In construction, concrete is used because of its abundance and low cost. Concret

Question

In construction, concrete is used because of its abundance and low cost. Concrete is strong when it is in compression but is weak when in tension. To circumvent this steel bars (known as rebar) are placed inside the concrete to create a composite material known as reinforced concrete. The goal of this assignment is to start the design process of reinforced concrete with some visualizations. At the end of the assignment, given geometric properties of the beam and rebar as well as physical properties of the materials an image should be produced similar to the one shown below. Create a function named plotReinforcedBeam which inputs the beam points, the centroid of the beam, the rebar points, and the total mass and plots the reinforced beam using the line function. Add the total mass at the centroid of the beam and print the results to a file named "rbeam.png". you can store a number into a string using the sprintf function (ex. s=spr tf('***Pi=%f***',pi); display(s)).

Explanation / Answer

f exist('fy','var') ==0; fy =input('Steel yeild strength fy (ksi): '); end if exist('fcp','var')==0; fcp =input('Concrete compressive strength fcp (ksi): '); end if exist('Mu','var') ==0; Mu =input('Maximum bending moment Mu (in-kips): '); end if exist('b','var') ==0; b =input('Beam width b (in): '); end if exist('h','var') ==0; h =input('Total beam depth h (in): '); end if exist('dt','var') ==0; dt =input('Effective depth to centroid of deepest tensile steel layer dt (in): ');end if exist('dp','var') ==0; dp =input('Effective depth to centroid of shallowest compression steel layer dp (in): ');end if exist('phi','var')==0; phi = 0.9; end if exist('strip','var')==0; strip = 'no'; end if exist('msgs','var') ==0; msgs = 'yes'; end if exist('d','var') ==0; d = dt; end if Mu < 0; Mu = abs(Mu); end Es=29000; % Min tensile strain of for beams by ACI orcementcode etmin = 0.005; % max permissible strain in concrete eu = 0.003; beta1= find_beta1(fcp); if strcmp(strip,'yes');Asmin = find_As_min_strip(fy,fcp,b,h,d); rho_min =Asmin/(b*d); else rho_min = find_rho_min_beams(fcp,fy); end rho_max = find_rho_max_beams(fcp,fy,eu,etmin,beta1,Es); Asmax = findAs(rho_max,b,d); a = find_a(Asmax,fy,fcp,b); ierror = 'none'; phitest = 0; count = 0; while abs (phi-phitest)>0.001 && count= Mu % SINGLY REINFORCED beam design [Ast ierror] = find_As_a(Mu, phi, fy, d, fcp, b); rho_web = find_rho(Ast, b,d); [rho_web cerror] = code_duct_reqs_chk(rho_web,rho_max,rho_min); if strcmp(cerror,'none')==0; Ast = rho_web* b* d; end a_actual = find_a(Ast,fy,fcp,b); c = findc(a_actual,beta1); phi= find_phi(c,dt); phiMn_total = find_phiMn_recbeam( phi,a_actual,d,0,Ast,0,0,fy ); Asc= 0; else % DOUBLY REINFORCED beam design % Part of Mu that steel (Ast) in tension would resist Mut = Mdmax; % Part of Mu that steel in compression would resist Muc = Mu - Mut; % Crossectional area of steel in compression assuming fsp = fy Asc = find_As(Muc,phi,fy,d,2*dp); Ast = Asmax; a_actual = a; cerror = 'none'; % reinf ratio of rebars in compression and tension respectively rho_c = find_rho(Asc,b,d); rho_t = find_rho(Ast+Asc,b,d); [Asc fsp] = rho_crit_check(rho_c,rho_t,beta1,fcp,fy,dp,dt,eu,Es,a_actual,Asc); Ast_t = Asc + Ast; phiMn_total = find_phiMn_recbeam( phi,a_actual,d,dp,Ast,Asc,fsp,fy ); end count = count+1; end if strcmp(msgs,'yes') if abs(phi-phitest)>0.01; ierror = 'results unreliable since iteration was forced to stop'; end if Mdmax >= Mu fprintf (' Singly reinforced beam ') fprintf ('Area of steel in compression = %.2f in^2 : ',Ast); else fprintf (' Doubly reinforced beam ') fprintf ('Area of steel in compression = %.1f in^2 : ',Asc); fprintf ('Area of steel in tension = %.2f in^2 : ',Ast_t); end fprintf ('Factored Moment capacity (phi*Mn) = %.1f in-kips ',phiMn_total); if strcmp(cerror,'none')==0; fprintf ('%s ',cerror); end if strcmp(ierror,'none')==0; fprintf ('%s ',ierror); end end end

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