak mengijinkan terjadinya translasi vertikal.

Selesaikan persamaan deformasi konsisten

Gunakan persamaan (1), untuk menghitung YB:

-63200 (k-ft3)/EI + 2666.67 (k-ft3)/EI * YB = 0YB = 23.7 Kalikan beban satuan Q pada YB dengan 23,7 untuk mendapatkan reaksi yang sebenarnya. Jika hasilnya positif, ini berarti reaksinya searah dengan arah beban satuan, demikian juga sebaliknya.

Tentukan reaksi perletakannya

Berikan nilai hasil perhitungan YB bersama-sama dengan beban luar aslinya.Hitung reaksi perletakan yang belum diketahui dengan persamaan kesetimbangan. (Fx = 0, Fy = 0 dan M = 0).

Gambar 7 Balok dengan reaksi perletakan

Gambar diagram M, D, N

Geser:

Gamabr 8 Diagram gaya lintangMomen:

Gamabre 9 Diagram MGaris elastis:

Gambar 10 Bentuk garis elatisConsistent Deformations - Force MethodIndeterminate Beam with Moment Reaction as Redundant

problem statement

Using the method of consistent deformations, determine the reactions, moments and shears under the loading conditions shown. The modulus of elasticity (E) and the moment of inertia (I) are constant for the entire beam.

Figure 1 - Beam structure to analyze

determine the degree of indeterminacy

The structure is statically indeterminate to the first degree (r = 4, e = 3, n = r-e = 4-3 = 1).

select redundant and remove restraint

To solve for this single degree of indeterminacy, the structure has to be reduced to a statically determinate and stable structure. This is done by removing a redundant support.

In this example, the moment reaction at support A is selected as a redundant to remove in order to obtain a primary determinate structure.

Figure 2 - Primary structure

determine reactions and draw moment diagram for the primary structure

Calculate the support reactions of the primary structure.

The resulting system, (A0 indicates the resulting deflection or deformation at the location of the removed redundant for the primary structure).

Figure 3 - Support reactions

Determine the moment diagram M0 due to the applied loads on the primary structure.

In this example, the cantilever method is used to develop the moment diagram. (See a Virtual Work Cantilever Example for a complete description of this step)

Figure 4 - M0 - Moment diagram due to applied loads

calculate deformation at redundant

Using the virtual work method, calculate the rotational translation of support A that correponds to the redundant MA. Remove all loads an apply a unit moment in the direction of the redundant, draw the moment diagram, ma, and sketch the deflected shape due to the unit moment.

The resulting system, (faa is the deformation caused by the unit load).

Figure 5 - Primary structure with unit load applied and resulting deflected shape

Figure 6 - Moment diagram ma with MA = 1 ft-kCalculate the rotational translation corresponding to the redundant MA at support A using the following equation:

Calculate the deformation at the redundant, A0. Use the method of virtual work, calculate the areas on the M0 diagram and multiply each area by the corresponding heights, hi, measured at the centroid of this area on the ma diagram:

Area No.Area/EI (A) (k-ft2)/EIHeight (h) on ma diagram (k-ft)Ai*hi (k2-ft3)/EI

A012/3 x 20 x 100/EI = 1333.33/EI-1/2-666.67/EI

A021/2 x 20 x -36/EI = -360/EI-1/3120/EI

A031/2 x 6 x -36/EI = -108/EI00

Total = Q(A0) = -546.67/EI

Therefore, with Q = 1 ft-k;Q(A0) = -546.67 (k2-ft3)/EIA0 = -546.67 (k-ft2)/EI

Calculate the flexibility coefficient, faa, by determining the deformation of the primary structure when subjected to the redundant load, MA = 1 ft-k.

