Question

Consider the beam shown in Figure 1). Assume the support at A is red and B is a roller. Take 20(10) ksi and 24k/ft. The moment of inertia for each segment is shown in the figure Determine they component of reaction at A Express your answer using three significant figures. AED Ivec ? Submit Best Am - Part Determine the moment of reaction A Express your answer using three significant figures. Enter positive value of the moment is counterclockwise and negative value of the moment is clockwise Figure ΠΛΣΦΙΠνος ? MA Suomi Request Part Determine they component of reaction with Express your answer using three significant figures VO Avec 2 AM

Answer 1

Since photo is not clear so it's tough to understand the length of AB it should be 16 or 18 so i take and calculated for both

For 18

5 kip 2.4 kip/ft (ft) 0 18 30

Moment Distribution Method Participating Joints Any joint which is connected by 2 beam segments is considered to be participating. A beam segment is the length of the beam between two joints or a joint and a free end. Joint Type Participating Position (ft) A fixed No 0 B 18 free No 30 Stiffness Factors To calculate the stiffness factors of each beam segment connected to each participating joint, use the following formula: SARET KAB= LAB Where: KAB = the stiffness factor of the beam segment AB SAB = the stiffness constant of the beam segment AB, which can be taken as O for free ends 3 for pinned or roller support ends and = 4 for fixed ends, or with a far joint which is participating E = the modulus of elasticity of the heam segment's material I = the moment of insttia of the beam segment's cross-section LAB = the length of the beam segment AB The total joint stiffness is the sum of the left-hand side and right-hand side stiffness factors K8 = BA + kno Distribution Factors The distribution factor of the sides of the joints is that side's stiffness factor divided the joint stiffness factor. Note that for beams with continuous El, there is no need to calculate these terms as they cancet KRA DFRA= KB Joint LS-1 B 18 Li+ S+1+1 12 0 0 DF:-1 DF.:+1 0 4 0.22222 0.22222 Members The beam segments are tabulated below with the left and right hand facity that will be used to determine end moments. Joints which areparticipating are considered fred as they will be locked during the analysis, otherwise joints are considered not to resist bending. Segment Left-hand Fixity Right-hand Fixity Length (ft) AB fixed fixed 18 BC fixed free 12 Fixed End Moments Explanatory note regarding distributed loads: for the ease of calculation, trapazoidal distributed loads will be split into two triangularly varying loads Segment 1 Distributed Loads 1 Loadw = 2.4 kip/ft Left Hand Moment LHM = 64.8 kipft wxt LHM= 12 2.4 lip/ft x 324 LHME 12 LHM = 64.8 leipf!

wx 12 RHM 12 RHM= 2.4 kip/ft x 324 / 12 RHM = 64.8 kipft Summation Load FEME FEMR 64.8 kipft 64.8 kipft Σ 64.8 kipft 648 kipft 1 Segment 2 Point Load 1 Distance between start of segment and load a = 12 ft Load P = 5 leip Left Hand Moment LHM = 60 kipft LHM-Pxa LHM = 5 kip x 12 LHM = 60 kipf! Right Hand Moment RHM =0 kipft Distributed Loads 1 Distance from start to end of distributed load a = 0 ft Distance from fred support to equivalent point load of distribute load b =0 ft Loadw = 2.4 lip/f! Equivalent point load P-wxa=0 kip Left Hand Moment LHM - Olip! LHM=PxD LHM=0 kip x0 ft LHM = 0 kipft Right Hand Moment RHM = 0 kipt Summation 1 Load FEM FEMR 60 kipft O kipft 2 O kipft O kipft 60 kipft O kipft Moment Distribution DF. DER 10 COF. COFR 0.50 0 BB -64.8 60 А 64.8 0 67.200 kipft -60 kipft 60 kipft O kipft Segment Reactions Variables used for this section Distance between start of segment and start of load a Distance between start of load and end of load Seament 1

Take the sum of the moments about the left support EM: = +(w1 x bx (a + bx 0.500))+ FEMR1 +FEM 1:2 EM. = +(-2.4 kip/ ft x 18 ft x (O ft +18 ft x 0.500)) +67.200 kipft +-60 kipft SML = -381.60 kripft Take the sum of the moments about the right support EMR= -(wy xbx(L-a-bx0.500)+FEMRI+FEM2 EMR=-(-2.4 kip/ ft x 18 ft x (18 ft - 0 ft - 18 ft x 0.500)) +67.200 kipft +-60 kipft EMR = 396.00 kipft Divide the sum of the moments about the right support by the length to get the reaction at left support SMR Ry = 22.000 kaip L Take the sum of the forces in the y-direction EF = +(bxw) - Ryl - RR EF, = +(18 ft x -2.4 kip/ft) - 22.000 kip - RR RyR = 212 kip Segment 2 Take the sum of the moments about the left support M. = +(Psa x 12 ft) + (wi xbx (a+bx0.500))+ FEMR2+FEM EM. = +(-5 kip x 12 ft)+(-2.4 kip/ ft x 0 ft x (0 ft+0 ft x 0.500)) +60 kipft +0 kipft M = 0 kipft Take the sum of the moments about the right support EMR=-(PA0 ft) - (W x x (L-a-bx0.500) + FEMR2 +FEM3 EMR=-(-5 kip x 0 ft) - (-2.4 kip/ft x 0 ft x (12 ft - 0 ft - 0 ft x 0.500)) + 60 kipft +0 kipft EMR = 60 kipft Divide the sum of the moments about the right support by the length to get the reaction at left support ΣΜΕ Ryt 5 kip No reaction at night support RR=0 Joint Reactions A RY 22.000 leip M67.200 leipft B 26.2 kip 0 0 0

