Question

Truss ABCD is simply supported at joints A and C. Force F = 100 N and acts downward at joint D. Length AB = BC = 4m and length BD = 3m. Determine the force (N) acting in member BD and its direction. (use +### for tension and -### for compression). D B A С F

Answer 1

COL VEN: 1) FT 1OON 2.) AB=BC= 4 m 3.) BD=3m SOLUTION: First of all, we will check will check the given truss is not. determinate have So, menor of members in tiga = 4 no. members = 5 m-aj-3 Thereve = 2 x4-) j= nor of joints 28-3 [No. of joint = 5 As in the given truss, of so, the truss is statically determinate, Now, Let us vertical reactions at A and C be (Raly and Re respectively at horizontal reaction voller joint, reaction be present and similarly hinged joint, so both horizontal and vertical reaction will be present. No. A be Let uy a (RA) Since, cis here only vertical s (Roln B J Rc FRnly מן Now, Equating forces in vertical direction: (Rely trc = f = 100 N ... (1)

Now, taking moment about EMA =0 = F x AB = Rc X AC AB=4m وا =) 100x 4 = Rex 8 ALE ABABL 50 = 4 + 4 =8m 400 RCE & = SON Now putting ean (ii) in is we have', (Rely = 100-RC= 100-50 = SON Now, Since, there is no horizontal external torre on given truss so (Ra) n method of joints. Now, We will use at point c, FBD will be. 80 S لی BE4- here to, here For te <DCB=37° FBC fcd= Force in section CD (Assumed the given FBC = Force in section BC direction Assumed) {fy=0 [Force in [Force in y direction F CD Sin 37 fRc =0 500S 2/25 3 N (E1 AdolfoD = FRE Frot 3 singh Sint ca (ve) sign means force is towards e. Now, In 30 T force in n-direction] =) FCD rosir & FBC

[cost-4 x y S 200 .. Fi = - FCD Cos 37° =-(-250) + 2 sexy 3x8 Consider the joint A. FBD of join A. 350) لي 3 Now, PAD 32 (Rady a 2300 cum A B FAB Here , FADE force in FAB So, ZFy= =) (PAly & FAD sin 37" =0 member AD force in member AB [force in y dierction ] FAD sin3y = -Ray soxs -250 3 ✓ 3 = 1..FAD Now, Sinto Efn = FAD (0337° +FAB=0 -o [ for equilibrium) FAB = -FAD Cos370 so -250 514 3 200 N - ) ( 3 Nowe, FAB I FOD Efnc I Consider the joint B. Z fy=0 Frp- F= 0 FBD=F=loon [FOD Force in memben BD F Ans.

Hence, & Force acting BD in member Loon its direction is upward. and Do

Answer 2

COL VEN: 1) FT 1OON 2.) AB=BC= 4 m 3.) BD=3m SOLUTION: First of all, we will check will check the given truss is not. determinate have So, menor of members in tiga = 4 no. members = 5 m-aj-3 Thereve = 2 x4-) j= nor of joints 28-3 [No. of joint = 5 As in the given truss, of so, the truss is statically determinate, Now, Let us vertical reactions at A and C be (Raly and Re respectively at horizontal reaction voller joint, reaction be present and similarly hinged joint, so both horizontal and vertical reaction will be present. No. A be Let uy a (RA) Since, cis here only vertical s (Roln B J Rc FRnly מן Now, Equating forces in vertical direction: (Rely trc = f = 100 N ... (1)

Now, taking moment about EMA =0 = F x AB = Rc X AC AB=4m وا =) 100x 4 = Rex 8 ALE ABABL 50 = 4 + 4 =8m 400 RCE & = SON Now putting ean (ii) in is we have', (Rely = 100-RC= 100-50 = SON Now, Since, there is no horizontal external torre on given truss so (Ra) n method of joints. Now, We will use at point c, FBD will be. 80 S لی BE4- here to, here For te <DCB=37° FBC fcd= Force in section CD (Assumed the given FBC = Force in section BC direction Assumed) {fy=0 [Force in [Force in y direction F CD Sin 37 fRc =0 500S 2/25 3 N (E1 AdolfoD = FRE Frot 3 singh Sint ca (ve) sign means force is towards e. Now, In 30 T force in n-direction] =) FCD rosir & FBC

[cost-4 x y S 200 .. Fi = - FCD Cos 37° =-(-250) + 2 sexy 3x8 Consider the joint A. FBD of join A. 350) لي 3 Now, PAD 32 (Rady a 2300 cum A B FAB Here , FADE force in FAB So, ZFy= =) (PAly & FAD sin 37" =0 member AD force in member AB [force in y dierction ] FAD sin3y = -Ray soxs -250 3 ✓ 3 = 1..FAD Now, Sinto Efn = FAD (0337° +FAB=0 -o [ for equilibrium) FAB = -FAD Cos370 so -250 514 3 200 N - ) ( 3 Nowe, FAB I FOD Efnc I Consider the joint B. Z fy=0 Frp- F= 0 FBD=F=loon [FOD Force in memben BD F Ans.