Again, using the method of virtual work, calculate the areas on the ma diagram and multiply by the corresponding heights, hi, measured at the centroid of each area:

Area No.Area/EI (A) (k-ft2)/EIHeight (h) on ma diagram (k-ft)Ai*hi (k2-ft3)/EI

A111/2 x 20 x -1/EI = -10/EI-2/36.67/EI

Total = Q(faa) = 6.67/EI

Therefore, with Q = 1 ft-k;Q(faa) = 6.67 (k2-ft3)/EIfaa = 6.67 (k-ft2)/EI

write consistent deformation equation

The consistent deformation equation that corresponds to the redundant MA (the moment reaction at support A) is:

A0 + faa * MA = 0 (1)

This equation is set equal to zero since the fixed support at A does not allow any rotation.

solve consistent deformation equation

Using equation (1), solve for MA:

-546.67 (k-ft2)/EI + 6.67 (k-ft2)/EI * MA = 0MA = 82

Multiply the unit moment, Q, at MA by 82 to get the final reaction. The positive answer indicates that the reaction is in the direction of the applied unit moment.

determine support reactions

Impose the value of the calculated MA along with the other applied loads on the original structure. Calculate the remaining reactions using the three static equilibrium equations, (Fx = 0, Fy = 0 and M = 0).

Figure 7 - Beam with support reactions

draw moment, shear, and axial load diagrams

Shear:

Figure 8 - Final shear diagramMoment:

Figure 9 - Final moment diagramDeflected Shape:

Figure 10 - Deflected shape

Consistent Deformations - Force MethodOnce Statically Indeterminate Frame with Horizontal Reaction as Redundant

problem statement

Using the method of consistent deformations, determine the reactions, draw moment, shear, and axial load diagrams for the frame in the accompanying figure. The modulus of elasticity (E) and the moment of inertia (I) are constant for the entire structure.

Figure 1 - Frame structure to analyze


The structure is statically indeterminate to the first degree (r=4, e=3, n = r-e = 4-3 = 1).

select redundants and remove restraints


In this example, the horizontal reaction at support D is selected as a redundant to remove in order to obtain a primary determinate structure.

Figure 2 - Primary determinate structure



Figure 3 - Support reactionsDetermine the moment diagram M0 due to the applied loads on the primary structure.

Figure 4 - Mo - Moment diagram of primary structurenote: colors represent one method of dividing the area in the moment diagram to be used in the visual integration.

sketch deflected shape

Sketch an approximate deflected shape for the primary structure, (see Fig. 4). Label the deflection that occurs in the direction of the redundant at the released support.

Figure 5 - Deflected shape

calculate deformations at redundants

Using the virtual work method, calculate the horizontal translation of support D that corresponds to the redundant XD. Remove all loads and apply a unit force in the direction of the redundant, draw the moment diagram, md, and sketch the deflected shape due to this unit load.

Figure 6(a) - Primary structure with unit load applied

Figure 6(b) - Moment diagram md with XD = 1 k

Figure 6(c) - Deflected shape with XD = 1 k

Calculate the translation, D0, at Support D using the following equation:

Using the method of virtual work, calculate areas on the M0 diagram and multiply each area by the corresponding heights, hi, measured at the centroid of this area on the md diagram:

Area No.Area/EI (A) (k-ft2)/EIHeight (h) on md diagram (k-ft)Ai*hi (k2-ft3)/EI

A011/2 x 15 x 206.25/EI = 1546.875/EI2/3 x -7.5 = -5-7734.375/EI

A021/2 x 15 x 206.25/EI = 1546.875/EI2/3 x -15 = -10-15468.75/EI

A031/2 x 15 x 112.5/EI = 843.75/EI5/6 x -15 = -12.5-10546.875/EI

A041/2 x 15 x 112.5/EI = 843.75/EI2/3 x -15 = -10-8437.5/EI

A052/3 x 15 x 28.125/EI = 281.25/EI1/2 x -15 = -7.5-2109.375/EI

Total = Q(D0) = -44296.875/EI

Therefore, with Q = 1 k;Q(D0) = -44296.875 (k2-ft3)/EID0 = -44296.875 (k-ft3)/EI

Calculate the flexibility coefficient, fdd, by determining the deformations of the primary structure when subjected to the redundant load, XD = 1 k.

Again, using the method of virtual work, calculate areas on the md diagram and multiply by the corresponding heights, hi, measured at the centroid of the area:


A111/2 x 30 x -15/EI = -225/EI2/3 x -15 = -102250/EI

A121/2 x 15 x -15/EI = -112.5/EI2/3 x -15 = -101125/EI

fdd = 3375/EI

Therefore, with Q = 1 k;Q(fdd) = 3375 (k2-ft3)/EIfdd = 3375 (k-ft3)/EI

write consistent deformation equations

The consistent deformation equation that corresponds to the redundant X1 (the horizontal reaction at support D) is:

D0 + fdd * X1 = 0 (1)

This equation is set equal to zero since the pinned support at D does not allow any translation in the direction of the redundant, i.e., in the horizontal direction.

solve consistent deformation equation

Using equation (1), solve for X1:

-44,296.875 (k-ft3)/EI + 3375/EI (k-ft3)/EI* XD = 0X1 = 13.125

Multiply the unit load, Q, at XD by 13.125 to get the final reaction. The positive answer indicates that the reaction is in the direction of the applied unit load.


