simonsay

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Questions

The following questions were asked by visitors named simonsay.

The truss structure ABCD is loaded with a concentrated force P = 2 kN applied at a 45° angle at joint D as indicated in the figure. All bars in the truss have equal cross sectional area and are made of the same homogeneous linear elastic material. The dim...

Answers

The following answers were posted by visitors named simonsay.

HW3_2A for 2L/3≤x≤L: (p_0*2*L/27)

1_A : -1.17, -1.17, 1.4142, 0.936, 2.35 1_B : -1.414, 0.702, 0.702

1_A : -1.17, -1.17, 1.4142, 0.936, 2.35 1_B : -1.414, 0.702, 0.702 1_C:-4.1, -4.13,2.647,6.646

1_A : -1.17, -1.17, 1.4142, 0.936, 2.35 1_B : -1.414, 0.702, 0.702 1_C:-4.13, -4.13,2.647,6.646

4_2 part IIA RxA=4, RyA=0 4_2 part IIB RxE=4, RyE=2

4_2 part IIA RxA=4, RyA=0 4_2 part IIB RxE=4, RyE=2

4_2 part IIA RxA=-4, RyA=0 4_2 part IIB RxE=4, RyE=2

N_AB N_BC N_BE N_BD N_CD N_ED -1 1 -0,7071 0 0 0 0 0 -0,7071 -1 0 0 0 -1 0 0 -0,7071 0 0 0 0 0 -0,7071 0 0 0 0 0 0,7071 -1 0 0 0 1 0,7071 0 b 0 0 0 2 0 0 X = a\b N AB 4,000 N BC 2,000 N BE -2,828 N BD 2,000 N CD -2,828 N ED -2,000

-1,1,-0.7071,0,0,0 0,0,-0.7071,-1,0,0 0,-1,0,0,-0.7071,0 0,0,0,0,-0.7071,0 0,0,0,0,0.7071,-1 0,0,0,1,0.7071,0 1/sqrt(2) = 1/1.4142 = 0.7071

you must type the numbers, curly N just to indicate where is aplied

q1_2_1 400, 300, 700, 565.685 q1_2_2 -800, -300, 0, 700

5.3 % DO NOT EDIT % % % % % % % % % % % % % % % % % % % % % % % % L=3; % in [m] t_0=8; % in [kN] x = linspace(0,L); % t_x=t_0*(L-x)/L; % in [kN.m] % DO NOT EDIT % % % % % % % % % % % % % % % % % % % % % % % % T =((t_0/L)*(L-x).^2)/2; % DO NOT EDIT % % % %...

HW6_1A 2*d_A^3*pi*t*G/(3*L )

6_2A= -t_0*x 6_2B= -t_0*(x^2-L^2)/(8*pi*G_0*R^4) 6_2C= -L^2*t_0/(8*pi*G_0*R^4) 6_2D= 5*t_0*L/(4*pi*R^3)

6_3 8*phi_m*G*I_p/(L^2)

6_3 = 8*phi_m*G*I_p/(L^2)

6_1B (2*Q*L^2)/(G*pi*t*d_A^2*(L+x)^2)

6_2C= -L^2*t_0/(8*pi*G_0*R^4)

HW7_1A L.Sup = hinge R.Sup= roller

HW7_1A: BENDING MOMENT DIAGRAMS AND STRESSES Length, L= a+b Point load a=a Point load b=P

HW7_1B: BENDING MOMENT DIAGRAMS AND STRESSES, PART II Cantilever A,B left sup=A (wall) L=2L Mo=PL/3, at B=2L P at L

HW7_1A_1 P*(1-a/(a+b))*x -(a/(a+b))*P*x+P*a HW7_1A_2 -(a/(a+b))*P*a+P*a a HW7_1A_3 2

HW7_1A_1 P*(1-a/(a+b))*x -(a/(a+b))*P*x+P*a HW7_1A_2 -(a/(a+b))*P*a+P*a a HW7_1A_3 2

HW7_1A_1 P*(1-a/(a+b))*x -(a/(a+b))*P*x+P*a HW7_1A_2 -(a/(a+b))*P*a+P*a a HW7_1A_3 2

HW7_1A_1 P*(1-a/(a+b))*x -(a/(a+b))*P*x+P*a HW7_1A_2 -(a/(a+b))*P*a+P*a a HW7_1A_3 2

HW7_1A_1 P*(1-a/(a+b))*x -(a/(a+b))*P*x+P*a HW7_1A_2 -(a/(a+b))*P*a+P*a a HW7_1A_3 2

HW7_3_1 PART 2 -P*x+P*(x-L_1)*(L_1+L_2)/L_2

HW7_3_1 PART 2 -P*x+P*(x-L_1)*(L_1+L_2)/L_2

7.1B1 Mmax- =-8000

HW 7_3_1 -P*x (P capital)

HW 7_2_2 576 -576 281.33 -281.33 HW 7_3_1 -P*x

HW 7_2_2 576 -576 281.33 -281.33 HW 7_3_1 -P*x

HW 7_2_2 576 -576 281.33 -281.33 HW 7_3_1 -P*x

HW7_3_2 part 2 0.5 part 3 0.05

HW7_3_2 part 2 0.5 part 3 0.05

HW7_3_2 part 2 0.5 part 3 0.05