Hence, & Force acting BD in member Loon its direction is upward. and Do

Answer 3

COL VEN: 1) FT 1OON 2.) AB=BC= 4 m 3.) BD=3m SOLUTION: First of all, we will check will check the given truss is not. determinate have So, menor of members in tiga = 4 no. members = 5 m-aj-3 Thereve = 2 x4-) j= nor of joints 28-3 [No. of joint = 5 As in the given truss, of so, the truss is statically determinate, Now, Let us vertical reactions at A and C be (Raly and Re respectively at horizontal reaction voller joint, reaction be present and similarly hinged joint, so both horizontal and vertical reaction will be present. No. A be Let uy a (RA) Since, cis here only vertical s (Roln B J Rc FRnly מן Now, Equating forces in vertical direction: (Rely trc = f = 100 N ... (1)

Now, taking moment about EMA =0 = F x AB = Rc X AC AB=4m وا =) 100x 4 = Rex 8 ALE ABABL 50 = 4 + 4 =8m 400 RCE & = SON Now putting ean (ii) in is we have', (Rely = 100-RC= 100-50 = SON Now, Since, there is no horizontal external torre on given truss so (Ra) n method of joints. Now, We will use at point c, FBD will be. 80 S لی BE4- here to, here For te <DCB=37° FBC fcd= Force in section CD (Assumed the given FBC = Force in section BC direction Assumed) {fy=0 [Force in [Force in y direction F CD Sin 37 fRc =0 500S 2/25 3 N (E1 AdolfoD = FRE Frot 3 singh Sint ca (ve) sign means force is towards e. Now, In 30 T force in n-direction] =) FCD rosir & FBC

[cost-4 x y S 200 .. Fi = - FCD Cos 37° =-(-250) + 2 sexy 3x8 Consider the joint A. FBD of join A. 350) لي 3 Now, PAD 32 (Rady a 2300 cum A B FAB Here , FADE force in FAB So, ZFy= =) (PAly & FAD sin 37" =0 member AD force in member AB [force in y dierction ] FAD sin3y = -Ray soxs -250 3 ✓ 3 = 1..FAD Now, Sinto Efn = FAD (0337° +FAB=0 -o [ for equilibrium) FAB = -FAD Cos370 so -250 514 3 200 N - ) ( 3 Nowe, FAB I FOD Efnc I Consider the joint B. Z fy=0 Frp- F= 0 FBD=F=loon [FOD Force in memben BD F Ans.

Hence, & Force acting BD in member Loon its direction is upward. and Do

Answer 4

COL VEN: 1) FT 1OON 2.) AB=BC= 4 m 3.) BD=3m SOLUTION: First of all, we will check will check the given truss is not. determinate have So, menor of members in tiga = 4 no. members = 5 m-aj-3 Thereve = 2 x4-) j= nor of joints 28-3 [No. of joint = 5 As in the given truss, of so, the truss is statically determinate, Now, Let us vertical reactions at A and C be (Raly and Re respectively at horizontal reaction voller joint, reaction be present and similarly hinged joint, so both horizontal and vertical reaction will be present. No. A be Let uy a (RA) Since, cis here only vertical s (Roln B J Rc FRnly מן Now, Equating forces in vertical direction: (Rely trc = f = 100 N ... (1)

Now, taking moment about EMA =0 = F x AB = Rc X AC AB=4m وا =) 100x 4 = Rex 8 ALE ABABL 50 = 4 + 4 =8m 400 RCE & = SON Now putting ean (ii) in is we have', (Rely = 100-RC= 100-50 = SON Now, Since, there is no horizontal external torre on given truss so (Ra) n method of joints. Now, We will use at point c, FBD will be. 80 S لی BE4- here to, here For te <DCB=37° FBC fcd= Force in section CD (Assumed the given FBC = Force in section BC direction Assumed) {fy=0 [Force in [Force in y direction F CD Sin 37 fRc =0 500S 2/25 3 N (E1 AdolfoD = FRE Frot 3 singh Sint ca (ve) sign means force is towards e. Now, In 30 T force in n-direction] =) FCD rosir & FBC

[cost-4 x y S 200 .. Fi = - FCD Cos 37° =-(-250) + 2 sexy 3x8 Consider the joint A. FBD of join A. 350) لي 3 Now, PAD 32 (Rady a 2300 cum A B FAB Here , FADE force in FAB So, ZFy= =) (PAly & FAD sin 37" =0 member AD force in member AB [force in y dierction ] FAD sin3y = -Ray soxs -250 3 ✓ 3 = 1..FAD Now, Sinto Efn = FAD (0337° +FAB=0 -o [ for equilibrium) FAB = -FAD Cos370 so -250 514 3 200 N - ) ( 3 Nowe, FAB I FOD Efnc I Consider the joint B. Z fy=0 Frp- F= 0 FBD=F=loon [FOD Force in memben BD F Ans.

Hence, & Force acting BD in member Loon its direction is upward. and Do