Impose the value of the calculated support reaction at X1 corresponding to the redundant XD, along with the other applied loads on the original structure. Calculate the remaining reactions using the three static equilibrium equations, (Fx = 0, Fy = 0 and M = 0). Figure 7 - Frame with support reactions


All reactions are now known and are used to draw the complete shear and moment diagrams for the structure.

Shear:


Figure 9 - Final moment diagramAxial Load:

Figure 10 - Final axial load diagramDeflected Shape:

Figure 11 - Deflected shapeConsistent Deformations - Force MethodOnce Statically Indeterminate Frame with Vertical Reaction as Redundant

problem statement




The structure is statically indeterminate to the first degree(r=4, e=3, n = r-e = 4-3 = 1).



In this example, the vertical reaction at support D is selected as a redundant to remove in order to obtain a primary determinate structure.





Figure 4 - Mo - Moment diagram of primary structure


Sketch an approximate deflected shape for the primary structure, (see Fig. 4). Label the deflection that occurs in the direction of the redundant at the released support.

Figure 5 - Deflected shape of primary structure


Using the virtual work method, calculate the vertical translation of support D that corresponds to the redundant YD. Remove all loads and apply a unit force in the direction of the redundant, draw the moment diagram, md, and sketch the deflected shape due to this unit load.

Figure 6(a) - Primary structure with unit load applied

Figure 6(b) - Moment diagram md with YD = 1 k

Figure 6(c) - Deflected shape with YD = 1 k

Calculate the translation, D0 (see Fig. 5), at Support D using the following equation:

Using the method of virtual work, calculate areas on the M0 diagram and multiply each area by the corresponding heights, hi, measured at the centroid of this area on the md diagram:


A0115 x -300/EI = -4500/EI30-135000/EI

A021/3 x 15 x -112.5/EI = -562.5/EI30-16875/EI

A031/2 x 15 x -300/EI = -2250/EI5/6 x 30 = 25-56250/EI

Total = Q(D0) = -208125/EI

Therefore, with Q = 1 k ;Q(D0) = -208125 (k2-ft3)/EID0 = -208125 (k-ft3)/EI

Calculate the flexibility coefficient, fdd, by determining the deformations of the primary structure when subjected to the redundant load, YD = 1 k .

Again, using the method of virtual work, calculate areas on the md diagram and multiply by the corresponding heights, hi, measured at the centroid of the area:


A1115 x 30/EI = 450/EI3013500/EI

A121/2 x 30 x 30/EI = 450/EI2/3 x 30 = 209000/EI

Total = Q(fdd) = 22500/EI

Therefore, with Q = 1 k ;Q(fdd) = 22500 (k2-ft3)/EIfdd = 22500 (k-ft3)/EI


The consistent deformation equation that corresponds to the redundant YD (the vertical reaction at support D) is:

D0 + fdd* YD = 0 (1)

This equation is set equal to zero since the pinned support at D does not allow any translation in the direction of the redundant, i.e., in the vertical direction.

solve consistent deformation equations

Using equation (1), solve for YD:

-208,125 (k-ft3)/EI + 22,500 (k-ft3)/EI* YD = 0YD = 9.25 Multiply the unit load, Q, at YD by 9.25 to get the final reaction. The positive answer indicates that this reaction is in the direction of the applied unit load.


Impose the value of the calculated support reaction at YD along with the other applied loads on the original structure. Calculate the remaining reactions using the three static equilibrium equations, (Fx = 0, Fy = 0 and M = 0).

Figure 7 - Frame with support reactions



Shear:




Figure 11 - Deflected shapeConsistent Deformations - Force MethodTwice Statically Indeterminate Frame with Horizontal and Vertical Reactions as Redundants

problem statement




The structure is statically indeterminate to the second degree (r=5, e=3, n = r-e = 5-3 = 2).