5 kip 2.4 kip/ft 67.2 kip-it 22 kip 20.2 kip 1 (ft) 0 18 30

B)22 kip

C)67.2 Kip-ft

D)26.2 kip

For AB=16

5 kip 2.4 kipin b I ) 0 16 28

0 B Moment Distribution Method Participating Joints Any joint which is connected by 2 beam segments is considered to be participating. A beam segment is the length of the beam between two joints or a joint and a free end. Joint Type Participating Position (ft) A fixed NO roller 16 free No 28 Stiffness Factors To calculate the stiffness factors of each beam segment connected to each participating joint, use the following formula: SABET KAB LAB Where: kAB = the stiffness factor of the beam segment AB SAB = the stiffness constant of the beam segment AB, which can be taken as = 0 for free ends 3 for pinned or roller support ends and 4 for fixed ends or with a far joint which is participating E = the modulus of elasticity of the beam segment's material 1 = the moment of inertia of the beam segment's cross-section LAB = the length of the beam segment AB The total joint stiffness is the sum of the left-hand side and right-hand side stiffness factors. K8 = ka+kac Distribution Factors The distribution factor of the sides of the joints is that side's stiffness factor divided the joint stiffness factor. Note that for beams with continuous El, there is no need to calculate these terms as they cancel KRA DFRA= KB Joint LS 1-1 Litl Si +1 B 16 4 0.250 12 0 0 K 0.250 DF:-1 DF,+1 1 0 Members The beam segments are tabulated below with the left and right hand facity that will be used to determine end moments. Joints which areparticipating are considered fixed as they will be locked during the analysis, otherwise joints are considered not to resist bending. Segment Left-hand Fixity Right-hand Fixity Length (ft) AB fixed fixer 16 BC fixed free 12 Fixed End Moments Explanatory note regarding distributed loads: for the ease of calculation, trapazoidal distributed loads will be split into two triangularly varying loads Segment 1 Distributed Loads 1 Loadw = 2.4 kip/ Left Hand Moment LHM = 51.200 kipft LHM= wXL 12 LHM 2.4 lip/ft x 256f 12 LHM = 51.200 kip

RHM= w x12 12 RHM= 2.4 kip//t x 256 ft? 12 RHM=51.200 kipfi Summation Load FEM FEMR 51.200 kipft 51.200 kipft 51.200 kipft 51200 kipft 1 Segment 2 Point Load 1 Distance between start of segment and load a = 12 ft Load P = 5 kip Left Hand Moment LHM = 60 kripft LHM=Pxa LHM = 5 kip x 12ft LHM - 60 kipft Right Hand Moment RHM=0 kipft Distributed Loads 1 Distance from start to end of distributed load a=0 ft Distance from fred support to equivalent point load of distribute load b = 0 ft Loadw = 2.4 kip/ft Equivalent point load P=wx a = 0 kip Left Hand Moment LHM = 0 kipft LHM=PxD LHM=0 kip x0 ft LHM=0 kipft Right Hand Moment RHM = 0 kipft Summation 1 Load FEMI FEMR 60 kipft Okipfi 2 O leipt O leift 60 kipft O kipft Moment Distribution DFDFR 1 0 COF COFR 0.50 0 в в 51.200 60 A 51.200 c 0 Σ 46.8 kipft -60 kipft 60 kipft O kipft Segment Reactions Variables used for this section Distance between start of segment and start of load a Distance between start of load and end of load Segment 1

Take the sum of the moments about the left support EM1 = +(wi xb x (a+hx0.500)) + FEMRI+FEM 1.2 EML = +(-2.4 kip/ft x 16 ft XO ft + 16 ft x 0.500))+ 46.8 kipft +-60 kipft EM = -320.4 kipft Take the sum of the moments about the right support EM:= -(wy xbx(L-a-bx0.500)+FEMR+FEM :2 EMR=-(-2.4 kip/ft x 16 ft x (16 ft-oft - 16 ft x 0.500)) +46.8 kipft +-60 kipft SMR= 294 kipft Divide the sum of the moments about the right support by the length to get the reaction at left support ΣΜΑ Ryl= L = 18.375 kip Take the sum of the forces in the y-direction EF, = +(bxw) - Ry - RR EF, = +(16 ft x -2.4 kip/ft) - 18.375 kip - RR Ryr = 20.025 krip Segment 2 Take the sum of the moments about the left support XML = +(P)1 x 12 ft) + (wi xbx (a+bx0.500))+ FEMR2+FEM3 EM. = +(-5 kip x 12 ft)+(-2.4 kip/ ft x 0 ft x (0 ft+0 ft x 0.500)) +60 kipft +0 kipft EM = 0 kipft Take the sum of the moments about the right support EMR=-(P x 0 ft) -(w xbx (L-a-bx0.500) + FEMR2 +FEM3 EMR=-(-5 kip x0 ft) - (-2.4 kip/ft x 0 ft x (12 ft - 0 ft-o ft x 0.500))+ 60 kipft +0 kipft ΣΜΑ = 60 leipft Divide the sum of the moments about the right support by the length to get the reaction at left support ΣΜΑ Ryt = 5 laip L No reaction at right support RR = 0 Joint Reactions A B Ry 18.375 kip 25.025 kip 0 M 46.8 kipf 0 0

B)18.375 kip

C)46.8 kip-ft

D)25.025 Kip

Hope it helpful to you for any doubt comment below i will solve your query
rate my answer your rating is important to us