To solve for two degrees of indeterminacy, the structure has to be reduced to a statically determinate and stable structure. This is done by removing two redundant supports.

In this example, the horizontal and vertical reactions at support D are selected as the redundants to remove in order to obtain a primary determinate structure.





Figure 4 - Mo - Moment diagram of primary structure


Sketch an approximate deflected shape for the primary structure, (see Fig. 4). Label the deflection that occurs in the direction of each redundant at the released support.

Figure 5 - Deflected shape of primary structure

calculate deformations at redundants

Using the virtual work method, calculate the vertical translation of support D that corresponds to the redundants X1 and X2. Remove all loads and apply a unit force in the direction of each redundant, draw the moment diagram, m, and sketch the deflected shape due to the unit load.

For the redundant X1 acting in the direction of the reaction XD:

Figure 6(a) - Primary structure with unit load applied at X1

Figure 6(b) - Moment diagram m1 with X1 = 1 k

Figure 6(c) - Deflected shape with X1 = 1 k

Calculate the horizontal translation corresponding to the redundant X1, D1, at support D using the virtual work equation:

Using the method of virtual work, calculate areas on the M0 diagram and multiply each area by the corresponding heights, hi, measured at the centroid of this area on the m1 diagram:

Area No.Area/EI (A) (k-ft2)/EIHeight (h) on m1 diagram (k-ft)Ai*hi (k2-ft3)/EI

A0115 x -300/EI = -4500/EI1/2 x 15 = 7.5-33750/EI

A021/3 x 15 x -112.5/EI = -562.5/EI3/4 x 15 = 11.25-6328.125/EI

A031/2 x 15 x -300/EI = -2250/EI00/EI

Total = Q(D1) = -40078.125/EI

Therefore, with Q = 1 k;Q(D1) = -40078.125 (k2-ft3)/EID1 = -40078.125 (k-ft3)/EI

For the redundant X2 acting in the direction of the reaction YD:

Figure 7(a) - Primary structure with unit load applied at X2

Figure 7(b) - Moment diagram m2 with X2 = 1 k

Figure 7(c) - Deflected shape with X2 = 1 k

Calculate the horizontal translation corresponding to the redundant X2, D2, at support D.

Again, using the method of virtual work, calculate areas on the M0 diagram and multiply each area by the corresponding heights, hi, measured at the centroid of this area on the m2 diagram:


A0115 x -300/EI = -4500/EI30-135000/EI

A021/3 x 15 x -112.5/EI = -562.5/EI30-16875/EI

A031/2 x 15 x -300/EI = -2250/EI5/6 x 30 = 25-56250/EI

Total = Q(D2) = -208125/EI

Therefore, with Q = 1 k;Q(D2) = -208125 (k2-ft3)/EID2 = -208125 (k-ft3)/EI

Calculate the flexibility coefficients, fdd, by determining the deformations of the primary structure when subjected to the redundant loads.

Again, using the method of virtual work, calculate areas on the m diagrams and multiply by the corresponding heights, hi, measured at the centroid of each area:

For f11, the coefficient at X1 due to the load applied at X1:


A111/2 x 15 x 15/EI = 112.5/EI2/3 x 15 = 101125/EI

Total = Q(f11) = 1125/EI

Therefore, with Q = 1 k;Q(f11) = 1125 (k2-ft3)/EIf11 = 1125 (k-ft3)/EI



A111/2 x 15 x 15/EI = 112.5/EI303375/EI

Total = Q(f12) = 3375/EI




A2115 x 30/EI = 450/EI3013500/EI

A221/2 x 30 x 30/EI = 450/EI2/3 x 30 = 209000/EI

Total = Q(f22) = 122500/EI




A2115 x 30/EI = 450/EI1/2 x 15 = 7.53375/EI

A221/2 x 30 x 30/EI = 450/EI00/EI

Total = Q(f21) = 3375/EI


note: f12 and f21 will always have the same result


The consistent deformation equations are:

D1 + f11* X1 + f21* X2 = 0 (1)

D2 + f12* X1 + f22* X2 = 0 (2)

These equations are set equal to zero since the pinned support at D does not allow translation in the direction of the redundants.


Using equation (1) and (2), simultaneously solve for X1 and X2:

-40,078.125 (k-ft3)/EI + 1125(k-ft3)/EI * X1 + 3375(k-ft3)/EI* X2 = 0 (1)-208,125 (k-ft3)/EI + 3375 (k-ft3)/EI* X1 + 22,500 (k-ft3)/EI* X2 = 0 (2)X1 = 14.32X2 = 7.10 The answers imply that the applied unit loads need to be multiplied, respectively, by the corresponding result in order to obtain the final reactions.

The positive answers indicate that the reactions are in the same direction as the applied unit loads.


Impose the new, correct values of XD and YD, along with the other applied loads on the original structure. Calculate the remaining reactions using the three static equilibrium equations, (Fx = 0, Fy = 0 and M = 0).

Figure 8 - Frame with support reactions



Shear:




Figure 12 - Deflected shapeConsistent Deformations - Force MethodIndeterminate Truss

problem statement

Using the method of consistent deformations, determine the vertical and horizontal reactions at A and E and the resulting member loads for the truss in the accompanying figure.

The member properties are A = 2 in2 and E = 29x103 ksi.

Figure 1 - Truss structure to analyze


The structure is indeterminate to the first degree (r=4, e=3, n=r-n=4-3=1).

select redundant and remove restraint

To solve for the single degree of indeterminacy, the structure has to be reduced to a statically determinate and stable structure. This is done by removing a redundant support.

In this example, the horizontal reaction at support E is selected as a redundant to remove in order to obtain a primary determinate structure.


determine reactions and member forces

Calculate the support reactions of the primary structure, then determine the individual member forces by using either the method of sections, or the method of joints.

Figure 3 - Support reactions and member forces


Use the virtual work method, calculate the horizontal translation of support E that corresponds to the removed redundant XE. Remove all loads and apply a unit force in the direction of the redundant.

Figure 4 - Primary structure with unit load appliedCalculate the member forces due to the unit load.

Figure 5 - Member forces due to unit loadSince the horizontal translation at E is equal to zero in the original structure, the horizontal translation of the released structure at support E must be countered by a force at E which causes an equal translation in the opposite direction.

Since the member forces are known in the released structure, and the virtual structure with a unit load applied at E, the deflection at support E (E0) caused by the applied loads on the released structure is determined by the following equation.

where m is equal to the number of members, n is the force in the member due to the virtual load, N is the force in the member due to the applied loads, L is the length, A is the area, and E represents Young's Modulus of Elasticity.

The force required to counter this deflection is found by determining the deflection caused by a unit load in the positive direction at the removed support, and multiplying this answer by the unknown support reaction.

write consistent deformation equation

The consistent deformation euqation that correspondes to the redundant XE (the horizontal reaction at support E) is:

E0 + fee*XE = 0 (1)

where E0 is the deflection at E of the primary structure, fee is the deflection of the released structure at E caused by the unit load, and XE is the value of the unknown redundant at E.


The deflection of the released structure under applied loads, E0. note: for trusses, only the members which are subjected to the unit load need to be included

Membern(k)N(k)L(ft)nNL/AE (k2-ft)/AE

AB1-25.834-103.33/AE

BC1-25.834-103.33/AE

CD1-44.1674-176.67/AE

DE1-44.1674-176.67/AE

Total = E0 = -560/AE

The deflection of the released structure at E caused by the unit load, fee.

Membern (k)N (k)L(ft)nNL/AE (k2-ft)/AE

AB1144/AE

BC1144/AE

CD1144/AE

DE1144/AE

Total = fee = 16/AE

Using equation (1), solve for XE:

-560 (ft2-k)/AE + 16 (ft2-k)/AE*XE = 0XE = 35

Multiply the unit load by this value to get the final reaction. The positive answer indicates that the reaction is in the direction of the applied unit load.


Impose the value of the calculated XE along with the other applied loads on the original truss. Calculate the remaining reactions using the trhee static equilibrium equations, (Fx = 0, Fy = 0 and M = 0).

Figure 6 - Truss with resultant support reactionsThe resulting member forces are now determined by using the method of sections, or method of joints. However, a much easier method is to use superposition and add the effects caused by the redundant load on the released structure.

In this example, the 35 k load will cause only the top chord of the truss to experience a 35 k tensile load in each member. This result can be added to those found in figure 3 above. The final results,

Figure 8 - Truss with final member forcesDeflections - Method of Virtual WorkRotation of a Beam - Cantilever

problem statement

Using the same structure as used in the Beam Deflection examples, determine the rotation at A of the beam shown in the figure below using the Cantilever Method. The modulus of elasticity (E) and the moment of inertia (I) are constant for the entire beam.

Note: The colors of the loads and moments are used to help indicate the contribution of each force to the deflection or rotation being calculated. The moment diagrams show the moments induced by a load using the same color as the load.

Figure 1 - Beam structure to analyze

calculate the support reactions

Calculate the support reactions (caused by the applied loads) using the following relationships:

Check these reactions by summing the vertical forces.

The resulting system,

Figure 2 - Beam structure with support reactions

draw moment diagrams (M) for the structure under applied "real" loads

Using the cantilever method, fix the structure at joint B and draw the resulting moment diagram induced by the applied "real" loads.

Plot the moment diagram for each applied load separately, i.e., by parts. The final results can then be obtained by utilizing the method of superposition i.e., by summing the contribution of each individual load to the displacement being calculated. This method is applicable since the structure is assumed to be elastic and the deflections are small.

Note: The centroid of each area is indicated by the numbered arrow and dot.

i) Moment diagram due to the 56 ft-k concentrated moment at A,

Figure 3 - Moment diagram due to 56 ft-k momentii) Moment diagram due to the 2 k/ft applied load,

Figure 4 - Moment diagram due to 2 k/ft applied loadiii) Moment diagram due to the 21 k support reaction at A,

Figure 5 - Moment diagram due to 21 k support reactioniv) Moment diagram due to the 6k applied load at end C

Figure 6 - Moment diagram due to 6 k applied loadNotice that the resultant moment diagram (figure 3 above) is the sum of these four diagrams.

Figure 7 - Resultant moment diagram

apply virtual load

Apply the virtual load at the point of interest in the desired direction. In this case, we want to know the rotation at point A. Therefore, apply a unit moment at point A in the positive (clockwise) direction.

Figure 8 - Beam with virtual unit load applied

solve the support reactions due to the virtual load

Following the same procedure used previously, calculate the support reactions (caused by the virtual load) using the following relationships:

Check these reactions by summing the vertical forces.

The resulting system,

Figure 9 - Support reactions due to virtual unit load

draw virtual moment diagram (m)

Determine the moment diagram due to the virtual load using the same method as used to find the moment diagrams for the applied loads.

Moment diagram due to the virtual load by using the cantilever method and fixing the structure at joint B.

Figure 10 - Moment diagram due to virtual unit load

calculate areas and centroids

Once the "real" moment diagrams are determined, calculate the area enclosed by each moment diagram and determine the location of the centroid of each of these areas.

Area No.Area/EI (ft2-k/EI)Location of centroidfrom support (ft)

1.-56x20/EI=-1120/EIX1 = 1/2x20 = 10

2.1/3x20x-400/EI=-2666.67/EIX2 = 3/4x20 = 15

3.1/2x20x420/EI=4200/EIX3 = 2/3x20 = 13.33

4.1/2x6x-36/EI=-108/EIX4 = 1/3x6 = 2

determine heights of virtual moment diagram at centroids

Determine the values - heights (hi) - on the virtual moment diagram (m) at the centroids of the moments due to the real loads. This is needed to carry out the integration by using the equation given in the introduction,

Heights (hi) and locations by the cantilever method.

Figure 11 - Heights on virtual moment diagram

integrate

Integrate the equation by using the visual integration approach.

Multiply the areas of the "real" moment diagram by the heights found in the virtual moment diagram and add them together.

Area No.Area (a)from M diagram (ft2-k/EI)Height (h)from m diagram (ft-k)Ai*hi (ft3-k2/EI)

1.-1120/EI1-1120/EI

1.-1120/EI-1/2560/EI

2.-2666.67/EI1-2666.67/EI

2.-2666.67/EI-3/42000/EI

3.4200/EI14200/EI

3.4200/EI-2/3-2800/EI

4.-108/EI00/EI

Total173.33/EI

Since EI is constant throughout the structure, the total rotation at A equals +173.33 ft3-k2/EI.

The positive sign indicates that the rotation is in the same direction as the unit moment applied at A - therefore the rotation is in the clockwise direction.

If values of E and I are specified, the vertical deflection at C in inches can be determined. For example, let E = 29,000 ksi, I = 144 in4, and Q = 1 ft-k, then

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